Degree Sequences, Forcibly Chordal Graphs, and Combinatorial Proof Systems

DEGREE SEQUENCES, FORCIBLY CHORDAL GRAPHS, AND COMBINATORIAL PROOF SYSTEMS

DISSERTATION

Presented in Partial Fulﬁllment of the Requirements for

the Degree Doctor of Philosophy in the Graduate

School of the Ohio State University

Christian Altomare, B.S.

Graduate Program in Mathematics

The Ohio State University 2009

Dissertation Committee:

Dr. G. Neil Robertson, Advisor

Dr. John Maharry

Dr. Akos Seress

ABSTRACT

We study the structure theory of two combinatorial objects closely related to graphs.

First, we consider degree sequences, and we prove several results originally motivated by attempts to prove what was, until recently, S.B. Rao’s conjecture, and what is now a theorem of Paul Seymour and Maria Chudnovsky, namely, that graphic degree sequences are well quasi ordered. We give a new, surprisingly non-graph the- oretic proof of the bounded case of this theorem. Next, we obtain an exact structure theorem of degree sequences excluding a square and a pentagon. Using this result, we then prove a structure theorem for degree sequences excluding a square and, more generally, excluding an arbitrary cycle. It should be noted that taking complements, this yields a structure theorem for excluding a matching.

The structure theorems above, however, are stated in terms of forcibly chordal graphs, hence we next begin their characterization. While an exact characterization seems difficult, certain partial results are obtained. Specifically, we first characterize the degree sequences of forcibly chordal trees. Next, we use this result to extend the characterization to forcibly chordal forests. Finally, we characterize forcibly chordal graphs having a certain path structure.

Next, we deﬁne a class of combinatorial objects that generalizes digraphs and partial orders, which is motivated by proof systems arising in mathematical logic. We

ii give what we believe will be the basic theory of these objects, including deﬁnitions, theorems, and proofs. We deﬁne the minors of a proof system, and we give two forbidden minors theorems, one of them characterizing partial orders as proof systems by forbidden minors.

iii To Moomar.

iv ACKNOWLEDGMENTS

First and foremost, I wish to thank Neil Robertson, my advisor. It is every student’s wish to have an advisor with such depth of understanding, breadth of knowledge, and raw intuition for his ﬁeld of expertise. I have gained from him not only knowledge, but an understanding of how research mathematics is carried out. His ability to ﬁnd the right generalization to prove, the right special case to consider, the right approach to try, and the right question to ask at all, has continually amazed me.

Second, I would like to thank S.B. Rao for a beautiful conjecture.

Third, I would like to thank Christopher McClain for his generous and patient help related to typesetting and document preparation, which are not my strong suits.

Fourth, I wish to thank Akos Seress and John Maharry for their time and eﬀort participating in my thesis committee.

Fifth, I would like to thank everyone in the Ohio State University Mathematics

Department who has helped me in my time since I started taking mathematics courses here as a high school student. In particular, in the order in which I met them, I am thankful to John Maharry, Alexander Dynin, Judie Monson, Yung-Chen Lu, Vitaly

Bergelson, Randall Dougherty, Tim Carlson, Cindy Bernlohr, Boris Pittel, and once again my advisor for the amount of time, eﬀort, and patience they were willing to spend toward my career and development. I thank the countless others in the

v department who have helped me as well. Without their help, this would not be possible.

Sixth, I thank my parents, Richard and Karen Altomare.

vi VITA

April 7, 1980 ...... Born - Columbus, OH

1998-2001 ...... Undergraduate, The Ohio State University

2001 ...... B.S. in Mathematics, The Ohio State University

2001-Present ...... Graduate Teaching Associate, The Ohio State University

FIELDS OF STUDY

Major Field: Mathematics

Specialization: Graph Theory

vii TABLE OF CONTENTS

Abstract ...... ii

Dedication ...... iv

Acknowledgments ...... v

Vita ...... vii

CHAPTER PAGE

1 Introduction ...... 1

1.1 Introduction to Degree Sequences ...... 1 1.2 Degree Sequence Basics, Notation, and Conventions . . . 5 1.3 Introduction to Combinatorial Proof Systems ...... 9

2 The Bounded Case of Rao’s Conjecture ...... 12

3 Excluding Matchings and Cycles ...... 24

4 Forcibly Chordal Trees ...... 42

5 Forcibly Chordal Forests ...... 54

6 Forcibly Chordal Graphs ...... 59

7 Combinatorial Proof Systems ...... 86

7.1 Introduction ...... 86 7.2 Proof Closure ...... 87 7.3 The Merge ...... 88 7.4 Preceding Set Proof Systems ...... 92 7.4.1 Examples ...... 94 7.4.2 Motivation for Deﬁnition of Proof ...... 96 7.4.3 Proof Deﬁnition and Basics ...... 97

viii 7.4.4 Autonomous Sets ...... 99 7.4.5 Axioms ...... 103 7.4.6 The Information in the Set of Proofs ...... 104 7.5 Autonomous Systems ...... 106 7.5.1 Introduction ...... 106 7.5.2 Deﬁnition of Proof Revisited ...... 107 7.6 The Canonical Orders ...... 107 7.6.1 Canonical Order Deﬁnition and Basics ...... 107 7.6.2 Descendability ...... 109 7.6.3 Canonical Order Basic Theorems And Examples ...... 116 7.7 Partial Orders As Ausyses ...... 120 7.8 Well Founded Autonomous Systems ...... 128 7.9 Blocking ...... 133 7.9.1 The Blocking Order ...... 140 7.9.2 Blocking In Posets ...... 143 7.10 Ausys Lexicographic Sum ...... 144 7.11 Subausys, Dot, Homomorphisms, and Minors ...... 150 7.12 Relations to Matching and Connectivity ...... 163

Bibliography ...... 165

ix CHAPTER 1

INTRODUCTION

This work studies two classes of objects. The ﬁrst class we study is the class of degree sequences of ﬁnite graphs. The second class we study is a class of combinatorial proof systems we call autonomous systems.

1.1 Introduction to Degree Sequences

We assume familiarity with basic graph theory. Deﬁnitions and conventions are as in

[3] unless otherwise stated. Our graphs are ﬁnite, simple, and undirected throughout unless otherwise stated.

Deﬁnition 1.1.1. Let G be a graph with vertices v1, . . . , vn, listed such that d(v1) ≥

· · · ≥ d(vn). Then the degree sequence of G, denoted by D(G), is the sequence

(d(v1), . . . , d(vn)).

We make no use of the fact that, according to our deﬁnition, the degree sequence is a decreasing sequence. It is rather simply the easiest way to make the degree sequence of a graph unique, so we can refer to the degree sequence D(G) of G, as opposed to a degree sequence of G.

We note that while a degree sequence does not technically have any vertices, it can be

1 very suggestive to think of the vertices of a degree sequence, which we sometimes do.

The degree sequence (2, 2, 2, 1, 1), for instance, would be said to have three vertices of degree 2 and two of degree 1.

Deﬁnition 1.1.2. Let D be a degree sequence and let G be a graph. We say that G realizes D, or that G is a realization of D, if D(G) = D. We denote by R(D) the set of realizations of D.

Myriad theorems in combinatorics, and in particular graph theory, study the graphs not “containing” a ﬁxed graph, for various notions of containment. It is fruitful to deﬁne a notion of containment for degree sequences as well so that similar questions may be asked and theorems proved.

Deﬁnition 1.1.3. Let D1 and D2 be degree sequences. We write D1 ≤ D2 if there is a graph G1 in R(D1) and a graph G2 in R(D2) such that G1 is an induced subgraph of G2.

The reader may check that ≤ is a reflexive, transitive relation. One motivation for making this definition is that the induced subgraph relation for graphs can be extremely difficult to work with, even for questions that are tractable if the induced subgraph relation is replaced with another containment relation. The relation ≤ for degree sequences is similar to, but more tractable in many cases than, the induced subgraph relation for graphs.

2 A discussion of claw free graphs and degree sequences best illustrates this point. A claw is the unique graph up to isomorphism with degree sequence (3, 1, 1, 1). Suppose we wish to ﬁnd the structure of claw free graphs. What claw free means of course depends on the containment relation used. If we work with the minor relation, we are asking which graphs have no claw as a minor. It is trivial that a graph is claw free in this sense iﬀ it has no vertices of degree three or more. The claw free graphs are trivially then exactly the disjoint unions of paths and cycles.

If instead of working with the minor relation, we rather work with the induced subgraph relation, the structure of claw free graphs is then a deep and diﬃcult theorem of Chudnovsky and Seymour, proved in a series of ﬁve papers totalling over 200 pages.

Now, if instead of working with graphs excluding a claw as an induced subgraph, we instead ask which degree sequences exclude the degree sequence of a claw, then the structure theorem given by Robertson and Song can be proved in under six pages.

Thus, in passing from induced subgraphs to the ≤ relation on degree sequences, we have a theorem that is motivated by induced subgraphs, yet still more amenable to analysis.

With this motivation, degree sequence analogues of questions asked for graphs are often asked for degree sequences. The celebrated Minor Theorem of Robertson and Sey- mour says that finite graphs are well quasi ordered under the minor relation. A well quasi order is a reflexive, transitive relation T on a set X such that if x1, x2, . . . , xn,... is an infinite sequence in X then there exist i and j with i < j such that xiT xj. Anal- ogous to the Minor Theorem, S.B. Rao’s famous conjecture, proved in 2008 by M.

Chudnovsky and P. Seymour and to appear in [1], says that degree sequences of

3 graphs are well quasi ordered under ≤. We refer to this theorem as Rao’s conjecture throughout.

In chapter 2, we give a proof that for each positive integer k, Rao’s conjecture holds for degree sequences of maximum degree at most k. Our proof was obtained inde- pendently of Chudnovsky and Seymour’s proof of Rao’s conjecture, and our proof makes no use of the structure theory for degree sequences of those authors. In fact, our proof has surprisingly little graph theory at all, which leads us to believe we may be able to obtain results in a far more abstract, general setting in future works.

Just as Rao’s conjecture is natural in light of the Minor Theorem, it is also natural, in light of the many graph theorems excluding minors, topological minors, and so on, to attempt to ﬁnd the structure of degree sequences excluding a given degree sequence.

In chapter 3, we characterize degree sequences excluding (the degree sequence of) certain matchings and cycles.

These exclusion results we obtain are stated in terms of pentagons, hexagons, the complete bipartite graph K3,3, the split graphs first defined in [5], a binary operation we call “plus” first defined in [13] to characterize degree sequences with at most one realization up to isomorphism, and in terms of forcibly chordal graphs. A graph is chordal if no induced cycle has four or more vertices. A graph is forcibly chordal if every graph with the same degree sequence is chordal.

While our exclusion theorems are exact, they are only valuable structure theorems to the extent we understand the structure of the pentagons, hexagons, K3,3, split graphs, and forcibly chordal graphs they are stated in terms of. Pentagons, hexagons, and

K3,3 may be considered well understood. The structure of split graphs has been found

4 by Chudnovsky and Seymour in their proof of Rao’s conjecture to appear in [1]. We may thus take the structure of split graphs as known. That leaves the forcibly chordal graphs.

Our partial characterization of forcibly chordal graphs occupies us for the next three chapters. In Chapter 4, we characterize the forcibly chordal trees. In Chapter 5, we use these results to extend the characterization to forcibly chordal forests. In Chapter

6, we characterize connected, forcibly chordal graphs having a path structure, in a sense to be deﬁned in that chapter. We believe these results can be extended in upcoming work to fully characterize forcibly chordal graphs.

1.2 Degree Sequence Basics, Notation, and Conventions

In order to make our presentation self contained and more eﬃcient, we give the basic notation, theorems, deﬁnitions, and conventions here for easy reference. First, we must eliminate any possibility of ambiguity in the containment relation we will use throughout chapters 2 through 6.

Deﬁnition 1.2.1. We say a graph G excludes a graph H if G contains no induced subgraph isomorphic to H. We say a degree sequence D2 excludes a degree sequence D1 if D1 6≤ D2. We say that a degree sequence D excludes a graph G if D excludes D(G).

Note that while the subgraph and minor relations are far more commonly used in graph theory than the induced subgraph relation, in light of the above deﬁnitions, we will work exclusively with the induced subgraph relation. In light of this fact, we

5 make certain conventions to simplify wording throughout. If we say G contains H, we mean as an induced subgraph, if we say G contains a hole, we mean an induced hole, and so on. As such, when it causes no confusion, we will often “forget” to say induced.

Another consequence of the fact that we work strictly with the induced subgraph relation is that we can often simplify presentation by identifying the set X and the induced subgraph G[X] of the graph G, which we often do when no confusion arises.

If we say a subset X of a graph G has a certain graph property, we mean that G[X] does. Moreover, if G is a graph and X is a subset of V (G), we permit ourselves to say X is a subset of G. Since we do not distinguish between X and G[X] in this work, the reader should note that in particular, when we write X ⊆ G, we always mean that X is an induced subgraph of G. ` We use the notation G1 G2 to denote the disjoint union of graphs G1 and G2. Sim- `n ilarly, i=1 Gi denotes the disjoint union of graphs G1,...,Gn. If k is a nonnegative integer, we use the notation k · G or kG to denote the disjoint union of k isomorphic copies of G.

Given subsets X and Y of a graph G, we say that X is complete to Y if each x in X is adjacent to each y in Y . We say that X is complete if all pairs of distinct vertices in X are adjacent. We say x in G is a universal vertex, or simply that x is universal, if x is complete to G − x.

If G is a graph, Gc denotes the complement. If the degree sequence D is realized by a graph G, we may speak of the complementary degree sequence Dc as the degree sequence of Gc. Though D may in general have more than one graph realizing it, it

6 is simple to check this deﬁnition of Dc does not depend on the choice of the realizing graph, and Dc is thus well deﬁned.

The set X is anti-complete in G if X is complete in Gc. The set X is anti-complete to Y in G if X is complete to Y in Gc. In general, a graph or set is said to be anti-P if property P holds on taking complements. An anti-hexagon, for instance, is the complement of a hexagon, an anti-forcibly chordal graph is the complement of a forcibly chordal graph, and so on.

Chapters 2 through 6 make extensive use of switchings, which we now deﬁne.

Deﬁnition 1.2.2. Let G be a graph. A switching is a tuple (a, b, c, d) of distinct vertices in G such that a and b are adjacent, b and c are nonadjacent, c and d are adjacent, and d and a are nonadjacent. The edges of the switching are ab and cd.

The nonedges of the switching are bc and da. If (a, b, c, d) is a switching in G then the graph G − ab + bc − cd + da is said to arise from G by a switching in G. If there is a sequence of graphs G1,...,Gn such that Gi+1 arises from Gi by a switching in Gi for each i with 1 ≤ i < n then Gn is said to arise from G1 by a sequence of switchings.

Another way to state that (a, b, c, d) is a switching in G is that ab and cd are edges of G while bc and da are nonedges of G. It is very important to note this deﬁnition says nothing about whether or not ac is an edge or nonedge of G, and similarly for bd. Moreover, we stress that if we say xy is not an edge of a switching, xy may or may not be an edge of G. Similarly, if we say e is not a nonedge of the switching, while it is tempting to see this statement as a double negation equivalent to e being

7 an edge of the switching, this is not the case. The edge e may or may not be an edge of the switching.

The reason for this behavior is simple. A switching has exactly two edges and two nonedges. This leaves two pairs of vertices in {a, b, c, d} that are either edges of G yet not edges of the switching, or nonedges or G yet not nonedges of the switching.

While care is needed on these points, no confusion arises if such care is taken, and we speak of switchings rather informally by listing the two edges and the two nonedges.

We are rarely so formal as to present a switching as a tuple as in the deﬁnition.

We call two graphs equivalent if they have the same degree sequence. The reader may note that if H arises from G by a switching in G then D(H) = D(G). Moreover, by induction on the number of switchings, one sees that if H arises from G by a sequence of switchings then D(H) = D(G). The following converse is a theorem ﬁrst proved in [6]. It is used at key points in chapter 2 and extensively throughout chapters 3 through 6 as our primary tool.

Theorem 1.2.3. Graphs G and H are equivalent iﬀ H arises from G by a sequence of switchings.

We now ﬁx notation and conventions regarding the most important types of graphs we use.

Deﬁnition 1.2.4. A graph G is called a split graph if V (G) can be partitioned into

(possibly empty) cells A and B such that G[A] is complete and G[B] is anti-complete.

The partition (A, B) is called a split partition.

8 We note the above deﬁnition allows for possibly empty split graphs. In general, we allow empty graphs, but in cases where no problems arise, we casually disregard empty graphs without comment if doing otherwise would unnecessarily complicate a statement with trivialities.

We let Ck denote a cycle on k vertices, Pk denote a path on k vertices (not k edges),

Mk denote the matching kP2, and Kk the complete graph on k vertices. We often say triangle for C3, square for C4, and so on. A hole in a graph is an induced cycle on at least four vertices. A graph is called chordal if it has no holes.

1.3 Introduction to Combinatorial Proof Systems

In the second part of this work, we deﬁne and study certain combinatorial proof systems that we call autonomous systems. No background in logic is required to understand this part, though a basic understanding of partial orders, linear orders, well foundedness and well orders, and transﬁnite induction and recursion is needed at some points. The necessary facts may be found in [11] and [9]. The fact that we need assume no previous exposure to logic from the reader arises from our abstract approach, which of necessity starts from scratch, diverges early from that typically studied by logicians, and soon far more closely resembles structural graph and partial order theory than classical proof theory.

Proof theory is one of the main branches of mathematical logic. While proof theory as understood by mathematical logicians does indeed study proofs, it is just as fair to say that proof theorists study syntax and semantics, for the statements of typical results

9 in proof theory would be impossible to formulate, let alone prove, without syntactic and semantic notions. While proof theory has many deep and diﬃcult results, they are deep and diﬃcult results for proof systems with a great deal of structure beyond the proofs themselves.

In 2001, the author had the goal of studying proofs in as general and abstract a setting as possible. A proof is considered a (not necessarily ﬁnite or even well founded) partial order such that for all x, the set of elements less than x is enough information to infer x. We take, “is enough information to infer”, as a primitive notion. More precisely, the proof system is a set together with a set of pairs (S, x), with S a subset of and x a point in the domain, which we take to mean that S implies x.

We explicitly note that in this context, we have no syntax or semantics. We have only implication and proofs. While it may seem a priori that this is too general to prove anything, we in fact obtain nontrivial results. It is fair to say we obtain no logical theorems. Our theorems are purely combinatorial. This was, in fact, a great surprise to the author, who intended to prove logical results and found himself instead working in structural combinatorics.

Roughly, just as there are rooted and unrooted trees, there are also rooted and unrooted proof systems. While rooted trees have singleton roots, rooted proof systems allow arbitrary root sets, which are in fact the axioms of the proof system.

Unrooted proof systems generalize directed graphs. Roughly speaking, if a directed edge from x to y is taken to mean x implies y, we can instead allow directed edges from an arbitrary set X to a point y to mean that set X of “formulas” implies y. An unrooted proof system could, therefore, be thought of as a directed hypergraph.

10 Rooted proof systems, on the other hand, generalize well founded partial orders.

This work focuses on a generalization of rooted proof systems, which we will call autonomous systems, and which we will deﬁne without reference to unrooted proof systems. The topic of unrooted proof systems will be addressed in future work of the author.

We give the basic definitions, theorems, and constructions related to these autonomous systems that have proved useful in their study. We give three distinct axiomatiza- tions of autonomous systems, give numerous characterizations of partial orders as autonomous systems, and define what we call the canonical orders that encode context dependent needing in a proof system and turn out to be an important structural tool in proving even statements making no mention of these orders. We define the notions of weak and strong aut descendability, two finiteness conditions on which many autonomous system theorems and proofs depend. We define homomorphisms and two containment relations that allow us to define the minors of a proof system.

We then use the canonical orders to prove two forbidden minors theorems that hold under the assumption of strong aut descendability. (In particular, they hold for finite and even finitary autonomous systems.) We also extend the definition of partial order lexicographic sum to autonomous systems and prove the basic properties of the lexicographic sum

11 CHAPTER 2

THE BOUNDED CASE OF RAO’S CONJECTURE

In this chapter, we answer a question posed by N. Robertson, who asked if graphic degree sequences of bounded degree can be realized as disjoint unions of graphs with bounded size components. Our answer in the aﬃrmative implies the bounded case of

S.B. Rao’s Conjecture, which we state now.

Theorem 2.0.1. Graphic degree sequences of bounded degree are well quasi ordered.

There is surprisingly little graph theory in our proof. In fact, the graph theory only comes in the initial lemmas constructing graphs with certain prescribed degree sequences. Though there is an existence proof of all these initial lemmas using the

Erd¨os-Gallaiinequalities proved in [4], our goal is to give a detailed construction from

ﬁrst principles. We therefore avoid using any outside results in this proof.

We now turn to the proof.

Deﬁnition 2.0.2. A graph G is called k-regular if every vertex has degree k. A graph is called regular if it is k-regular for some k.

Lemma 2.0.3. Let k be an even integer. Then there is an integer Lk such that for all L ≥ Lk there is a k-regular graph G on L vertices.

12 Proof. k is even, so let k = 2l, and let Lk = k + 1 = 2l + 1. For each L ≥ Lk, we deﬁne a graph G on the integers 0, 1,...,L − 1 by letting

E(G) = {xy : 1 ≤ |x − y| mod L ≤ l|}

Obviously, G has L vertices, and since L is at least 2l + 1, it follows that for all x, the

2l vertices x − l, x − l + 1, . . . , x − 1, x + 1, . . . , x + l are parwise distinct. Therefore each x in G has degree 2l = k. Therefore G is a k-regular graph on L vertices as needed, thus completing the proof.

The graphs in the above proof are called circulants.

Lemma 2.0.4. Let k be an odd integer. Then there is an integer Lk such that for all even L ≥ Lk there is a k-regular graph G on L vertices.

Proof. Let Lk = 2k. It is enough to construct, for each even L ≥ Lk, a k-regular bipartite graph G on L vertices. So take an even L ≥ Lk and let L = 2l. Note then that l ≥ k. Take disjoint sets A = {v1, . . . , vl} and B = {w1, . . . , wl}. Deﬁne G as the graph with vertex set A ∪ B and edge set

E(G) = {viwj : 0 ≤ wj − vi mod l ≤ k − 1}

G is then a k-regular graph on L vertices, and the proof is complete.

Lemma 2.0.5. Given a positive integer k and a nonnegative integer j, there is a graph G with exactly 2j vertices of degree k − 1 and all other vertices of degree k.

13 Proof. Let m = max{k, j}. Take disjoint sets A = {v1, . . . , vm} and B = {w1, . . . , wm}. Deﬁne G0 as the graph with vertex set A ∪ B and edge set

E(G) = {viwj : 0 ≤ wj − vi mod m ≤ k − 1}

0 Let G = G − v1w1 − v2w2 − · · · − vjwj. Then dG(vi) = dG(wi) = k − 1 for 1 ≤ i ≤ j and dG(vi) = dG(wi) = k for i > j. The claim follows.

Lemma 2.0.6. Given a positive integer k and nonnegative integer j such that 2j 6= k, there is a graph G with exactly one vertex of degree 2j and all other vertices of degree k.

Proof. By the previous lemma, there is a graph G0 with exactly 2j vertices of degree k − 1 and all other vertices of degree k. (Note the G0 of this lemma is the G of the previous lemma.) Let v be a point not in G0. Let G be the graph on V (G0) ∪ {v} such that xy ∈ E(G) iﬀ one of the following conditions holds:

i xy ∈ E(G0).

ii x = v and dG0 (y) = k − 1.

It follows by deﬁnition of G that v is adjacent in G to exactly the vertices of degree

0 0 k − 1 in G . By choice of G , there are exactly 2j of these. So dG(v) = 2j. It is enough to show dG(x) = k for all other vertices in G, so take x 6= v. Then dG0 (x) is k or k − 1. If dG0 (x) = k then it follows by deﬁnition of G that for all y in V (G),

0 the edge xy is in G iﬀ it is in G . Therefore dG(x) = dG0 (x) = k. If dG0 (x) = k − 1 then it follows from the deﬁnition of G that NG(x) = NG0 (x) ∪ {v}. Therefore dG(x) = dG0 (x) + 1 = k − 1 + 1 = k.

14 Lemma 2.0.7. Given nonnegative integers j and k, there is a graph G with exactly

2j + 1 vertices of degree 2k and all others of degree 2k + 1.

Proof. Let m = max{2k + 1, k + j}. Let A = {v1 . . . , vm} and B = {w1, . . . , wm} be disjoint sets. Deﬁne a graph G00 on A ∪ B by letting A and B be anti-complete and

00 00 letting viwj ∈ E(G ) iﬀ 0 ≤ (wj − vi) mod m ≤ 2k. Note G is a 2k + 1-regular graph.

0 00 Let G = G − v1w1 − v2w2 − · · · − vk+jwk+j. Then for 1 ≤ i ≤ k + j, we have dG0 (vi) = dG00 (vi)−1 = (2k+1)−1 = 2k. Similarly, dG0 (wi) = 2k. For i > k+j, we see

0 from the deﬁnition of G that NG0 (vi) = NG00 (vi) therefore dG0 (vi) = dG00 (vi) = 2k + 1.

Similarly, for i > k + j, we see that dG0 (wi) = 2k + 1. Let z be a point not in G0. Let G be the graph on V (G0) ∪ {z} such that E(G) =

0 E(G ) ∪ {zvi|1 ≤ i ≤ k} ∪ {zwi|1 ≤ i ≤ k}. We show G has the desired properties. First, note that since z is not in G0, it follows directly from the deﬁnition of G that dG(z) = 2k. Now consider vi with 1 ≤ i ≤ k. Then NG(vi) = NG0 (vi) ∪ {z} therefore dG(vi) = dG0 (vi) + 1 = 2k + 1. Similarly, dG(wi) = 2k + 1 for 1 ≤ i ≤ k.

By similar reasoning, the reader may check that dG(vi) = dG(wi) = 2k+1 if k+j+1 ≤ i ≤ m and that dG(vi) = dG(wi) = 2k if k + 1 ≤ i ≤ k + j. Therefore G has exactly 2j + 1 vertices of degree 2k and the rest of degree 2k + 1 as claimed.

Lemma 2.0.8. Given distinct, nonnegative integers integers j and k, there is a graph G with exactly one vertex of degree 2j +1 and all other vertices of degree 2k +1.

15 Proof. By the previous lemma, there is a graph G0 with exactly 2j + 1 vertices of degree 2k and all other vertices of degree 2k+1. (The G0 of this lemma is the G of the previous lemma.) Take y not in G0. Deﬁne G as the graph with vertex set V (G0)∪{y}

0 0 and edge set E(G ) ∪ {yx|x ∈ V (G ) and dG0 (x) = 2k}. We show G has the desired property by showing dG(y) = 2j + 1 and all other vertices have degree 2k + 1 in G. First, by choice of G0, we know there are exactly 2j + 1 elements x in G0 such that dG0 (x) = 2k. Since NG(y) consists, by deﬁnition of G, of exactly these elements, we see that dG(y) = |NG(y)| = 2j +1. We show all other vertices of G have degree 2k +1 in G.

So take x ∈ V (G) − y. Then dG0 (x) is 2k or 2k + 1. If dG0 (x) = 2k then by deﬁnition of G, we see that NG(x) = NG0 (x) ∪ {y}. Therefore dG(x) = dG0 (x) + 1 = 2k + 1. If dG0 (x) = 2k +1 then it follows by the deﬁnition of G that NG(x) = NG0 (x). Therefore dG(x) = dG0 (x) = 2k + 1. Therefore all vertices other than y have degree 2k + 1 in G, as was to be shown.

We note that in the following lemma, i and j may or may not be distinct. The possibility that i = j must be allowed for use in a later proof.

Lemma 2.0.9. Let i and j be nonnegative integers. Let k be a positive, even integer.

Then there is a graph G with vertices v 6= w such that dG(v) = 2i + 1, dG(w) = 2j + 1 and dG(x) = k for all other vertices x in G.

0 Proof. We know there is a graph G with a vertex y of degree 2i+2j+2 and dG0 (x) = k for all other x. Let G be obtained from G0 by splitting the vertex y into nonadjacent

16 vertices v and w such that v is adjacent in G to 2i + 1 of the G0 neighbors of y and w is adjacent in G to the remaining 2j + 1 G0 neighbors of y.

Deﬁnition 2.0.10. Let U be a class of graphs. U is called productive if the following conditions hold:

(i) For every odd, nonnegative integer k, there is an integer LU,k such that for all

even L ≥ LU,k, there is a k-regular graph G in U of cardinality L.

(ii) For every even, nonnegative integer k, there is an integer LU,k such that for all

L ≥ LU,k, there is a k-regular graph G in U of cardinality L.

(iii) Given positive integers j, k, with 2j 6= k, there is a graph in U with exactly one

vertex of degree 2j and all other vertices of degree k.

(iv) Given distinct, nonnegative integers j and k, there is a graph with exactly one

vertex of degree 2j + 1 and all other vertices of degree 2k + 1.

(v) Given nonnegative integers i and j, and a positive, even k, there is a graph G in

U with vertices v 6= w such that dG(v) = 2i + 1, dG(w) = 2j + 1 and dG(x) = k for all other vertices x in G.

Corollary 2.0.11. The class of ﬁnite graphs is productive.

Proof. This is a restatement of the lemmas thus far proved.

17 Deﬁnition 2.0.12. We call a class U of graphs ﬁnitely representable if there is a

ﬁnite subset F of U such that for every graph G in U, there is a graph G0 such that

D(G0) = D(G) and G0 is the disjoint union of graphs in F .

Lemma 2.0.13. The finite union of finitely representable classes is also finitely representable.

Proof. Let U1,...,Un be ﬁnitely representable and let

n [ U = Ui i=1

Since each Ui is ﬁnitely representable, there is for each i a ﬁnite subset Fi of Ui such that every graph in Ui has the same degree sequence as a disjoint union of graphs in Fi. Let n [ F = Fi i=1

Then given a graph G in U, there is i such that G is in Ui. Therefore G has the same degree sequence as the disjoint union of some graphs in Fi. Since F contains Fi, we see G has the same degree sequence as the disjoint union of some graphs in F . Since G is an arbitrary graph in U and F is ﬁnite, we see that U is ﬁnitely representable, as was to be shown.

Lemma 2.0.14. If U is a ﬁnite set of graphs then U is ﬁnitely representable.

18 Proof. Let F = U.

We make use of the following basic fact from number theory.

Lemma 2.0.15. Let S be a nonempty set of positive integers and let g be its greatest common divisor. Then there is a ﬁnite subset FS of S and a positive integer n such

0 0 that for all n ≥ n, we can write n g as a1s1 + ··· + apsp for some positive integer p, some nonnegative integers a1, . . . , ap, and some elements s1, . . . , sp of S.

Theorem 2.0.16. Let U be a class of graphs and let Uk be the class of k-regular graphs in U. Then Uk is ﬁnitely representable.

Proof. If Uk is empty then it is vacuously true that Uk is ﬁnitely representable, so suppose Uk is nonempty.

Let S be the set of cardinalities of graphs in Uk. Then S is nonempty. Let g be its greatest common divisor. By the previous lemma, there exists n such that for

0 0 all n ≥ n, we can write n g as a1s1 + ··· + apsp for some p positive integer p, some positive integers a1, . . . , ap, and some elements s1, . . . , sp of S. Since each si is in S, it follows by the deﬁnition of S that there are graphs G1,...,Gp in Uk such that

|Gi| = si for each i. Let F = {G1,...,Gp} ∪ {H ∈ Uk : |H| < ng}. Note that since there are only ﬁnitely many graphs on less than ng vertices, F is a ﬁnite set.

By deﬁnition of ﬁnite representability, it is enough to show that given G in Uk, there is a graph G0 with the same degree sequence as G such that G0 is the disjoint union of graphs in F . So take a graph G in Uk.

19 If |G| < ng then G is in F by deﬁnition of F , so we see that G itself is a graph with the same degree as G that is the disjoint union of elements of F . So suppose

|G| ≥ ng. Then we may write |G| = a1s1 + ··· + apsp as in the previous lemma. Consider the graph p 0 a G = aiGi i=1 0 Pp Pp 0 Then |G | = i=1 ai|Gi| = i=1 aisi = |G|. Also note that G and G are both k- regular. Since G and G0 are k-regular graphs of the same cardinality, they have the same degree sequence. Clearly, we have expressed G0 as the disjoint union of graphs in F . This completes the proof of the lemma.

Deﬁnition 2.0.17. A degree class sequence C is an inﬁnite sequence c0, c1, c2, c3,... with values in {1, 2, 3,..., ∞} such that ci is eventually 1.

∞ Deﬁnition 2.0.18. Let U be a class of graphs and C = (ci)i=1 a degree class sequence.

Then UC denotes the class of graphs G in U such that for all i, the graph G has less than ci vertices of degree i.

Definition 2.0.19. Let X = {xi}i≥0 be a sequence. We define the support S(X) of X as {i : xi 6= 1}. We define the infinity support S∞(X) of X as {i : xi = ∞}.

Lemma 2.0.20. Let U be a productive class. Let C be a degree class sequence such that S(C) is ﬁnite. Then UC is ﬁnitely representable.

20 Proof. The proof is by induction on |S∞(C)|. If |S∞(C)| = 0 then UC is finite, so we know by Lemma 2.0.14 that UC is finitely representable. Suppose the result is true for all degree class sequences with infinity support of cardinality at most N. We must prove that UC is finitely representable for each degree class sequence C such that |S∞(C)| = N + 1.

For every proper subset X of S∞(C) and every positive integer M, let XM be the degree class sequence such that XM (i) = C(i) for all i except that XM (i) = M for all i in S∞(C) − X. Since the inﬁnity support of XM is a proper subset of

the infinity support S∞(C), we know by the induction hypothesis that UXM is finitely representable for each such XM . Since the finite union of finitely representable classes is finitely representable, we see that for each M, the class

[ UXM X(S∞(C) is finitely representable. To show UC is finitely representable, it is therefore enough to show that [ WM := UC − UXM X(S∞(C) is finitely representable for some M.

Note that WM is the class of graphs G in UC such that if C(i) = ∞ then G has at least M vertices of degree i. We have only to show this class is ﬁnitely representable for some large enough M. This is immediate from the deﬁnition of productivity and

Theorem 2.0.16. If M is large enough, we simply take out vertices in G whose degree d is in S(C) − S∞(C) by using the almost regular graphs whose vertices all have the same degree except possibly one or two. More precisely, we subtract the degree

21 sequence of these almost regular graphs from the degree sequence D of G. Call the remaining degree sequence D0, which we do not yet know is graphic.

The degree sequence D0 has an even number of vertices of odd degree. We may pair them up. (More formally, we partition the set of vertices of odd degree into double- tons.) For each such pair {2i+1, 2j+1} in turn, by condition (iv) of Deﬁnition 2.0.10, we may choose a graph Gi,j with all vertices of degree 2i + 1 except one of degree

00 2j + 1. Let Di,j be the degree sequence of Gi,j. Let D be the degree sequence re-

0 sulting from subtracting each Di,j with i paired to j from the degree sequence D . It is again important to note that, at this point, we have not yet shown that D0 or D00 is realizable.

However, by choosing M large enough, D0 and D00 are indeed realizable. Our degree sequence remaining has an even number of vertices of each odd degree and at least M

00 vertices of each degree in S∞(C). By Theorem 2.0.16, D may therefore be realized as the disjoint union of finitely many regular graphs. Letting G00 realize G0, it is clear from the definitions of D0 and D00 that we may unite G00 with almost regular graphs to obtain a graph graph H with the same degree sequence as G. Since S(C) is finite, only finitely many such almost regular graphs are used. WM is therefore finitely representable as needed.

Theorem 2.0.21. For any ﬁxed bound k, degree sequences bounded by k are ﬁnitely realizable.

22 Proof. The class of degree sequences with all degrees at most k is simply UC where U is the class of ﬁnite graphs and C is the degree class sequence satisfying C(i) = ∞ if i ≤ k and C(i) = 1 for i > k. Since U is productive, we may apply the previous result.

We now prove Theorem 2.0.1.

Proof. We know that degree sequences of degree at most k can be realized as disjoint unions from a ﬁnite set F of graphs. Let G1,G2,... be a sequence of graphs each of which is a disjoint union of graphs in F = {F1,...,Ft}. Then for each i, we may write a a Gi = ci,1F1 ··· ci,tFt, for some nonnegative integers ci,t. We may choose a strictly increasing sequence

(in)n≥0 such that cin,j is an increasing sequence of n for each j. Then Gin is an increasing sequence of graphs under the induced subgraph relation. Since the sequence

G1,G2,... was an arbitrary sequence of disjoint unions in F , this shows ﬁnite disjoint unions in F are well quasi ordered under induced subgraph. In particular, their degree sequences are well quasi ordered under ≤.

23 CHAPTER 3

EXCLUDING MATCHINGS AND CYCLES

In this chapter, we derive structure theorems for some classes of degree sequences excluding the matching M2 and/or cycles. More precisely, we ﬁrst recall the characterization of split graphs by forbidden induced subgraphs. Next, we use this to characterize the degree sequences that exclude the matching M2 and a square. We next use this result to characterize degree sequences excluding only a square and, more generally, degree sequences excluding an arbitrary cycle. For each theorem one proves characterizing the degree sequences having a property X by excluding graphs in the set S, one may also prove the complementary theorem that the degree sequences whose complementary degree sequence has property X are exactly those that exclude graphs whose complement is in S. Taking the complementary theorem to the result on excluding a square, we characterize degree sequences excluding the matching M2. However, each of these theorems is stated in terms of another class: split graphs, forcibly chordal graphs, and, generalizing forcibly chordal graphs, the class of graphs which forcibly have all chordless cycles of length at most k. This leads naturally to the problem of characterizing forcibly chordal graphs, which we address in the following chapters.

24 We make use of the following propositions, the ﬁrst of which is a folklore theorem that may be taken as an exercise.

Proposition 3.0.22. The following are equivalent for a graph G:

(i) G excludes M2, C4, and C5.

(ii) G excludes M2 and all holes.

(iii) G is a split graph.

Corollary 3.0.23. Split graphs are chordal.

Proof. By Proposition 3.0.22, split graphs contain no holes. Therefore, they are chordal.

Proposition 3.0.24. The following are equivalent for a degree sequence D:

(i) D excludes the degree sequences (1, 1, 1, 1), (2, 2, 2, 2), and (2, 2, 2, 2, 2).

(ii) D excludes the degree sequence (1, 1, 1, 1) and the degree sequences of all cycles

on at least 4 vertices.

(iii) D is the degree sequence of a split graph.

25 Proof. D satisfies condition (i) of this theorem iff it is realized by a graph that satisfies condition (i) of Proposition 3.0.22. Similarly for conditions (ii) and (iii). Since the three conditions of Proposition 3.0.22 are equivalent, it thus follows that the three conditions of this theorem are equivalent.

The following proposition follows from the well known characterization of split graphs as those graphs for which at least one of the Erd¨os-Gallaiinequalities is equality.

Proposition 3.0.25. Let D be the degree sequence of a split graph. Then every realization of D is a split graph.

In other words, the above proposition states that every split graph is forcibly split.

In particular, we know the following.

Corollary 3.0.26. Every split graph is forcibly chordal.

Proof. If a graph is split then, by Proposition 3.0.22 it has no holes and is thus chordal. Every split graph is forcibly split, therefore every realization of the degree sequence of a split graph is split, and hence chordal. Since every realization of the degree sequence of a split graph is chordal, it follows that every split graph is forcibly chordal.

Our next lemmas will make use of the notion of half join, which we deﬁne below.

Informally, the half join is obtained by joining an arbitrary graph H completely to

26 the complete part of a split graph S and anti-completely to the anti-complete part of S.

Deﬁnition 3.0.27. Let S be a split graph with partition into a complete part A and anti-complete part B. Let H be an arbitrary graph. Then the half join (S, A, B, H) of

S and H with respect to the split partition (A, B) is deﬁned as the graph with vertex set V (S) ∪ V (H) and edge set

E(S) ∪ E(H) ∪ {xy : x ∈ H and y ∈ A}

The above deﬁnition of half join arises naturally and often when working with split graphs, and is used, for instance, in [13] to state a decomposition theorem for split graphs, though we make no use of this theorem. Tyshkevich does not use the word half join, or any other word, simply using notation to denote the operation, but we

ﬁnd it convenient to have a word denoting it, so we choose half join.

We also note that A and B are mentioned in addition to S because a split graph may have more than one split partition, but in practice, when talking about half joins, we are usually far less formal, and simply say “the half join of S and a pentagon” and similar. We will permit this abuse of language when it causes no confusion.

We will use the following lemma several times in proving the structure theorems of this chapter.

Lemma 3.0.28. Let S be a split graph with split partition (A, B) and let H be an arbitrary graph. Let G be a (not necessarily connected) graph on at least three vertices

27 with no induced triangles, no isolated vertices, and which is not a star. If G is an induced subgraph of the half join (S, A, B, H) then G is an induced subgraph of S or H.

Proof. Since (A, B) is a split partition, A is by deﬁnition complete. Therefore A ∩ G is complete and thus any three vertices of A ∩ G comprise an induced triangle in G.

Since G is triangle free by assumption, it follows that A ∩ G is empty, has exactly one vertex, or has exactly two vertices. We consider these three cases.

First, assume A ∩ G is empty. B is anti-complete to H and B itself is anti-complete, therefore every vertex of B ∩ G is an isolated vertex of G. Since G has no isolated vertices by assumption, it follows that B ∩ G is empty. Therefore G is an induced subgraph of H as the lemma claims.

Second, assume A∩G contains exactly one vertex x. Consider any other two vertices y and z in G. We show that y and z are non-adjacent. First, suppose one of y or z is in B. Without loss of generality, we may assume y is in B. Note that z is in B or H since x is the only element of A∩G. Since B is anti-complete and anti-complete to H, and since z is in either B or H, it follows that y and z are not adjacent. Now suppose neither y nor z is in B. Then both y and z are in H. Since H is complete to A, it follows that y and z are both adjacent to x. Since G has no induced triangles by assumption, it therefore follows that y and z are not adjacent.

This shows that for all choices of y and z in G distinct from x, the vertices y and z are non-adjacent. Now take any element y of G distinct from x. If y is in H then y is adjacent to x since H is complete to A. Otherwise, y is in B. Since G has no isolated vertices, we see that y must be adjacent to some vertex of G. Since vertices of B are

28 at most adjacent to vertices of A, we see that y is adjacent to some element of A ∩ G.

Since x is the only such vertex by assumption, we see that y is adjacent to x.

We have thus shown that x is complete to G − x. Since we have also shown G − x is anti-complete, we see that G is a star, contrary to assumption. This contradiction shows that A ∩ G can not have exactly one vertex.

Finally, assume A ∩ G contains exactly two vertices. Call them x and y. Since A is complete, x and y are adjacent. Suppose H ∩ G contains a vertex z. Since A is complete to H, it follows that z is adjacent to x and y and hence x, y, z comprise a triangle, contrary to choice of G as triangle free. Therefore H ∩ G is empty, which means G is an induced subgraph of S as claimed.

In all three cases, G is an induced subgraph of H or S, thus completing the proof.

In the following proposition, we characterize graphs excluding M2 and C4. This proposition is notable in two ways. First, we prove the result for graphs, which is stronger than simply proving the analogous result for degree sequences, and it is somewhat surprising a nice characterization exists for graphs at all. Later in the chapter, for instance, when we exclude M2 alone, it will be quite necessary to use degree sequences rather than graphs.

Second, we have pointed out each exclusion theorem has a complementary theorem, but the following proposition is self complementary. The results later in the chapter lose the property of self complementarity as well.

Theorem 3.0.29. The following are equivalent for a graph G:

29 (i) G excludes M2 and C4.

(ii) G is a split graph or the half join of a split graph and a pentagon.

Proof. To see that (ii) implies (i), note that by Proposition 3.0.22, we know that if

G is a split graph then G has no induced M2 and no induced holes, and in particular no induced square. Now suppose G is the half join of a split graph and a pentagon.

Note that since M2 and C4 have at least three vertices, have no induced triangles, no isolated vertices, and are not stars, it follows from Lemma 3.0.28 that if M2 or C4 is an induced subgraph of the half join of a split graph and a pentagon then M2 or C4 must be an induced subgraph of a split graph or an induced subgraph of a pentagon.

It is easy to see a pentagon contains no induced M2 or C4, and we have already noted a split graph contains no induced M2 or C4, therefore the half join of a split graph and a pentagon has no induced M2 or C4.

For the other direction, suppose G has no induced M2 and no induced C4. If G also has no induced pentagon then by Proposition 3.0.22, we know that G is a split graph as desired. So suppose G has an induced pentagon C. We show that G[V (G) − C] is a split graph and that G is the half join of G[V (G) − C] and C.

Toward this end, we ﬁrst show that every vertex x of V (G) − C is either complete or anti-complete to C. So let x be in V (G) − C. Let C = {a, b, c, d, e} with the vertices in that cyclic order. Suppose x is adjacent to at least one vertex in C. Without loss of generality, x is adjacent to a. If x has degree 1 in G[C ∪ x] then {x, a} and {c, d} are independent edges and G thus has an induced M2, contrary to hypothesis. Suppose

30 x has degree 2 in G[C ∪ x]. Then x is either adjacent to a vertex of C adjacent to a or a vertex of C at distance 2 from a in C. Suppose x is adjacent to a vertex adjacent to a. Without loss of generality, x is adjacent to b. Then {x, a} and {c, d} are again independent edges, contrary to hypothesis that G has no induced M2. So suppose x is adjacent to a vertex at distance 2 from a in C. Without loss of generality, x is adjacent to c. Then x, a, b, c is an induced 4 cycle in G, contrary to hypothesis. Both possible graphs in which x has degree 2 in G[C ∪ x] result in a contradiction, thus x has degree greater than 2 in this graph.

Now suppose x has degree 3 or 4 in G[C ∪x]. Consider the complement K of G[C ∪x].

The complement of a pentagon is a pentagon, so K consists a pentagon together with a vertex adjacent to either 1 or 2 vertices of that pentagon. By the previous paragraph, such a graph contains an induced M2 or C4, so K has an induced M2 or C4. If K has an induced C4 then by taking complements, G[C∪x] has an an induced M2. Similarly, if K has an induced M2 then by taking complements, G[C ∪ x] has an induced C4.

Both these possibilities are contrary to assumption that G has no induced M2 or C4. Thus x can not have degree 3 or 4 in G[C ∪ x] either.

The only remaining possibility is that x has degree 5 in G[C ∪ x], or in other words, x is complete to C. We have thus shown that if x is not anti-complete to C then x is complete to C. Since x was arbitrary, we have shown that every vertex outside C is either complete or anti-complete to C.

Let A be the set of vertices outside C and complete to C, and let B be the set of vertices outside C and anti-complete to C. We know that A ∪ B ∪ C = G. We show that A is complete and B is anti-complete.

31 First, to show A is complete, let x and y be vertices of A. Suppose they are not adjacent. Then letting a and c be non-adjacent vertices of C, we see that x, a, y, c is an induced square, contrary to choice of G. Thus x and y are adjacent. Since x and y are arbitrary vertices of A, it follows that A is complete.

We have thus shown that in a graph excluding M2 and C4 but containing a pentagon, the set of vertices complete to the pentagon is itself complete. By taking complements, and noting the complement of a pentagon is a pentagon, we see that the set of vertices anti-complete to the pentagon is itself anti-complete. Thus B is anti-complete.

We now know that G = A ∪ B ∪ C, that A is complete, B is anti-complete, A is complete to C, and B is anti-complete to C. It thus follows directly from the deﬁnition of half join that G is the half join of a split graph and a pentagon, as was to be shown.

Lemma 3.0.30. Consider the half join (S, A, B, H) of a graph H and split graph S with complete part A and anti-complete part B. Let x, y, z, w be a switching operation in the half join. Then {x, y, z, w} ⊆ H or {x, y, z, w} ⊆ S. In other words, the switching either lies entirely in H or entirely in the split graph.

Proof. Let x, y, z, w be a switching operation in the half join (S, A, B, H). Simple checking shows G = (S, A, B, H)[x, y, z, w] is a graph on four vertices with no isolated vertices, no induced triangle, and which is not a star. By lemma Lemma 3.0.28, it follows that G is contained in S or H, which proves our claim.

32 Lemma 3.0.31. If a degree sequence D is realized by the half join of a graph H and a split graph S then every realization of D is the half join of a graph with the same degree sequence of H and a split graph with the same degree sequence as S.

Proof. By assumption, D is realized by a half join of a graph H and a split graph S.

Call this graph K. Let K0 be some other realization of D. Then K0 can be obtained from K by a sequence of switching operations, so take K1,...,Kn such that K1 = K,

0 Kn = K , and Ki+1 is obtained from Ki by some single switching operation.

We show by induction on i that Ki satisﬁes the conclusion of the corollary for each i. For i = 1, K1 = K satisﬁes the conclusions of the corollary by choice of K. Now suppose the conclusion of the corollary holds for i and consider Ki+1. By the inductive hypothesis, Ki is the half join of a split graph Si with the same degree sequence as

S and a graph Hi with the same degree sequence as H. Ki+1 arises from Ki by a switching operation. By the Lemma 3.0.30, the vertices of this switching operation must lie entirely in Si or entirely in Hi. If the switching operation lies in Si then let

Si+1 be the graph obtained from Si by performing the switching and let Hi+1 = Hi.

If the switching operation lies in Hi then let Hi+1 be the graph obtained from Hi by performing the switching and let Si+1 = Si. Then Hi+1 has the same degree sequence as Hi, Si+1 is a split graph with the same degree sequence as Si, and Ki+1 is the half join of Hi+1 and Si+1, completing the induction. Thus Ki satisﬁes the conclusion

0 of the corollary for each i. In particular, K = Kn satisﬁes the conclusion of the corollary. Since K0 was an arbitrary realization of D, it follows that every realization of D satisﬁes the conclusion of the corollary, as was to be shown.

33 Theorem 3.0.32. The following are equivalent for a degree sequence D:

(i) D excludes (1, 1, 1, 1) and (2, 2, 2, 2) as degree sequences.

(ii) D is the degree sequence of a split graph or D is the degree sequence of the half

join of a split graph and a pentagon.

Proof. Assume (i) and let G realize D. Then G excludes M2 and C4 as induced subgraphs. Therefore G is a split graph or the half join of a pentagon and a split graph by Theorem 3.0.29. Therefore D is realized by a graph as required in condition

(ii).

For the other direction, assume condition (ii) holds. Then D is realized by a split graph or the half join of a split graph and a pentagon. We consider each possibility.

If D is realized by a split graph then every realization is a split graph since split graphs are forcibly split. Split graphs have no induced C4 or M2, therefore every realization of D excludes M2 and C4 as induced subgraphs. Therefore D excludes (1, 1, 1, 1) and (2, 2, 2, 2) as degree sequences, as was to be shown.

Similarly, if D is realized by the half join of a split graph and a pentagon then by

Lemma 3.0.31, every realization is a half join of a split graph and a pentagon. By

Theorem 3.0.29, it follows that every realization excludes M2 and C4 as induced subgraphs, and thus D excludes (1, 1, 1, 1) and (2, 2, 2, 2) as degree sequences. Thus in each case, D excludes both degree sequences, as was to be shown.

34 Lemma 3.0.33. Let k ≥ 5. Suppose D excludes Ck−1, but D does not exclude Ck.

Let G be a realization of D containing a cycle C isomorphic to Ck. Then each vertex of G − C is complete or anti-complete to C.

Proof. Assume not. Then there is a vertex x outside of C adjacent to some vertex y of C and non-adjacent to some other vertex z of C. Let v be a neighbor of z in C distinct from y. Let K = G[C ∪ x]. Deﬁne K0 as the graph obtained from K/{v, z} by subdividing the edge {x, y}. Simple checking shows that K and K0 have the same

0 0 degree sequence. But K − x is isomorphic to Ck−1. Therefore K contains Ck−1 as an induced subgraph. Therefore K does not exclude D(Ck−1), and hence G does not exclude D(Ck−1) either. This implies that D(Ck−1) ≤ D, contrary to hypothesis. This contradiction shows that every vertex outside C is complete or anti-complete to

C as claimed.

Lemma 3.0.34. Let k ≥ 5. Suppose D excludes Ck−1, but D does not exclude Ck.

Let G be a realization of D containing a cycle C isomorphic to Ck, and let A be the set of vertices of G − C that are complete to C. Then A is complete.

Proof. Assume there are non-adjacent vertices x and y, both complete to C. Write C in cyclic order as c1, c2, . . . , ck. Then we can use c1, c3, x, y as a switching to obtain a graph G0 with the same degree sequence as G. The reader can check that

0 0 G [c1, c3, c4, . . . , ck] is a cycle in that cyclic order. Therefore G contains an induced

35 Ck−1. We thus see that D does not exclude Ck−1, contrary to hypothesis. This contradiction shows x and y must be adjacent. Since x and y are arbitrary elements of A, it follows that A is complete as claimed.

Lemma 3.0.35. Let k ≥ 5. Suppose D excludes Ck−1, but D does not exclude Ck.

Let G be a realization of D containing a cycle C isomorphic to Ck, and let B be the set of vertices of G − C that are anti-complete to C. Then B is anti-complete.

Proof. Let x and y be distinct vertices in B. It is enough to show x and y are not ` adjacent. Suppose they are adjacent. Then G[C ∪ {x, y}] is isomorphic to Ck P2, ` which has the same degree sequence as Ck−1 P3. Therefore D does not exclude Ck−1, contrary to assumption. This contradiction completes the proof.

Lemma 3.0.36. Let k ≥ 5. Suppose D excludes Ck−1, but D does not exclude Ck.

Let G be a realization of D containing a cycle C isomorphic to Ck, and let A and B be the sets of vertices of G−C that are complete and anti-complete to C, respectively.

Then G is the half join (G[A ∪ B], A, B, C) of C and G[A ∪ B].

Proof. We have only to show that every vertex of G − C is complete or anti-complete to C, that vertices complete to C are pairwise adjacent, and that vertices anti- complete to C are pairwise non-adjacent. This is exactly the content of Lemma 3.0.33,

Lemma 3.0.34, and Lemma 3.0.35.

36 Theorem 3.0.37. A degree sequence D excludes (2, 2, 2, 2) and (2, 2, 2, 2, 2) iﬀ D is forcibly chordal or D is the degree sequence of the half join of a split graph and a hexagon.

Proof. First, we prove the only if. Assume D is forcibly chordal. Then by deﬁnition, every realization of D has no holes, and thus D excludes the degree sequence of every hole, in particular (2, 2, 2, 2) and (2, 2, 2, 2, 2).

The next part of the only if direction is to prove if D is the degree sequence of the half join of a split graph and a hexagon then D excludes (2, 2, 2, 2) and (2, 2, 2, 2, 2).

It is enough to show every realization of D excludes the square and the pentagon as induced subgraphs, so let G realize D. Since the degree sequence of a hexagon can only be realized as a hexagon or two disjoint triangles, it follows from Lemma 3.0.31 that G is either the half join of a split graph and a hexagon or the half join of a split graph and the disjoint union of two triangles. Suppose G has an induced square or pentagon C. Then by Lemma 3.0.28, it follows that C is an induced subgraph of a hexagon, an induced subgraph of the disjoint union of two triangles, or an induced subgraph of a split graph, which is in all three cases a contradiction. This contradiction completes the proof of the only if direction of the lemma.

For the other direction, let D be a degree sequence excluding (2, 2, 2, 2) and (2, 2, 2, 2, 2).

We must show D is forcibly chordal or D is the degree sequence of the half join of a split graph and a hexagon.

First, we note no realization of D contains a hole with 7 or more vertices, for assume some realization G contains such a cycle Cn with n ≥ 7 as an induced subgraph. This

37 graph has the same degree sequence as the disjoint union of Cn−4 and a square, and this disjoint union obviously contains the square as an induced subgraph. Therefore

(2, 2, 2, 2) ≤ D, contrary to assumption.

Either no realization of D contains an induced hexagon or some realization does.

We consider these two cases. First, assume no realization of D contains an induced hexagon. Since D excludes (2, 2, 2, 2) and (2, 2, 2, 2, 2), we also see that no realization contains an induced square or pentagon. By the last paragraph, no realization contains an induced cycle on 7 or more vertices. Together, this implies that no realization contains a hole. In other words, every realization is a chordal graph, and D is forcibly chordal as claimed.

Next, assume some realization G of D contains a hexagon C. Then D excludes D(C5) but not D(C6). Let A and B be the sets of vertices not in C that are complete and anti-complete, respectively, to C. Then by Lemma 3.0.36 with k = 6, we see that G is the half join of the split graph G[A∪B] and the cycle C, thus completing the proof of the lemma.

Theorem 3.0.38. A degree sequence D excludes (2, 2, 2, 2) iﬀ one of the following conditions holds:

(i) D is forcibly chordal.

(ii) D is the degree sequence of the half join of a split graph and a pentagon.

(iii) D is the degree sequence of the half join of a split graph and a hexagon.

38 Proof. We know by Theorem 3.0.29 and Theorem 3.0.37 that if D is one of the three above mentioned types of degree sequences then D excludes (2, 2, 2, 2). We show the converse.

D excludes (2, 2, 2, 2, 2) or D does not exclude (2, 2, 2, 2, 2). Suppose it does. Then by Theorem 3.0.37, it follows that D is forcibly chordal or D is the degree sequence of the half join of a split graph and a hexagon. The lemma is thus proved in this case.

We therefore assume D does not exclude (2, 2, 2, 2, 2). Then there is a realization G of D containing a pentagon C as an induced subgraph. Let A and B be the vertices of G − C that are complete and anti-complete to C, respectively. Since D excludes

(2, 2, 2, 2) = D(C4) by hypothesis, we see from Lemma 3.0.36 with k = 5 that G is the half join (G[A ∪ B], A, B, C). This completes the proof.

By noting the complement of M2 is C4, we get a theorem characterizing degree sequences excluding M2 as a corollary.

Theorem 3.0.39. A degree sequence D excludes (1, 1, 1, 1) iﬀ one of the following conditions holds:

(i) D is forcibly anti-chordal.

(ii) D is the degree sequence of the half join of a split graph and a pentagon.

(iii) D is the degree sequence of the half join of a split graph and K3,3.

39 Proof. Just take complements, use Theorem 3.0.38, and note anti-chordal graphs are the complements of chordal graphs by deﬁnition, the pentagon is self-complementary, and the complement of a hexagon has the same degree sequence as K3,3.

We next wish to generalize the characterization of degree sequences excluding a square found in Theorem 3.0.38 to longer cycles.

Theorem 3.0.40. Let n ≥ 4. A degree sequence D excludes the degree sequence of

Cn iﬀ one of the following conditions holds:

(i) No realization of D has a chordless cycle on n or more vertices.

(ii) D is the degree sequence of the half join of a split graph and Cn+1.

(iii) D is the degree sequence of the half join of a split graph and Cn+2.

Proof. We prove this essentially by generalizing the proof of Theorem 3.0.38. By similar reasoning to that in previous proofs, it is easy to see degree sequences in the above three classes exclude the degree sequence of Cn. We prove the converse.

So, let D exclude the degree sequence of Cn. We must show D falls into one of the above three classes as claimed.

First, note D excludes the degree sequence of Cn+k for all k ≥ 3. To see this, assume not. The degree sequence of Cn+k is the same as the degree sequence of the disjoint union of Cn and Ck. Ck exists since k ≥ 3 by assumption. So D does not exclude the degree sequence of this disjoint union. Therefore D has a realization G containing

40 the disjoint union as an induced subgraph. In particular, G contains an induced Cn, contrary to assumption that D excludes the degree sequence of Cn. This contradiction proves our claim.

Next, we break into cases. The ﬁrst case we consider is that D excludes the degree sequence of Cn+1 and Cn+2. D excludes the degree sequence of Cn by hypothesis, and by the previous paragraph, D excludes the degree sequence of Cn+k for k at least three. Therefore D excludes the degree sequence of all cycles on at least n vertices.

Therefore, as claimed, no realization has a chordless cycle on n or more vertices.

The other case is that D does not exclude both the degree sequence of Cn+1 and Cn+2.

Then D has a realization G containing either Cn+1 or Cn+2 as an induced subgraph.

Let k = n + 1 if G contains an induced Cn+1 and let k = n + 2 otherwise. In either case, G contains an induced Ck but no induced Ck−1. Let Ck = C. Then it follows by Lemma 3.0.36 that G is the half join (G[A ∪ B], A, B, C) of G[A ∪ B] and C, thus completing the proof.

41 CHAPTER 4

FORCIBLY CHORDAL TREES

In the last chapter, we proved, among other things, a structure theorem for the degree sequences excluding the degree sequence of a square. However, this structure theorem is not sharp, for it is stated in terms of the class of forcibly chordal graphs, and it is in no way obvious a priori what this class of graphs is.

As such, the next natural step is to give a precise characterization of the forcibly chordal graphs. As the ﬁrst major step in this process, in this chapter, we characterize the forcibly chordal trees.

The following lemma is used freely without comment.

Lemma 4.0.41. Let H be an induced subgraph of G and let H0 have the same degree sequence as H. Then there is a graph G0 with the same degree sequence as G such that H0 is an induced subgraph of G0.

Proof. Enumerate the vertices of G as v1, . . . , vn, and assume, without loss of generality, that there is some i with 1 ≤ i ≤ n such that H = G[v1, . . . , vi]. Since H and H0 have the same degree sequence, it follows that there is a sequence of switchings in

0 the vertices v1, . . . , vi that transforms H into H . This same sequence of switchings,

42 applied with G instead of H as the starting graph, produces a graph G0 that contains

H0 as an induced subgraph by construction.

The following lemma is trivial, but worth explicitly stating as we freely use it without comment.

Lemma 4.0.42. If G and G0 have the same degree sequence then G is forcibly chordal iﬀ G0 is.

Proof. Immediate from the deﬁnition of forcible chordality.

To ﬁx terminology before stating the next lemma, we note that for us, Pn denotes an n vertex path, not an n edge path. We will only use parts (i) and (ii) of the following lemma in this chapter, and the reader may skip the other parts until they are later needed, but we include all parts listed here for easy reference.

Lemma 4.0.43. Forcibly chordal graphs exclude the following graphs as induced subgraphs:

(i) P6

(ii) Trees such that at least three neighbors of some vertex are nonleaves.

` (iii) P5 P3

(iv) 4P3

43 (v) 2P4

` (vi) P4 2P3

(vii) 2C3

Proof. We prove the contrapositives of these statements.

For (i), suppose G contains an induced P6. P6 has the same degree sequence as the disjoint union H of a square and an edge, therefore by Theorem 4.0.41, there is a graph G0 with the same degree sequence as G such that G0 contains H, and hence a square, as an induced subgraph. Therefore G0 is not forcibly chordal, so G is not forcibly chordal either, thus proving (i).

For (ii), suppose G contains such an induced tree T . Let r be a vertex of T with three nonleaf neighbors. Consider T as a rooted tree with root r. Let x, y, and z be three nonleaf neighbors of r. Let x0, y0, and z0 be successors of x, y, and z, respectively.

Then G[r, x, y, z, x0, y0, z0] = H has degree sequence (3, 2, 2, 2, 1, 1, 1). But this is also the degree sequence of a graph H0 isomorphic to a 6 point path with a pendant vertex adjoined in the middle. If G contains T then by Theorem 4.0.41, there is a graph G0 containing H0 such that G and G0 have the same degree sequence. Since H0 contains

0 0 P6, it follows that G contains P6 as well. By part (i), it follows that G is not forcibly chordal. Therefore G is not forcibly chordal, proving part (ii). ` For (iii), suppose G contains H = P5 P3. H has the same degree sequence as

0 ` 0 H = P6 P2, therefore there is a graph G with the same degree sequence as G such

44 0 0 0 0 that G contains H , and hence G contains P6. By part (i), it follows that G is not forcibly chordal. Therefore G is not forcibly chordal.

0 For (iv), suppose G contains H = 4P3. H has the same degree sequence as H =

` 0 C4 4P2, therefore there is a graph G with the same degree sequence as G such

0 0 0 0 that G contains H , and hence G contains C4. Therefore G is not forcibly chordal. Therefore G is not forcibly chordal.

0 For (v), suppose G contains H = 2P4. H has the same degree sequence as H =

` 0 C4 2P2, therefore there is a graph G with the same degree sequence as G such

0 0 0 0 that G contains H , and hence G contains C4. Therefore G is not forcibly chordal. Therefore G is not forcibly chordal. ` ` For (vi), note P4 2P3 has the same degree sequence as P6 2P2.

For (vii), note 2C3 has the same degree sequence as C6.

In order to state the characterization of forcibly chordal trees, we ﬁrst need to deﬁne the term X-stars. Intuitively, an X-star is obtained from a graph X by letting a star grow out of each vertex.

Deﬁnition 4.0.44. Let X and G be graphs. G is called an X-star if there is a subset

Y of V (G) such that G[Y ] is isomorphic to X and such that each vertex of G − Y is adjacent to exactly one vertex of Y and no other vertices. The set Y is called the root set, and the vertices of Y will be referred to as the root vertices. Edges with exactly one end in Y will be called pendant edges. A P1-star is called a star.

The following lemmas will be used in characterizing the forcibly chordal trees.

45 Lemma 4.0.45. A switching in a P3-star yields a P3-star or the disjoint union of a

C3-star and an edge.

Proof. Take a P3-star G with root vertices u, v, and w such that v is adjacent to u and w, and hence u and w are non-adjacent. Now consider an arbitrary switching in

G, and let G0 denote the graph obtained from G by this switching. We must show

0 G is a P3-star or the disjoint union of an edge and a C3-star. Suppose ﬁrst that neither edge of the switching is a pendant edge. Then one of the edges is uv and the other edge is vw. But these edges are not disjoint, hence can not determine a switching. Therefore either one or two edges of the switching are pendant edges. We consider these two cases.

First, assume exactly one edge determining the switching is a pendant edge. Then we may assume, without loss of generality, that the nonpendant edge determining the switching is uv. The pendant edge of the switching must then be disjoint to uv, therefore the root vertex of the pendant edge of the switching is w. Let wx be the pendant edge of the switching. Then either wu or wv is one of the nonedges determining the switching, and v and w are adjacent by hypothesis, therefore wu is one of the nonedges of the switching, which implies vx is the other nonedge of the switching.

We claim this switching yields a P3-star with R = {u, v, w} as the root set. To see

0 0 this, we must show ﬁrst that G [R] is isomorphic to P3, and second, that in G , every vertex outside R is adjacent to exactly one vertex of R and no other vertices. Now, the edge vw remains unaﬀected by the switching, u and v become nonadjacent after

46 switching, and v and w become adjacent after switching. Hence R is isomorphic to

P3 after switching. Now take a vertex y outside R. Note that by deﬁnition of an X-star, before the switching each such y is adjacent to exactly one vertex of R and no other vertices. If y 6= x then the neighborhood of y is unchanged by the switching.

If y = x then after switching, y is adjacent to v and only v, and v is in R. Therefore in either case, after switching, y is adjacent to exactly one vertex of R and no other vertices, as required. This proves a switching with exactly one pendant edge yields a

P3-star. Next, assume both edges of the switching are pendant edges. Since the edges of the switching are disjoint, they can not have the same root vertex. Let us call the root vertices of the pendant edges of the switching r and r0. rr0 is a nonedge of the switching or it is not. We consider these two cases.

First, suppose rr0 is a nonedge of the switching. Then rr0 is in particular a nonedge in G, therefore neither r nor r0 equals v, so without loss of generality, r = u, r0 = w, the edges of the switching are rx and r0y for some x, y, and the nonedges of the

0 switching are rr and xy. We claim switching yields the disjoint union of a C3-star and an edge with R = {u, v, w} as the root set.

Now, x and y lie outside the root set R of G, and hence have degree 1 in G. Switching does not aﬀect the degree of any vertex, therefore x and y have degree 1 in G0 as well. Since x and y are adjacent to one another in G0, it follows that G0 is the disjoint

0 0 union of the edge xy and G − xy. We show G − xy is a C3-star. Toward that end, note that uw is an edge of G0, and uv and vw are unaﬀected by the

0 switching, so R is indeed isomorphic to C3 in G − xy. Now take any vertex of z in

47 G0 − xy − R. Then z lies outside R and hence is a pendant vertex of G. Therefore in

G, z is adjacent to exactly one vertex of R and no other vertices. Since no vertices outside R ∪ {x, y} have their neighborhoods changed by the switching, it follows that in G0 too, z is adjacent to exactly one vertex of R and no other vertices. Therefore

0 G − xy is a C3-star as claimed. Second, suppose rr0 is not a nonedge of the switching. Then the edges of the switching are rx and r0y and the nonedges of the switching are ry and r0x. In this case, switching

0 0 essentially just renames x and y, and G is isomorphic to G, hence G is a P3-star.

0 Thus in all cases, G is a P3-star or the disjoint union of an edge and a C3-star, completing the proof.

Lemma 4.0.46. A switching in the disjoint union of a C3-star and an edge yields a

P3-star or the disjoint union of a C3-star and an edge.

Proof. Let G be the disjoint union of an edge xy and a C3-star with root set R = {u, v, w}. Take a switching S in G and let G0 be the graph arising from G after

0 applying S. We show G is a P3-star or the disjoint union of an edge and a C3-star. We break into cases by which edges and nonedges of G determine the switching S.

First, suppose xy is not in the switching S. Then both edges of S lie in the C3-star

G−xy. The root set R is isomorphic to C3, so any two edges of the root set intersect. The two edges of S, on the other hand, are disjoint, therefore at most one edge of S is contained in the root set R. If some edge of S does lie in R then without loss of generality it is uv. The other edge of S must be disjoint to uv, so its root vertex is w.

48 But then wu or wv must be a nonedge of S, contrary to the fact that w is adjacent to both u and v in G. Therefore neither edge of S lies in R.

Since neither edge of S lies in R then both edges of S are pendant edges, and since they are disjoint, they must have diﬀerent root vertices. Without loss of generality, we may assume the edges of the switching are ua and vb for some a and b. Since u and v are adjacent in G, uv is not a nonedge of S, therefore the nonedges of S are ub and va. Therefore G0 essentially arises from G by renaming a and b, and hence G and

0 0 G are isomorphic. In particular, G is the disjoint union of an edge and a C3-star as desired.

Next, suppose xy is an edge of the switching S. The other edge of the switching is either a pendant edge or an edge contained in R. We consider these two cases.

First, assume the other edge of the switching is a pendant edge, without loss of generality ua for some a. Without loss of generality, ux and ay are the nonedges of the switching. This switching essentially just renames a and x, therefore G and G0

0 are isomorphic, and in particular, G is the disjoint union of an edge and a C3-star, as desired.

Finally, assume the edge of S other than xy is a nonpendant edge, which hence lies in R. Without loss of generality, this edge is uv. Then xy and uv are the edges of S and without loss of generality, xu and yv are the nonedges. Trivial checking shows

0 that G is a P3-star with root set {u, v, w}.

0 Thus G is a P3-star or the disjoint union of an edge and a C3-star in all cases, completing the proof.

49 0 Corollary 4.0.47. If G is a P3-star and G has the same degree sequence as G then

0 G is a P3-star or the disjoint union of a C3-star and an edge.

Proof. Since two graphs have the same degree sequence iﬀ one can be obtained from the other by a sequence of switchings, it is enough to show that every switching in a

P3-star yields a P3-star or the disjoint union of a C3-star and an edge, and that every switching in the disjoint union of a C3-star and an edge yields either a P3-star or the disjoint union of a C3-star and an edge. These facts are exactly the last two lemmas.

Corollary 4.0.48. P3-stars are forcibly chordal.

Proof. Let G be a P3-star. We must show every realization of the degree sequence of G is chordal. Let G1,...,Gn be a sequence of graphs such that G = G1 and Gi+1 arises from Gi by a switching in Gi for 1 ≤ i < n. It is enough to show by induction that each Gi is chordal. Noting that P3-stars and the disjoint union of an edge and a C3-star are both chordal, it is enough to show by induction that each Gi falls into one of these two classes. G1 = G is a P3-star or the disjoint union of an edge and a C3-star by assumption. If Gi is a P3-star or the disjoint union of an edge and a

C3-star then, by Theorem 4.0.45 and Theorem 4.0.45, it follows that Gi+1 is as well, thus completing the proof.

Finally, we can use the results just proved to characterize the forcibly chordal trees.

50 Theorem 4.0.49. The forcibly chordal trees are exactly the stars, P2-stars, and P3- stars.

Proof. We must show all forcibly chordal trees are of one of these three types, and conversely that graphs of these three types are in fact forcibly chordal.

First, we show every forcibly chordal tree T is in fact a star, P2-star, or P3-star. Choose a root r for T . We note that by part (ii) of Theorem 4.0.43, r has at most two nonleaf neighbors. Since r has zero, one, or two nonleaf neighbors, we consider these three cases.

First, if r has no nonleaf neighbors, then every neighbor of r is a leaf, therefore T consists only of r and its leaves, and T is a star.

Second, if r has exactly one nonleaf neighbor, call it v, let r = w, and then T consists of two adjacent vertices v and w together with pendant vertices adjoined to each. By deﬁnition, T is thus a P2-star with root set {v, w}. Third, assume r has exactly two nonleaf neighbors. Call them u and v and let r = w.

Then T [u, v, w] is a three point path. T consists of u, v, w, and pendant vertices adjoined to them. Therefore by deﬁnition, T is a P3-star as claimed. We have thus shown all forcibly chordal trees are stars, P2-stars, or P3-stars as claimed.

For the converse, let G be a star, a P2-star, or a P3-star. We must show G is forcibly chordal. Note that in all three cases, G is an induced subgaph of a P3-star H. H is forcibly chordal by Theorem 4.0.49, and an induced subgraph of a forcibly chordal graph is forcibly chordal, therefore G is forcibly chordal as claimed.

51 It is worth restating the previous theorem in two equivalent ways.

Corollary 4.0.50. The forcibly chordal trees are exactly the X-stars for which the graph X is a tree on at most three elements.

Corollary 4.0.51. The forcibly chordal trees are exactly the trees with at most three vertices of degree at least two.

Proof. Let T be a tree. If T has at most three vertices of degree at least two then the same is true of every graph G with the same degree sequence as T . Such a graph G does not have enough vertices of degree at least two to contain a hole. Therefore G is chordal. Since G is an arbitrary graph with the same degree sequence as T , we see that T is forcibly chordal.

Conversely, let T be a tree with at most three vertices of degree at least two. If T is a single vertex or an edge then T is trivially forcibly chordal as needed, so suppose not. Let X be the set of vertices v in T such that dT (v) ≥ 2, so that X has at most three vertices by hypothesis. Take any vertex y not in X. Then y must have degree at least one since T is a tree that is not itself an isolated vertex, and y must have degree at most one since y is not in X, so dT (y) = 1. Let z be the unique vertex of T adjacent to y. If z has degree 1 also then {y, z} is a component of T , contrary to assumption that T is not an edge. Therefore z has degree at least two, which means z is in X. Since y is an arbitrary vertex of V (T ) − X, we see that T is an X-star.

We need only show T [X] is a tree. T [X] is acyclic since T is acyclic. To show T [X] is connected, let v and w be disntinct vertices of X, and let let v1, . . . , vn be the unique

52 T -path from v to w, with v = v1 and w = vn. Since all interior points of this path have degree at least two, they must all lie in X. Since v and w are arbitrarily chosen distinct vertices of X, we see that T [X] is connected. Therefore T [X] is a tree, and the proof is complete.

53 CHAPTER 5

FORCIBLY CHORDAL FORESTS

In the last chapter, we characterized the forcibly chordal trees. In this chapter, we use this characterization to characterize the forcibly chordal forests. It should be noted this is not as simple as saying, “the forests whose components are forcibly chordal trees”. That condition is obviously necessary, but it is not suﬃcient. For instance, the forest 2P4 is the disjoint union of two forcibly chordal trees, but has the same ` degree sequence as 2P2 C4, and is therefore not a forcibly chordal forest.

Deﬁnition 5.0.52. Let m ≥ 0 and let G be a graph. Then mG denotes the disjoint union of m copies of G.

` 0 Lemma 5.0.53. Let G = H mP2 where H is an X-star with root set R. Let G have the same degree sequence as G. Then there is m0 ≥ 0 and graphs H0,X0 such

0 0 ` 0 0 0 that G = H m P2 and H is an X -star with the same root set R.

Proof. Since G and G0 have the same degree sequence, we may assume without loss of generality that V (G) = V (G0). It is immediate from the deﬁnitions of matching

0 and root set that dG(x) = 1 for all vertices x in V (G) − R. Since G and G have the

54 0 same degree sequence, we see that dG0 (x) = 1 for all x in V (G ) − R as well. Let M be the set of all x in V (G0) − R whose unique neighbor is also in V (G0) − R. Then

G[M] is a matching. Moreover, it then follows that the unique neighbor of every x in V (G0) − R − M is in R. The lemma then follows with m0 = |M|/2, X0 = G0[R], and H0 = G0[V (G) − M].

`n `n Lemma 5.0.54. For 1 ≤ i ≤ n, let Gi be an Xi-star. Then i=1 Gi is a i=1 Xi star.

Proof. Immediate from the deﬁnition of X-star.

Theorem 5.0.55. A forest is forcibly chordal iﬀ it is of one of the following types:

` ` (i) H iP2 jP1, where H is a P3-star and i, j ≥ 0.

` ` (ii) H iP2 jP1, where H is a P2-star, and i, j ≥ 0.

` ` (iii) H iP2 jP1, where H is the disjoint union of a P1-star and a P2-star, and i, j ≥ 0.

Sm ` ` (iv) n=1 Hn iP2 jP1, where i, j ≥ 0, each Hn is a P1-star, and 0 ≤ m ≤ 3.

Proof. First, we show that if G is of one of these four types then G is a forcibly chordal forest. Obviously, they are all forests. Note that since adding or removing isolated

55 vertices does not aﬀect the forcible chordality of a graph, we may assume j = 0. Note ` also that by the last lemma, a graph as in part (iii) is the disjoint union of a P1 P2- star and a matching, and a graph as in (iv) is the disjoint union of an mP1-star and a matching for some m with 0 ≤ m ≤ 3. Therefore in cases (i), (ii), (iii), and (iv), the graph is the disjoint union of a matching and an X-star, where X is a graph on at most 3 vertices. Moreover, by Lemma 5.0.53, it follows that any graph with the same degree sequence as one of these four types is the disjoint union of a matching and such an X-star as well. It is easy to see that if |X| ≤ 3 then the disjoint union of an X-star and a matching is chordal. Since all realizations of the four classes of graphs listed in the lemma arise as just such a disjoint union with |X| ≤ 3, it follows that these forests are all in fact forcibly chordal.

For the converse, we show that if G is a forcibly chordal forest then G is of one of the four types claimed in the theorem, so take a forcibly chordal forest G. Note each component of G is a forcibly chordal tree. Therefore by Theorem 4.0.49, we know each component of G is a star, a P2-star, or a P3-star. (It is worth noting at this point that an X-star is not assumed to have pendant vertices, so by our deﬁnition, an isolated vertex is a star.)

First, assume one of the components C is a P3-star that is not a P2-star. Then

C contains P5. If G has another component B such that B is not an edge or an ` isolated vertex then B contains P3. Therefore G contains P5 P3. By part (iii) of Theorem 4.0.43, we see that G is not forcibly chordal, contrary to assumption. This contradiction shows B must be an edge or isolated vertex. Since B was an arbitrarily chosen component of G other than C, it follows that G is the disjoint union of C, a

56 (possibly empty) set of edges, and a (possibly empty) set of isolated vertices, as in part (i) of this lemma, thus the lemma is proved in the case that G has a component that is a P3-star but not a P2-star.

So, we may now assume all components of G are either stars or P2-stars. Suppose G has at least two P2-star components that are not stars. Then each such component contains P4, therefore G contains 2P4. Therefore by part (v) of Theorem 4.0.43, it follows that G is not forcibly chordal, contrary to assumption. This contradiction shows G has only one P2-star component C that is not a star. Suppose G contains two star components other than C that are not edges or isolated vertices. Each ` such star component contains P3. Therefore G contains P4 2P3, so by part (vi) of Theorem 4.0.43, it follows that G is not forcibly chordal, contrary to assumption.

This contradiction shows that G has at most one component other than C that is not an edge or an isolated vertex, and if there is one, then it is a star. If there is no such star component other than the edges and isolated vertices, we are in case (ii) of this theorem. Otherwise, we are in case (iii). Hence in either case, we are done.

So we may now assume all components of G are stars. If G has at least 4 components that are not edges or isolated vertices then G contains 4P3, therefore by part (iv) of Theorem 4.0.43, it follows that G is not forcibly chordal, which is contrary to assumption. This contradiction shows there are at most three components that are not edges or isolated vertices. Therefore G is as claimed in part (iv) of this lemma, and the proof is complete.

Just as we stated the characterization of forcibly chordal trees in two equivalent ways

57 after the main theorem of the last chapter, we similarly give two equivalent statements of the above theorem now.

Corollary 5.0.56. The forcibly chordal forests are exactly the graphs of the form

a a H iP2 jP1 where H is an X-star for which the graph X is a forest on at most three elements.

Corollary 5.0.57. The forcibly chordal forests are exactly the forests with at most three vertices of degree at least two.

58 CHAPTER 6

FORCIBLY CHORDAL GRAPHS

In this chapter, we continue the partial characterization of forcibly chordal graphs.

First, we prove results concerning the structure of G[X ∪ Y ], where X and Y are maximal cliques of the forcibly chordal graph G. We will show that G[X ∪ Y ] has one of two structures, depending on whether or not there are edges from X − Y to

Y − X.

These structures, in addition to yielding interesting results in their own right, will be used to motivate a definition of adjacency of maximal cliques in a forcibly chordal graph. We believe that this definition of adjacency can, with further research, be used to give a complete characterization of connected, forcibly chordal graphs. As a partial result toward this end, we characterize forcibly chordal graphs that have a path structure to be defined below.

First, we note some basic deﬁnitions and results regarding chordal graphs to be used in the course of the chapter. The following deﬁnition and Proposition 6.0.59 may be found in [3].

Deﬁnition 6.0.58. Let G and H be graphs such that G ∩ H is a complete graph K, and such that there are no edges between V (G) − V (H) and V (H) − V (G). Then the graph G∪H is said to arise from G and H by pasting along the complete subgraph K.

59 When it is not necessary to make K explicit, we simply say G ∪ H arises from G and H by pasting along a complete subgraph.

Proposition 6.0.59. A graph G is chordal iﬀ it is a clique, or can be obtained from two smaller chordal graphs by pasting along a complete subgraph.

A universal vertex in a graph is a vertex that is adjacent to all other vertices. Our next two lemmas show that addition or deletion of universal vertices does not aﬀect the forcible chordality of a graph. This will allow us the convenience of ignoring the intersection of two cliques in certain cases, by allowing us to ﬁrst prove a statement in the case of disjoint cliques, and then add universal vertices to show the statement is still true, mutatis mutandis, in the case that the cliques are not necessarily disjoint.

Lemma 6.0.60. Let G be a graph and suppose v in G is complete to G − v. Then G is forcibly chordal iﬀ G − v is forcibly chordal.

Proof. First, note that induced subgraphs of forcibly chordal graphs are forcibly chordal so if G is forcibly chordal then so is G − v. For the converse, suppose G − v is forcibly chordal. We must show G is also.

So let H have the same degree sequence as G. We must show H is chordal. Since G and H have the same degree sequence, we see that v has the same degree in H as it does in G. Therefore v is complete to H − v. Note G − v and H − v have the same degree sequence, and G − v is forcibly chordal by assumption, therefore H − v

60 is chordal. So suppose H has a chordless cycle C on at least 4 vertices. Then v is in C. Therefore v is nonadjacent to at least one other vertex of C, contrary to the fact that v is complete to H − v. This contradiction proves H has no chordless cycle.

H is therefore chordal, which completes the proof.

Lemma 6.0.61. Let G be a graph, A ⊆ G, and suppose v is complete to G − v for all v in A. Then G is forcibly chordal iﬀ G − A is forcibly chordal.

Proof. Let |A| = n. Apply the last lemma n times.

Now, we begin determining the structure of G[X ∪Y ] for maximal cliques X and Y in a forcibly chordal graph G. The ﬁrst step in doing so is studying the neighborhoods in Y of vertices in X, and the neighborhoods in X of vertices in Y . In the next lemma, it is enough to assume that G is chordal, and we need not assume the cliques X and Y are maximal. Eventually, we will need to assume both forcible chordality of G, and maximality of the cliques X and Y .

Lemma 6.0.62. Let X and Y be two disjoint cliques in a chordal graph G, and choose v and w in X. Then NG(v)∩Y and NG(w)∩Y are comparable under inclusion.

Proof. Suppose not. Let NG(v) ∩ Y = Yv and NG(w) ∩ Y = Yw. Then Yv and Yw are incomparable under inclusion, so take s in Yv − Yw and t in Yw − Yv. Simple checking from the deﬁnitions shows that s, v, w, t comprise a four cycle, in that cyclic order.

61 Lemma 6.0.63. Let X and Y be disjoint, nonempty cliques in a chordal graph G.

Then there is n ≥ 1 and a partition of X into nonempty sets X1,...,Xn such that, given s in Xi and t in Xj, the following conditions hold:

(i) If i = j then Y ∩ NG(s) = Y ∩ NG(t).

(ii) If i < j then Y ∩ NG(s) ( Y ∩ NG(t).

Proof. Deﬁne a relation ≤ on X by letting s ≤ s0 iﬀ

0 Y ∩ NG(s) ⊆ Y ∩ NG(s )

If s ≤ s0 ≤ s00 then

0 00 Y ∩ NG(s) ⊆ Y ∩ NG(s ) ⊆ Y ∩ NG(s )

Therefore

00 Y ∩ NG(s) ⊆ Y ∩ NG(s )

Therefore s ≤ s00. This proves ≤ is a transitive relation. It follows by deﬁnition of ≤ that ≤ is reﬂexive as well.

A transitive, reflexive relation is known as a pre-order. It is a well known fact, and a simple exercise to show, that if we define s ∼ s0 iff s ≤ s0 and s0 ≤ s for a pre-order

≤ on X then ∼ is an equivalence relation on X, and that ≤ induces a partial order

0 ≤2 on the set X/ ∼ of ∼ equivalence classes. Namely, S ≤2 S for equivalence classes S and S0 iﬀ s ≤ s0 for some s ∈ S and for some s0 ∈ S0.

62 Now take two equivalence classes S and S0, and choose representatives s and s0,

0 0 respectively. By Lemma 6.0.62, we know that Y ∩NG(s) ⊆ Y ∩NG(s ) or Y ∩NG(s ) ⊆

0 0 0 0 Y ∩ NG(s). Therefore s ≤ s or s ≤ s. Therefore S ≤2 S or S ≤2 S. This shows ≤2 is in fact a total order on the ∼ classes.

Since X is finite and nonempty, there are finitely many nonempty ∼ classes. Let X1 be the ≤2 least ∼ class. Having defined X1,...,Xi for some i less than n, let Xi+1 be the ≤2 least ∼ class not already chosen as some Xj. The result follows with this choice of X1,...,Xn by definition of ≤ and ≤2.

Lemma 6.0.64. Let X and Y be disjoint, maximal cliques in a forcibly chordal graph G. If X is partitioned into sets X1,...,Xn as in the previous lemma and Y is similarly partitioned into sets Y1,...,Ym then m = n ≤ 2, and if n = 1 then there are no edges from X to Y .

Proof. We know the stated partition into Xi’s exists by the previous lemma. For each i, choose si in Xi. Then for i < j, the neighborhood of si in Y is a proper subset of the neighborhood of sj in Y .

Suppose n ≥ 3. Then there is t1 in Y adjacent to sn−1 but not sn−2, and there is t2 in Y adjacent to sn but not sn−1 or sn−2. Since Y is a maximal clique, it follows that no vertex of X is complete to Y . In particular, sn is not complete to Y . Therefore there is t3 in Y such that t3 and sn are not adjacent. Since sn has the largest possible neighborhood in Y under inclusion of all vertices of X, it follows that t3 is not adjacent to sn−1 or sn−2 either.

63 Consider H = G[sn−2, sn−1, sn, t1, t2, t3]. Simple checking shows that dH (sn−2) =

2, dH (sn−1) = 3, dH (sn) = 4, dH (t1) = 4, dH (t2) = 3, and dH (t3) = 2. Thus D(H) = (4, 4, 3, 3, 2, 2), which by simple checking, is also the degree sequence of

0 H = C6 +v1v3 +v1v4 +v2v4, where C6 is taken to be the cycle on vertices v1, v2, . . . , v6

0 in that cyclic order. Trivial checking shows H [v1, v4, v5, v6] is a four cycle in that cyclic order. Therefore H0 is not chordal. Since H and H0 have the same degree sequence, it follows that H is not forcibly chordal. Since H is an induced subgraph of G, it follows G is not forcibly chordal, contrary to assumption. This contradiction shows that n ≤ 2 as the lemma claims.

We must now show that m = n, and that if n = 1 then there are no edges from X to Y . We consider the cases n = 1 and n = 2.

First, suppose n = 1. Then X = X1, and any two vertices of X therefore have the same neighborhood in Y . Suppose this neighborhood contains a vertex t in Y . Then t is adjacent to every vertex of X, so X ∪ {t} is a clique. Since X and Y are assumed disjoint, it follows that t is not in X, therefore X ∪ {t} is a clique properly containing the maximal clique X, a contradiction. This contradiction shows the neighborhood in Y of each vertex of X is empty. It then follows that the neighborhood in X of each vertex of Y is empty, and thus, as the lemma claims, there are no edges from X to Y . Moreover, since all vertices of Y have the same neighborhood in X, namely the empty set, we see that m = 1 as claimed.

Second, assume n = 2. We must show m = 2. The same argument we used to show n ≤ 2 also shows that m ≤ 2. If m = 1 then by symmetry, the argument used

64 in the last paragraph shows that n = 1, contradiction. Therefore m = 2 as claimed, and the proof is complete.

Lemma 6.0.65. Let X and Y be disjoint, maximal cliques of a forcibly chordal graph G, let X1,...,Xn and Y1,...,Yn be as in the previous lemma, and 1 ≤ i, j ≤ n.

Then Xi is complete to Yj or anti-complete to Yj.

Proof. Suppose Xi is not anti-complete to Yj. It is enough to show Xi is complete to Yj. Since Xi is not anti-complete to Yj, we can choose x in Xi and y in Yj

0 0 0 0 such that x is adjacent to y. Now take x in Xi and y in Yj. Since x and y are arbitrary, it is enough to show x0 is adjacent to y0. The vertex x is adjacent to y by assumption. By deﬁnition of Xi, any two vertices of Xi have the same neighborhood in Y . Therefore x0 is adjacent to y as well. Since y is adjacent to x0 and since any two

0 0 vertices of Yj have the same neighborhood in X, it follows that y is adjacent to x as well. This completes the proof.

Corollary 6.0.66. Let X and Y be disjoint, maximal cliques of a forcibly chordal graph G with at least one edge between them. Then X can be partitioned into cells A and B, and Y into cells C and D, such that the following conditions hold:

(i) A is anti-complete to C

(ii) A is anti-complete to D

65 (iii) B is anti-complete to C

(iv) B is complete to D

(v) A is complete to B

(vi) C is complete to D

Proof. We have two disjoint, maximal cliques X and Y as in the previous lemma, so take n = m as in the statement of Lemma 6.0.63. Lemma 6.0.64 shows that n ≤ 2, and that if n = 1 then there are no edges from X to Y . Since we have at least one edge from X to Y by assumption, we see that n = 2. Let A = X1, B = X2, C = Y1, and D = Y2, so that {A, B} is a partition of X and {C,D} is a partition of Y . We show this choice of A, B, C, and D satisﬁes the desired properties.

First, note that A is trivially complete to B since they are both subsets of the clique X, and C is trivially complete to D since they are both subsets of the clique Y . These two facts were simply stated for easy reference. We now prove the remaining four.

We know by the previous lemma that A is complete or anti-complete to D. Suppose A is complete to D. Since the neighborhood of B in Y contains the neighborhood of A in Y , it follows that B is complete to D as well. Therefore D is complete to X.

Since D is nonempty, it follows that X ∪ D is a clique strictly containing X, contrary to choice of X as maximal. This contradiction shows that A is anti-complete to D.

By symmetry, we also know that C is anti-complete to B. Since C is anti-complete to B, and since the neighborhood of A in Y is contained in the neighborhood of B in Y , it follows that A is anti-complete to C.

66 Suppose B is anti-complete to D. Since we’ve already shown A is anti-complete to C, A is anti-complete to D, and C is anti-complete to B, it follows that X is anti-complete to Y , contrary to assumption. This contradiction shows B is not anti- complete to D. Therefore B is complete to D by the previous lemma, thus completing the proof.

Corollary 6.0.67. Let X and Y be disjoint, maximal cliques of a forcibly chordal graph G such that there is at least one edge between them. Then there is a in X and c in Y such that the following conditions hold.

(i) a is not adjacent to c.

(ii) a is anti-complete to Y − c

(iii) X − a is anti-complete to c

(iv) X − a is complete to Y − c

(v) a is complete to X − a

(vi) c is complete to Y − c

Proof. The hypotheses of this corollary are the same as those of the one preceding it. First, let A, B, C, and D be as in the previous corollary. We show A and C have exactly one vertex.

First, note that there is an edge from X to Y by hypothesis, and by the last lemma, this edge must be from B to D. Therefore B and D are nonempty. Suppose A is

67 empty. Then X = B. Since B is complete to D, it follows that X is complete to D.

Since D is nonempty, it follows that X ∪ D is a clique strictly containing X, contrary to maximality of X. This contradiction shows that A is nonempty. By symmetry, C is nonempty as well.

Suppose A has at least two vertices. Call them a and a0. We have already noted that B, C, and D are nonempty, so take b in B, c in C, and d in D. The reader may check that a, a0, and b are pairwise adjacent, d is adjacent to b and c, and there are no other edges between these 5 vertices. Therefore G[a, a0, b, c, d] has degree sequence

(3, 2, 2, 2, 1), which is also the degree sequence of a 4 cycle with a pendant vertex.

A 4 cycle with a pendant verex is of course not chordal, therefore G is not forcibly chordal, contrary to hypothesis. This contradiction shows A does not have at least two vertices. Since A is nonempty, it follows that A has exactly one vertex as claimed.

By symmetry, C has exactly 1 vertex as well.

Since A and C are singletons, it follows that A = {a} and C = {c} for some a and c.

Then B = X − a and D = Y − c. The reader may check that the six conclusions of this corollary are exactly the six conclusions of the previous corollary for these choices of A, B, C, and D. The result follows.

The following remark is key to what follows. The previous lemma has a converse as well, in the sense that if vertices a and c and cliques X and Y in a graph G satisfy the six conclusions of the lemma then G[X ∪ Y ] is a forcibly chordal graph. To see this, note that those six statements imply G[X ∪ Y ] is a split graph, and is thus forcibly

68 chordal. Therefore the above lemma can be considered the full structure theorem for two disjoint, maximal cliques in a forcibly chordal graph G.

The next proposition, a corollary of the previous, generalizes this structure to the case that the cliques X and Y are not necessarily disjoint, and can similarly be considered the full structure theorem in that somewhat more general setting.

Proposition 6.0.68. Let X and Y be maximal, not necessarily disjoint cliques of a forcibly chordal graph G such that there is at least one edge between X − Y and

Y − X. Then there is a in X and c in Y such that the following conditions hold.

(i) a is not adjacent to c.

(ii) a is anti-complete to Y − X − c

(iii) X − Y − a is anti-complete to c

(iv) X − Y − a is complete to Y − X − c

(v) a is complete to X − Y − a

(vi) c is complete to Y − X − c

Proof. This follows immediately from the previous lemma by noting that X − Y and

Y − X are disjoint, maximal cliques of the graph G − (X ∩ Y ).

Note that in the previous lemma, we referred to edges from X − Y to Y − X. This comes up often enough that the following deﬁnition will simplify the wording of a number of statements to come.

69 Deﬁnition 6.0.69. Let X and Y be cliques in a graph G. We say that X and Y have extra edges between them if there is at least one edge from X − Y to Y − X.

Otherwise, the cliques are said to have no extra edges between them.

So, the previous lemma gave the structure theorem for two maximal cliques in a forcibly chordal graph in the case that there is at least one extra edge between them.

Now that we have a structure theorem for the union of two cliques in this case, we wish to do the same in the case that there are no extra edges between them.

Proposition 6.0.70. Let X and Y be two distinct, maximal cliques of a graph G such that V (G) = X ∪Y , and such that there are no extra edges from X to Y . Then G is forcibly chordal iﬀ |X − Y | ≤ 2 or |Y − X| ≤ 2.

Proof. Note that every vertex v of X ∩ Y is complete to G − v. Since addition or deletion of universal vertices does not aﬀect the forcible chordality of a graph, we may assume without loss of generality that X ∩ Y = ∅. So X and Y are disjoint, maximal cliques of G. Therefore X − Y = X and Y − X = Y , so we may simply refer to X and Y throughout instead of X − Y and Y − X, and refer simply to edges from X to Y instead of extra edges from X to Y .

Suppose ﬁrst that |X| = 2. Then G is the disjoint union of a clique Kn and an edge. We must show G is forcibly chordal. We do this by showing there are in fact only two graphs up to isomorphism with the same degree sequence as G, and noting that each of these graphs is chordal. So let H be a graph with the same degree sequence as G.

70 Then H has exactly two vertices x and y of degree 1. Either x and y are adjacent or they are not. We consider these two cases. ` First, assume x and y are adjacent. Since D(H) = D(Kn P2), we see the remaining n vertices of H each have degree n−1. Since x and y have degree 1 in G and they are adjacent to one another, we see they are adjacent to no other vertices. Therefore for any vertex z in H − {x, y}, the neighborhood of z in H is contained in the n − 1 vertex set H − {x, y, z}. Since z has degree n − 1, it follows that z is complete to

H −{x, y, z}. Since z is an arbitrary vertex of H −{x, y}, it follows that H −{x, y} is ` complete. Therefore H is isomorphic to Kn P2, and this graph is trivially chordal. Second, assume x and y are not adjacent. Each has degree 1. Suppose ﬁrst that x and y are adjacent to the same vertex z. We know dH (z) = n − 1. Therefore z is adjacent to exactly n − 3 vertices of the n vertex set H − {x, y}. Therefore there is some vertex w that is not adjacent to z. Then w is adjacent to at most n − 2 vertices of H −{x, y}. Since x and y are adjacent to z and have degree 1, we know also that w is not adjacent to x or y. Therefore dH (w) ≤ n − 2, contrary to the fact that each vertex of H has degree 1 or n − 1. This contradiction shows that x and y can not be adjacent to the same vertex z.

So let x be adjacent to z and y be adjacent to w. Then z and w are each adjacent to exactly n − 2 other vertices of H − {x, y}, and every other vertex of H − {x, y} is adjacent to n−1 vertices of H −{x, y}. Therefore H −{x, y} = Kn −zw. Therefore H is uniquely determined up to isomorphism in the case that x and y are not adjacent.

It is trivial to check that this graph H is chordal.

71 This shows that each of the two possible realizations of the degree sequence of G is chordal in the case that |X| = 2. Therefore G is forcibly chordal in this case.

Suppose now that the cardinality of X less than 2. Then G is contained in the disjoint union of a clique and an edge, which is, by the previous paragraphs, forcibly chordal.

Since induced subgraphs of forcibly chordal graphs are forcibly chordal, it follows that G is forcibly chordal in this case as well. Similarly if |Y | ≤ 2. This shows that G is forcibly chordal if X or Y has at most two vertices, as claimed.

For the converse, we must show that G is not forcibly chordal if both X and Y have at least three vertices. So suppose both X and Y have at least 3 vertices. Then both X and Y contain C3. Therefore G contains 2C3 as an induced subgraph, since there are no edges from X to Y . Therefore by part (vii) of Theorem 4.0.43, we see that G is not forcibly chordal. This completes the proof.

By Proposition 6.0.59, every chordal graph can be obtained by starting with cliques and repeatedly pasting along complete subgraphs. In particular, the same is true for forcibly chordal graphs. It is therefore natural to study the structure of forcibly chordal graphs by viewing a forcibly chordal graph as a tree of cliques.

Viewing a chordal graph as a tree of cliques is by no means new. It is well known [3] that every chordal graph has a tree decomposition whose parts are cliques. While a discussion or deﬁnition of tree decompositions is outside the scope of this writing, suf-

ﬁce it to say such a tree decomposition of a chordal graph is certainly most naturally thought of as a tree of cliques.

However, in the case of forcibly chordal graphs, we can do better, and pick out certain

72 trees as most natural. If X and Y are maximal cliques of a forcibly chordal graph G such that X and Y have extra edges between them, we see that the graph G[X ∪ Y ] actually has three maximal cliques. These cliques are X, Y , and Z = X ∪ Y − {a, c}, in the notation of Proposition 6.0.68. Now, Z ∩ X and Z ∩ Y both strictly contain

X ∩ Y . Thus intuitively, Z seems more connected to X and Y than X and Y do to one another. We make this intuition precise in the following deﬁnition.

Deﬁnition 6.0.71. Let G be a graph. Deﬁne the graph MG of maximal cliques of G as the graph such that

(i) V (MG) is the set of maximal cliques of G.

(ii) Maximal cliques in G are adjacent in MG iﬀ they intersect nontrivially and have no extra edges between them.

We wish to view each connected, forcibly chordal graph as a tree of cliques in such a way that adjacent cliques have no extra edges between them. Such a tree exists, as a (not necessarily induced) spanning subtree of MG, if and only if MG is connected. Our next two lemmas show this is indeed the case.

Lemma 6.0.72. Let G be a connected, forcibly chordal graph and let X and Y be distinct, maximal cliques with extra edges between them. Then X and Y are in the same component of MG.

73 Proof. We choose a and c as in Proposition 6.0.68. X ∪Y −{a, c} is a clique. Enlarge this to a maximal clique M in G. Suppose there is an extra edge from M to X. M contains all vertices of X −a, so the extra edge must be from some vertex m of M −X to a. But then X ∪ {m} is a clique in G, contradicting choice of X as maximal. This shows there are no extra edges from M to X. By symmetry, there are no extra edges from M to Y .

But we know M intersects both X and Y . Therefore M is adjacent to both X and Y in MG. Therefore X and Y are in the same component of MG, completing the proof.

Lemma 6.0.73. Let G be a connected, forcibly chordal graph. Then MG is connected.

Proof. Suppose not. Then MG contains at least two components. We know by the previous lemma that any two maximal cliques in G with extra edges between them lie in the same component of MG, so that by the contrapositive, any two maximal cliques of G in distinct components of MG have no extra edges between them. Suppose that X and Y are disjoint for every two maximal cliques X and Y of G in distinct components of MG. Since any such pair also has no extra edges between them, and since MG has at least two components, it follows that G is disconnected, contrary to hypothesis. This contradiction shows that there must be some cliques X and Y in distinct components of MG such that X ∩ Y is nonempty. Since X and Y are not adjacent in MG, it follows that there are extra edges between them. Therefore

74 they lie in the same component of MG by the previous lemma, contrary to hypothesis. This contradiction proves the lemma.

Since we know that MG is connected for each forcibly chordal graph G, we may simply take a spanning tree of MG in order to view G as a tree of cliques in such a way that adjacent cliques have no extra edges between them. It is our belief that this line of reasoning will lead to a complete characterization of connected, forcibly chordal graphs. As a partial result, we characterize forcibly chordal graphs whose cliques have a path structure. First, we rigorously deﬁne this path structure.

Deﬁnition 6.0.74. Let G be a graph, and let P be an (induced) path in MG. Then G is said to have a P path structure if every pair of nonadjacent points X,Y in P satisﬁes X ∩ Y = ∅ when considered as subsets of G.

We say that G has a path structure if G has a P path structure for some P . We now characterize the forcibly chordal graphs with a path structure. Any path P is isomorphic to Pn for some nonnegative integer n, so it is enough to characterize, for each n, the forcibly chordal graphs with path structure Pn. Toward this end, we ﬁrst state a trivial lemma.

Lemma 6.0.75. A graph has a P1 structure iﬀ it is a clique.

The above lemma, in its triviality, did not require the assumption of forcible chordality. For n ≥ 2, that assumption is needed.

75 Theorem 6.0.76. The forcibly chordal graphs G with a P2 structure are exactly the graphs G such that V (G) is the disjoint union of sets X1, X2, and X3 such that the following conditions hold:

(i) Each Xi is nonempty.

(ii) X2 is complete to X1 and X3.

(iii) X1 is anti-complete to X3.

(iv) Either X1 or X3 has at most two vertices.

Proof. There are two directions to show. First, suppose there are three such sets X1,

X2, and X3 in a graph G. We must show G is a forcibly chordal graph with a P2 structure.

First, we show G is forcibly chordal. Note that X2 is complete to G − X2. Therefore, by Lemma 6.0.61, we see that G is forcibly chordal iﬀ G − X2 = G[X1 ∪ X3] is. But

G[X1 ∪ X3] is the disjoint union of two cliques with no edges between them. By Proposition 6.0.70, we know such a graph is forcibly chordal iﬀ one of the two cliques has at most two elements. But this is exactly condition (iv), therefore G is indeed forcibly chordal.

Next, we show G has a P2 structure. Simple checking shows the maximal cliques of G are exactly X1 ∪ X2 and X2 ∪ X3. Since X2 is nonempty, we see these two maximal cliques have nontrivial intersection. Moreover, since X1 is anti-complete to X3, we see the maximal cliques have no extra edges between them. So consider the path P

76 whose vertex set is the doubleton {X1 ∪X2,X2 ∪X3}, and such that these two vertices are adjacent. Then G has a P path structure by deﬁnition, as was to be shown.

This completes the proof of one direction of the lemma. For the converse, we suppose G is a forcibly chordal graph with a P2 structure. We must show three sets

X1, X2, and X3 with the desired properties exist. Let X and Y be the two vertices of P . Letting X1 = X − Y , X2 = X ∩ Y , and X3 = Y − X, we see that V (G) is the disjoint union of X1, X2, and X3. Since X and Y are adjacent in P , we see that

X2 = X ∩ Y 6= ∅. Moreover, X and Y are distinct, maximal cliques of G. Therefore

X1 = X − Y and X3 = Y − X are nonempty, so condition (i) holds.

For condition (ii), note that since X is a clique, in particular X1 = X −Y is complete to X2 = X ∩ Y . Similarly, X3 is complete to X2. For condition (iii), note that since X and Y are adjacent in P , it follows that the cliques X and Y have no extra edges between them. By deﬁnition, this means there are no edges from X1 = X − Y to X3 = Y − X. Therefore X1 is anti-complete to X3, as was to be shown.

For condition (iv), note that since G is forcibly chordal, so is G[X1 ∪ X3]. But

G[X1 ∪ X3] is the disjoint union of two cliques with no edges between them. By

Proposition 6.0.70, this can only be forcibly chordal if X1 or X3 has at most two elements. This completes the proof of the lemma.

Next, we characterize the forcibly chordal graphs with a P3 structure. Intuitively, we do this by thinking of P3 as two P2’s spliced together, and noting that each of those P2’s must yield the structure described in the previous lemma.

77 Theorem 6.0.77. The forcibly chordal graphs G with a P3 structure are exactly the graphs G such that V (G) is the disjoint union of sets X1,...,X5 such that the following conditions hold:

(i) X1, X2, X4, and X5 are nonempty.

(ii) Xi is complete to Xi+1 for 1 ≤ i ≤ 4 and X2 is complete to X4.

(iii) All other pairs of Xi and Xj are anti-complete.

(iv) X1 and X5 are singletons.

Proof. There are two directions to show. First, suppose there are ﬁve such sets

X1,...,X5 in a graph G. We must show G is a forcibly chordal graph with a P3 structure.

First, we show G is forcibly chordal. Note that X2 ∪ X3 ∪ X4 is a complete graph.

Also, the singletons X1 and X5 are anti-complete to each other. Therefore G is a split graph, and in particular forcibly chordal.

Next, we show G has a P3 structure. Simple checking shows the maximal cliques of G are exactly X = X1 ∪ X2, Y = X2 ∪ X3 ∪ X4, and Z = X4 ∪ X5. Let P be the path with vertex set {X,Y,Z} such that X and Y are adjacent, Y and Z are adjacent, but X and Z are not adjacent. Obviously, P is isomorphic to P3. We show G has a P structure.

Since each Xi except possibly X3 is nonempty, we see that both X ∩ Y = X2 and

Y ∩ Z = X4 are nonempty. It is also immediate from the deﬁnitions of X, Y , and Z

78 that there are no extra edges from X to Y , and no extra edges from Y to Z. Finally, by deﬁnition of X and Z, we see that X ∩ Z = ∅. Together, these facts imply, by deﬁnition of path structure, that G has path structure P , a P3 path structure, as was to be shown.

This completes the proof of one direction of the lemma. For the converse, we suppose G is a forcibly chordal graph with a P3 structure P . We must show ﬁve sets

X1,...,X5 with the desired properties exist. Let X, Y , and Z be the three vertices of P , where P is a three point path such that G has path structure P . Let

X1 = X − Y , X2 = X ∩ Y , X3 = Y − X − Z, X4 = Y ∩ Z, and X5 = Z − Y . We show that X1,...,X5 have the desired properties. First, note that since X, Y , and Z are the three vertices of P , it follows that X,

Y , and Z are exactly the maximal cliques of G. In particular, V (G) = X ∪ Y ∪ Z.

Moreover, simple checking shows that

5 [ X ∪ Y ∪ Z = Xi i=1

S5 Therefore V (G) = i=1 Xi. Simple checking also shows Xi ∩ Xj = ∅ for i 6= j, so

V (G) is the disjoint union of the Xi’s.

For condition (i), we must show each Xi is nonempty except possibly X3. Since X and Y are distinct, maximal cliques of G, it follows that X1 = X − Y is nonempty.

Similarly, X5 = Z −Y is nonempty. X and Y are adjacent in P , therefore X2 = X ∩Y is nonempty. Similarly, X4 = Y ∩ Z is nonempty. This proves condition (i). Next, we prove condition (ii). For each i with 1 ≤ i ≤ 4, the reader can check that

Xi ∪ Xi+1 is contained in one of the three cliques X, Y , or Z. Therefore Xi is

79 complete to Xi+1. Similarly, X2 ∪ X4 is contained in the clique Y . Therefore X2 is complete to X4. This proves condition (ii).

For condition (iii), we must show Xi is anti-complete to Xj for all other pairs i and j.

In other words, we must show X1 is anti-complete to X3, X4, and X5, and that X2 and X3 are anti-complete to X5. That X1 is anti-complete to X3 and X4 follows from the fact that there are no extra edges from X to Y . That X2 and X3 are anti-complete to X5 follows from the fact that there are no extra edges from Y to Z.

Suppose X1 is not anti-complete to X5. Then there is an edge from x1 to x5 for some x1 ∈ X1 and x5 ∈ X5. Choose x2 in X2 and x4 in X4. By the pairs we have shown to be complete and anti-complete, respectively, it follows that x1, x2, x4, x5 comprise a cycle in that cyclic order, contrary to choice of G as forcibly chordal. Therefore X1 is anti-complete to X5. This completes the proof of (iii).

For (iv), we note that by symmetry, it is enough to show that X1 is a singleton.

To see that X1 is a singleton, suppose not. We know X1 is not empty, so if it is not a singleton then it has at least two elements. Call them v and w. Choose xi in Xi for i = 2, 4, 5, and consider H = G[u, v, x2, x4, x5]. Simple checking shows

{u, v, x2} is a triangle, x2 is adjacent to x4, x4 is adjacent to x5, and H has no other edges. Therefore D(H) = (3, 2, 2, 2, 1). Simple checking shows this is also the degree sequence of a square with a pendant edge, which of course is not chordal. Therefore H is not forcibly chordal. Since H is an induced subgraph of G, we see that G is not forcibly chordal, contrary to assumption. This contradiction shows that X1 is in fact a singleton, thus proving (iv), and completing the proof of the lemma.

80 Theorem 6.0.78. The forcibly chordal graphs G with a P4 structure are exactly the graphs G such that V (G) is the disjoint union of sets Y1,...,Y5 such that the following conditions hold:

(i) Each Yi is nonempty.

(ii) Yi is complete to Yi+1 for 1 ≤ i ≤ 4.

(iii) All other pairs of Yi and Yj are anti-complete.

(iv) Y1, Y2, Y4, and Y5 are singletons.

(v) Y3 is a clique.

Proof. There are two directions to show. First, suppose there are ﬁve such sets

Y1,...,Y5 in a graph G. We must show G is a forcibly chordal graph with a P4 structure.

First, we show G is forcibly chordal. For i = 1, 2, 4, 5, let the singleton Yi equal {yi}.

Then y1 and y2 are adjacent, y2 and y4 are not, y4 and y5 are adjacent, and y5 and y1 are not. Therefore we may apply the switching operation y1, y2, y4, y5 to obtain a new

0 0 0 graph G . The reader can check that G [y1, y5] is an edge and G [Y2 ∪ Y3 ∪ Y4] is a

0 clique. Also, the reader may check that G has no edges from {y1, y2} to Y2 ∪ Y3 ∪ Y4. Therefore G0 is the union of two disjoint cliques, one of which has only two vertices.

By Proposition 6.0.70, we know G0 is forcibly chordal. Since G and G0 have the same degree sequence, it then follows that G is forcibly chordal.

81 Next, we show G has a P4 structure. Simple checking shows the maximal cliques of G are exactly X = Y1 ∪Y2, Y = Y2 ∪Y3, Z = Y3 ∪Y4, and W = Y4 ∪Y5. Let P be the path with vertex set {X,Y,Z,W } such that X and Y are adjacent, Y and Z are adjacent,

Z and W are adjacent, and no other pairs of vertices are adjacent. Obviously, P is isomorphic to P4. We show G has a P structure.

Since each Yi is nonempty, we see the sets X ∩ Y = Y2, Y ∩ Z = Y3, and Z ∩ W = Y4 are nonempty. It is also immediate from the deﬁnitions of X, Y , Z, and W that there are no extra edges from X to Y , from Y to Z, or from Z to W . Finally, by deﬁnition of X,Y,Z,W we see that

X ∩ Z = X ∩ W = Y ∩ W = ∅

Together, these facts imply, by deﬁnition of path structure, that G has path structure P , a P4 path structure, as was to be shown. This completes the proof of one direction of the lemma. For the converse, we suppose G is a forcibly chordal graph with a P4 structure P . We must show ﬁve sets

Y1,...,Y5 with the desired properties exist. Let X, Y , Z, and W be the four vertices of P , where P is a four point path such that G has path structure P .

Note that G[X ∪ Y ∪ Z] has path structure P [X,Y,Z]. Since P [X,Y,Z] is a three point path, we may use to deﬁne sets X1,...,X5 as in that lemma. Namely, we let

X1 = X − Y , X2 = X ∩ Y , X3 = Y − X − Z, X4 = Y ∩ Z, and X5 = Z − Y , and we know by the last lemma that these ﬁve sets satisfy that lemma’s conditions.

Letting Y1 = X1, Y2 = X2, Y3 = X4, Y4 = X5, and Y5 = W − Z, we show that

Y1,...,Y5 have the desired properties.

82 First, we show each Yi is nonempty. Note that X1,X2,X4,X5 are nonempty by the previous lemma, so Y1,Y2,Y3,Y4 are nonempty. Since W and Z are distinct, maximal cliques, we know neither is a subset of the other. Therefore Y5 = W − Z is nonempty as well. This completes the proof of condition (i).

Next, we show condition (ii), namely, that Yi is complete to Yi+1 for i = 1, 2, 3, 4. By the previous lemma, we know that Y1 = X1 is complete to Y2 = X2, that Y2 = X2 is complete to Y3 = X4, and that Y3 = X4 is complete to Y4 = X5.

It remains to show Y4 is complete to Y5. To do this, we ﬁrst need to show that Z − Y = Z ∩ W . Let v be a vertex of Z ∩ W . We know Y and W are disjoint by the path structure of G and non-adjacency of Y and W in P . Therefore v is in Z − Y .

Since v was chosen arbitrarily, we see that Z ∩ W ⊆ Z − Y . But Z − Y = X5 is a singleton by the previous lemma. Moreover, Z ∩ W is nonempty since Z and W are adjacent in P . Since the nonempty set Z ∩ W is a subset of the singleton Z − Y , it follows that these two sets are equal as claimed.

We now note that Y4 = X5 = Z − Y = Z ∩ W and Y5 = W − Z are both subsets of the clique W . In particular, Y4 is complete to Y5 as condition (ii) claims.

Next, we must prove condition (iii), that all other pairs Yi and Yj are anti-complete.

Without loss of generality, i < j. If j 6= 5 then the fact that Yi and Yj are anti- complete follows immediately from the last lemma and the deﬁnition of each Yi and Yj as an Xk for some k. It remains to show that Y5 is anti-complete to Y1,Y2,Y3. Note that Y1 = X1 = X − Y ⊂ X, Y2 = X2 = X ∩ Y ⊂ X, and Y3 = X4 = Y ∩ Z.

Therefore Yi is a subset of X ∪ Y for i = 1, 2, 3, and Y5 = W − Z is a subset of W . X ∪ Y and W are anti-complete by the path structure of G and the fact that W is

83 not adjacent to X or Y in P . In particular, Y5 is anti-complete to Y1, Y2, and Y3. This completes the proof of condition (iii).

Now, we must show that Y1, Y2, Y4, and Y5 are singletons. Y1 = X1 and Y4 = X5 are singletons by the previous lemma. The previous lemma shows, among other things, that if G has a three point path structure such that the path consists of the vertices

A, B, C in that order, then A − B and C − B are singletons. We can apply this also to G[Y ∪ Z ∪ W ], which has path structure P [Y,Z,W ], to see that W − Z = Y5 is a singleton, and that Y − Z is a singleton. Now, Y2 = X2 = Y ∩ X. Moreover, if v is a vertex of Y ∩ X then v is not in Z since X and Z are disjoint. Therefore v is in Y − Z. Since v is arbitrary, we see that the nonempty set Y ∩ X is subset of the singleton Y − Z. Therefore Y2 = Y ∩ X = Y − Z is a singleton, proving condition (iv).

To show (v), note that if Y3 is not a clique then Y3 must contain two nonadjacent vertices v and w. The reader may then check that G[y2, v, w, y4] is a four cycle, contrary to forcible chordality of G. Therefore Y3 is a clique as claimed.

We have only to show that V (G) is the disjoint union of Y1,...,Y5. We know by the previous lemma that X ∪ Y ∪ Z is the disjoint union of X1,...X5. Since W is disjoint to X and Y , it follows that W − Z = Y5 is disjoint to X ∪ Y ∪ Z. Therefore

V (G) = X ∪ Y ∪ Z ∪ W is the disjoint union of X1,...,X5 and Y5. Since Y1 = X1,

Y2 = X2, Y3 = X4, and Y4 = X5, the proof will thus be complete if we show that X3 is empty.

Suppose not. Then choose xi in Xi for 2 ≤ i ≤ 5 and y5 in Y5. Having shown which sets are complete and anti-complete to one another, respectively, it is simple

84 to check which of these vertices are adjacent. Doing this, one can check that the degree sequence of H = G[x2, x3, x4, x5, y5] is (3, 2, 2, 2, 1), which is also the degree sequence of a square with a pendant edge. As noted previously, this graph is not chordal. From this we see that H is not forcibly chordal, and therefore G is not forcibly chordal.

This contradiction shows that in fact X3 is empty, which completes the proof.

Corollary 6.0.79. Let n ≥ 5. Then there are no forcibly chordal graphs with a Pn structure.

Proof. It is enough to show there are no forcibly chordal graphs with a P5 structure.

We give a proof by contradiction. Suppose a forcibly chordal graph G has a P5 structure P . Call the vertices of the path X, Y, Z, W, V , in that order. We then get a P [X,Y,Z,W ] structure for G[X ∪ Y ∪ Z ∪ W ]. This path structure gives us sets

Y1,...,Y5 as in the previous lemma. Choose yi in Yi for 1 ≤ i ≤ 5. By the path structure P [Y, Z, W, V ] for G[Y ∪ Z ∪ W ∪ V ], we know that V − W is a singleton.

Without loss of generality, V − W = {v}. It is a matter of checking, using the complete and anti-complete pairs of the previous lemma, to see that G[y1, . . . , y5, v] is a six point path with the vertices in that order. A six point path is not chordal.

In particular, G is not forcibly chordal, a contradiction.

85 CHAPTER 7

COMBINATORIAL PROOF SYSTEMS

7.1 Introduction

This chapter begins the study of proof systems. While this may seem more related to logic than combinatorics, we study proof systems in an abstract, general setting. In this general setting, proof systems provide a common, yet natural generalization of graphs and partial orders, and their study is essentially combinatorial in nature. For instance, our characterization of ﬁnite partial orders by forbidden subdots in Corol- lary 7.11.17 is similar to the many forbidden minor theorems of graph theory, and we will also see that proof systems yield a common framework in which to simultaneously treat questions on graph matching and graph connectivity.

Our eventual formalization of “proof system” is what we will call an autonomous system, or ausys, which is essentially what one obtains by axiomatizing partial orders in terms of their downward closed sets then discarding the assumption that downward closed sets are closed under arbitrary intersection. Mathematically, we could make this deﬁnition now and immediately start proving ausys theorems. Pedagogically, the reader would have little to no intuition for what is going on. To address this issue, we start with motivation and the deﬁnition of what we call preceding set proof

86 systems. Preceding set proof systems are easily motivated, and we then deﬁne the mathematically more convenient ausyses in terms of them.

7.2 Proof Closure

We will be looking at abstract proof systems in which proofs are partial orders such that the set of predecessors of any element in the partial order is considered enough information to “infer” that element.

Given a set T of partial orders of subsets of a set P , we want to know what other partial orders <0 of subsets of P are to be considered proofs if every partial order in

T is. We want a way to say for various T and ≤ that if every partial order in the set

T is considered a proof then we must consider ≤ to be a proof also.

First, we make a deﬁnition that facilitates the discussion and makes things easier to understand.

Deﬁnition 7.2.1. Let (P, ≤P ) be a partial order and x in P . Then {≤P x} denotes the set {y ∈ P : y ≤P x}. Similarly for {

P x}.

Now, suppose we consider every partial order in a set T of partial orders to be a proof. Let x be in the domain of <, where < is in T . The set {< x} is considered enough information to infer x since < is in T and we are considering everything in

T to be a proof. So if {<0 x} contains {< x} for some partial order ≤0, then {<0 x} should be considered enough information to infer x too.

87 By this logic, if for every x in the domain of <0 there is some order < in T such that

{<0 x} contains {< x}, then for every x in the domain of <0, the set {<0 x} should be considered enough information to infer x. So <0 should be considered a proof.

This leads us to the following deﬁnition:

Deﬁnition 7.2.2. Let T be a set of partial orders of subsets of the set P . A partial order <0 whose domain is a subset of P is in the proof closure of T if for every x in the domain of <0 there is < in T such that the set {<0 x} contains the set {< x}. T is proof closed if T equals the proof closure of T .

It is simple to check the name proof closure is justiﬁed in the sense that it is indeed a closure operator on the set of partial orders of a given set.

To illustrate the definition of proof closure, note that for example, every downward closed subset of a partial order in T is in the proof closure of T . Once the merge is defined in Definition 7.3.1, we will see the merge of a sequence in T is also in T .

Given two partial orders ≤1 and ≤2 on a set P , we say that ≤2 lengthens ≤1, or ≤1 fattens ≤2, if x ≤1 y implies x ≤2 y for all x and y in P . With this convention, we may note a lengthening of any partial order in T is in the proof closure of T .

7.3 The Merge

The merge is a partial order construction useful for constructing new proofs from old.

We will use this construction at several key points in enlarging the domain of a proof

88 while maintaining certain properties, combining proofs {(Pi, ≤i)}i∈I to obtain a proof on the union of their domains, and so on.

Deﬁnition 7.3.1. Let T be a set of partial orders and let

(i) L(x)

(ii) L(x) = L(y) and x

We will sometimes say we are merging the partial orders in T , “forgetting” that

To use the merge to get new proofs from old, we will need to know the merge is in the proof closure of the orders being merged.

Lemma 7.3.2. Let T be a set of partial orders and let

Proof. Let D be the union of the domains of partial orders in T (that is, D is the domain of the merge) and let x in D. You must show there is a partial order <0 in

0 0 T such that every < predecessor of x is a

89 0 y < x then y is in L(x) so L(y) ≤ L(x). If L(y) < L(x) then y

It is worth noting the following proposition.

Lemma 7.3.3. Let T be a set of well founded partial orders and ≤T a well order of

T . Then the merge ≤m of ≤T is well founded.

Proof. Suppose not. Then there is an inﬁnite strictly decreasing ≤m sequence

x1 >m x2 >m x3 >m ...

of elements in the union of the domains of the posets in T . Call xi and xj equivalent if L(xi) = L(xj). There are either ﬁnitely many or inﬁnitely many eqivalence classes. We consider each case.

If there are only ﬁnitely many, then some class contains inﬁnitely many elements

y1 >m y2 >m ...

and L(yi) = L(yj) for all i, j. By deﬁnition of merge this means that

y1 >L(y1) y2 >L(y1) y3 >L(y1) ...,

contrary to the fact that L(y1) is a partial order in T and thus well founded by assumption.

90 So there must be inﬁnitely many equivalence classes. Picking from each class one member of the sequence of xi’s we get a sequence

y1 >m y2 >m y3 ...

such that for all i, j we have L(yi) 6= L(yj). Since L(yi) 6= L(yj) it follows from the deﬁnition of the merge that

L(y1) >T L(y2) >T L(y3) >T ...,

contrary to the assumption that ≤T is well founded.

The following two lemmas are two of the primary properties used when merging proofs to obtain a new proof.

Lemma 7.3.4. If (Pi, ≤i) for i in I is a set of partial orders then there is a partial S order in the proof closure of {Pi : i ∈ I} with domain i∈I Pi.

Proof. This is trivial from what has already been said. Take a well order ≤ of I. Now take the merge of the sequence of partial orders. We have proved the merge is in the proof closure of the Pi’s, and by deﬁnition of the merge the domain is the union of the Pi’s.

Before stating the next lemma, we need the following deﬁnition.

91 Deﬁnition 7.3.5. Let (A, ≤A) be a partial order. An end extension of (A, ≤A) is a partial order (B, ≤B) such that A is a ≤B downward closed subset of B and ≤B |A is the same partial order as ≤A.

Lemma 7.3.6. Let (A, ≤A) and (B, ≤B) be partial orders with A ⊆ B. Then there

0 is a partial order (B, ≤ ) of B in the proof closure of (A, ≤A) and (B, ≤B) that is an

0 end extension of (A, ≤A). If ≤A and ≤B are well founded then ≤ may be chosen as well founded as well.

Proof. Well order the set {≤A, ≤B} by choosing ≤A to be less than ≤B. Now take the merge of these two partial orders . We choose ≤0 to be the merge. We know the merge is a partial order on the union of the domains of A and B, which in this case is B itself since A ⊆ B. We also know the merge is in the proof closure of ≤A and

≤B.

We now show the merge is an end extension of ≤A. This is equivalent to showing two things. First, we have to show A is a downward closed subset of B in the merge ≤0.

0 Second, we must show ≤A and ≤ |A are the same partial order . To show A is ≤0 downward closed subset of B, let y ∈ A and x ≤0 y. We have to show x ∈ A. Well, x ≤0 y which means x precedes y in the merge, by deﬁnition of ≤0.

This implies L(x) ≤ L(y). We can not have L(x) < L(y) since y ∈ A and A is the

ﬁrst set being merged. So L(x) = L(y). So x ∈ A. It follows that A is ≤0 downward closed .

92 0 To show ≤A is the same partial order as ≤ |A is to show that for all x, y ∈ A, you

0 have x ≤ y iﬀ x ≤A y. Since both x, y are in A, we see that L(x) = L(y). Hence, by deﬁnition of merge (and the fact that we chose ≤0 as the merge) it follows that

0 x ≤ y iﬀ x ≤L(x) y iﬀ x ≤A y.

0 That ≤ is well founded if ≤A and ≤B are well founded is simply Lemma 7.3.3.

7.4 Preceding Set Proof Systems

In every abstraction, some information is kept, and much is discarded. For instance, though the complex numbers have an interesting algebraic structure, the topology of the plane simply has nothing to say about this. Topology speaks of open sets and anything that can be defined from them. So the first point to address is which information we want to capture in defining proof systems abstractly, and which is outside the scope of our subject.

We state now, just as topology takes “open set” as a basic notion and studies that which can be defined in terms of them, our preceding set proof systems essentially take “T ⇒ x” as a basic notion, and studies that which can be defined in terms of these implications. Our proof systems have nothing to say about syntax, semantics, truth or falsehood, models, and so on. While we may intuitively think of points of a proof system as “formulas”, in general, the points of our proof systems have no more structure than points of graphs or topological spaces. We now make this precise in the following definition.

93 Deﬁnition 7.4.1. A preceding set proof system is a set S, called the set of formulas, together with a set I of ordered pairs (T, x) with T ⊆ S and x ∈ S, called the rules of inference, such that the following property hold:

(i) Let T1,T2 be subsets of S and x ∈ S. If (T1, x) ∈ I and T1 ⊂ T2 then (T2, x) ∈ I.

Intuitively, every element in S is a formula. The ordered pair (T, x) says “from T infer x”. We infer a formula from a set of formulas. The condition states that we are dealing with monotonic proof systems, which in short means if you assume more you can prove more. Though there are exceptions, the vast majority of naturally occurring proof systems are monotonic, and we assume this monotonicity throughout.

When I is clear from the context, we will make the agreement that statements such as, “T implies x”, “T is enough information to infer x”, etc, means the pair (T, x) is in I. For convenience, we deﬁne T ⇒I x to mean the pair (T, x) is in I, and we may write T ⇒ x when I is clear from context.

7.4.1 Examples

First, in a typical proof system you might have modus ponens as a rule of inference.

In other words, given formulas φ and φ ⇒ ψ, from {φ, φ ⇒ ψ} to infer ψ. In the notation of our deﬁnition, {φ, φ ⇒ ψ} would be our set T and ψ would be our element x. Similarly for any other rule of inference found in a typical proof system arising in mathematical logic.

Second, given a partial order (P, ≤), we can let I be the set of pairs of the form

(T, x) such that T ⊆ P , x ∈ P , and T contains all y < x. It is essentially this

94 definition, when made for ausyses (to be defined below) instead of preceding set proof systems, that allows generalizing partial order notions to proof systems. Note this definition works equally well with infinite partial orders and even non-well founded partial orders, which is part of the motivation for allowing the very general setting we will allow, as opposed to making restrictions of finiteness or well foundedness in our general definitions.

Third, let S be the set of Borel sets of reals. We want to make S into a preceding set proof system that captures how Borel sets are constructed from the open sets.

The Borel sets are defined by starting with the open sets and allowing (transfinite iteration of) complementation and countable union. If we think of a rule (T, x) as saying, “once you have constructed everything in T , you may now construct x”, then our definition of preceding set proof system can capture the common mathematical theme of certain starting objects and certain construction rules. The starting objects, such as the open sets of the Borel hierarchy, are analogous to the axioms of proof systems of mathematical logic. To say we may “start” with an open set O, we simply put pairs of the form (∅,O) in I. To say once we have a set B, we may now construct its complement Bc, we put pairs of the form ({B},Bc) into I. Similarly, to allow S countable union as a construction rule, we put pairs of the form ({Bn}n≥0, n≥0 Bn) into I. Note that by convention, when we say a rule (T, x) goes into I, we implicitly mean that (T 0, x) goes into I for all T 0 containing T as well, to maintain monotonicity.

The previous example should be seen as far more general than the Borel sets. Such deﬁnitions and theorems, with starting objects and construction rules, are extremely common througout mathematics, and all yield natural examples of preceding set proof

95 systems. Also, the previous example should be noted as another natural example of an inﬁnitary proof system, further motivating the level of generality we maintain throughout.

Finally, we note that graphs can be made into preceding set proof systems in two natural ways. Informally, given a graph (V,E), one associated preceding set proof system with domain V ∪ {x} (for some element x not in V ) has all vertices in V as axioms, and allows infering the nonaxiom x from each doubleton in V corresponding to an edge. Another associated preceding set proof system chooses a vertex v in V as an axiom, and allows inferring elements adjacent to any vertex already proved.

While these definitions may at first seem unnatural, in fact, the former is related to graph matching and the latter to graph connectivity in such a way that graph matching and graph connectivity questions can be be simultaneously asked by asking the corresponding question regarding the proof systems. In particular, we give a problem whose solution would specialize to König’sDuality Theorem for the former type of proof systems and to (even the infinite version of) Menger’s Theorem for the latter.

While we do not need syntax or semantics, in order to develop an abstract theory of proofs we certainly need to deﬁne the notion of proof in a preceding set proof system.

For brevity, preceding set proof systems will simply be called proof systems for the rest of this section. First, we give the motivation for our deﬁnition.

96 7.4.2 Motivation for Deﬁnition of Proof

What is a proof of a theorem? It is a sequence of claims such that for any claim x, the set of claims which were given before x is considered enough information to infer x. So we want all of these to be considered proofs. Namely, given any finite sequence such that for every x in the sequence, the set of predecessors of x is enough information to infer x, this sequence should be considered a proof. Making, “is enough information to infer”, precise by saying “(T, x) is in I”, this would yield a rigorous definition of proof. We could say a proof is a finite sequence such that for all x in the sequence, we have {< x} ⇒I x. However, there is no reason to insist a proof is a finite sequence. A finite sequence is just a finite total order. The reason we think of proofs as being totally ordered is because that is the way they always appear on the pages of a book. For the general theory, it is actually more convenient and natural to allow partial orders in general to be considered proofs also, so long as the set of predecessors of any element is enough information to infer that element. For the general theory, we do not require these proofs to be finite or even well founded.

We are now ready to give the deﬁnition of proof in a proof system.

7.4.3 Proof Deﬁnition and Basics

Deﬁnition 7.4.2. Let (S,I) be a preceding set proof system. An I proof is a partial order (P, ≤) such that {< x} ⇒I x for all x ∈ P .

The astute reader may notice this does not in fact give the “right” deﬁnition of proof

97 for naturally arising proof systems in mathematical logic, as the rules of inference given there would, if non-well founded proofs are allowed, cause unprovable formulas to become provable. While this is easily remedied, we do not yet have the necessary definitions to state the remedy. For now, suffice it to say that the general theory is much smoother with the above definition, and roughly, for specific proof systems arising in mathematical logic, the solution is to define a first proof system as stated above, define a second proof system that is intuitively a well founded version of the

ﬁrst, and then apply the general theory to the well founded version. Though this well founded version of the proof system still allows non-well founded proofs, the non-well founded proofs in this proof system no longer yield proofs that “should not be” proofs. We make this precise in Lemma 7.8.5.

After deﬁning proof systems, the following question is natural. Given a set S, for which sets T of partial orders of subsets of S can rules of inference I be found such that T is exactly the set of I proofs? It is not hard to see that T has to be proof closed, and that this is enough is the content of the next proposition.

Proposition 7.4.3. Let T be a set of partial orders of subsets of the set S. Then there is a preceding set proof system (S,I) with T the set of I proofs iﬀ T is proof closed.

Proof. That the set of proofs of a preceding set proof system is proof closed is immediate from the deﬁnition of proof and proof closure.

98 For the converse, suppose T is proof closed. If {< x} ⊆ C ⊆ S then let C ⇒ x be a rule in I. Moreover, let rules of inference arising in this way be the only ordered pairs in I. It is enough to show (S,I) is a preceding set proof system that has T as its set of proofs.

That (S,I) is a preceding set proof system is just trivial checking of the deﬁnition.

What we have to show is (S,I) has T as its set of proofs. We have to show every partial order in T is a proof in (S,I) and every proof in (S,I) is a partial order in T .

First, let (P, ≤) be a partial order in T . We have to show it is a proof in (S,I).

Equivalently, we must show {< x} ⇒I x for all x ∈ P . This is trivial from the deﬁnition of I. We can choose any subset C of S containing {< x}, in particular

{< x} itself.

Now, let (P, ≤) be a proof in (S,I). We have to show this partial order is in T . T is proof closed by hypothesis, so it is enough to show ≤ is in the proof closure of T . By deﬁnition of proof closure, this means we have to show for all x ∈ P , there is some partial order (Q, ≤0) in T such that the set of < predecessors of x contains the set of

<0 predecessors of x. Again, this is clear from the deﬁnition of I.

7.4.4 Autonomous Sets

In these proof systems, we now wish to deﬁne the autonomous sets. The autonomous sets are, roughly speaking, sets that are self contained. Everything in an autonomous set can be proved without using anything outside that set. Their role in the theory of proof systems is analogous to the role of open sets in metric spaces. They do not

99 carry all of the information of preceding sets, but all the information we want for many cases, and the study of their structure is quite interesting.

The set of open sets of a metric space satisﬁes certain properties, which are then used as the deﬁning axioms of a topological space as a set together with a set of subsets, called open sets, such that certain properties hold.

We will show the set of autonomous sets satisﬁes certain properties and then deﬁne an autonomous system, or ausys, to be a set together with a set of subsets, called autonomous sets, such that certain properties hold.

There is one important way in which the relation of metric spaces to topological spaces diﬀers from that of proof systems to autonomous systems. Not every topological space is metrizable, but every ausys is induced by a preceding set proof system, as we will show.

Now, we give a deﬁnition that is basic to all that follows.

Deﬁnition 7.4.4. Let (S,I) be a preceding set proof system. Let P ⊆ S. P is autonomous if there is a proof (P, ≤) with domain exactly P .

By the above deﬁnition, the autonomous sets of a proof system are obviously uniquely determined by the set of proofs. The next lemma shows the set of proofs is in fact uniquely determined by the set of autonomous sets as well, so that the set of proofs and the set of autonomous sets contain the same information.

100 Proposition 7.4.5. Let (S,I) be a preceding set proof system. Let (P, ≤) be a partial order of a subset of S. Then ≤ is a proof iﬀ every ≤ downward closed set is autonomous.

Proof. If ≤ is a proof, then for every ≤ downward closed set D, ≤ |D is a proof with domain D, which shows D is autonomous.

Conversely, suppose every ≤ downward closed subset of P is autonomous. We must show ≤ is a proof. That is, we must show for all y in the domain P of ≤, that the rule {< y} ⇒ y is in I.

Well, {≤ y} is a ≤ downward closed subset of P and therefore autonomous. Since it is autonomous it has a proof ≤0. Since ≤0 is a proof it follows that the rule {<0 y} ⇒ y is in I. Now, {<0 y} ⊂ {< y}, so by upward closure of the rules of a preceding set proof system, the rule {< y} ⇒ y is in I too, and we are done.

So we now know the proofs and the autonomous sets carry the same information.

From either you can determine the other uniquely. And we know exactly when a set

T of proofs is the set of proofs of some preceding set proof system, namely, when T is proof closed. So it is natural to ask the corresponding question for the autonomous sets. The answer is given by the following lemma. This lemma is extremely important, for it gives us the axiomatization we will use in the deﬁnition of autonomous system.

The role of this lemma in studying autonomous systems is completely analogous to the role played in topology by the lemma that says in a metric space, the whole space

101 and null set are open, the open sets are closed under arbitrary union, and the open sets are closed under ﬁnite intersection.

Proposition 7.4.6. Let S be a set and T a set of subsets. Then T is the set of autonomous sets of a preceding set proof system on S iﬀ the following two conditions hold:

(i) T is closed under arbitrary union.

(ii) (The Order Property) If A ∈ T then there is a total order ≤0 with domain A

such that every ≤0 downward closed subset B of A is in T .

Proof. First, we prove if T is the set of autonomous sets of a preceding set proof system on S then (i) and (ii) are satisfied. Equivalently, we prove that if T is the set of I autonomous sets of some set I of rules of inference on S, then (i) and (ii) are satisfied. So take any such I. S First, we show (i). So let Aj ∈ T for j in an index set J and let A = j∈J Aj be their union. We must show A is in T . By choice of I we know each Aj is I autonomous which means each Aj has an I proof ≤j. Let ≤w be a well order of J and merge all of the ≤j’s in that order to get ≤m. By the definition of the merge we know ≤m S has domain A = Aj and we proved in Lemma 7.3.2 that the merge is in the proof closure of the orders being merged. Hence ≤m is in the proof closure of all the ≤j’s.

But each ≤j is an I proof therefore ≤m is in the proof closure of a set of I proofs.

By Proposition 7.4.3, the set of I proofs is proof closed, and hence must contain ≤m.

102 In other words, ≤m is an I proof with domain A, hence A is I autonomous, and (i) follows.

For (ii), suppose A is in T . We must show there is a total order ≤0 of A such that every

≤0 downward closed subset B of A is in T . Well, A is I autonomous and therefore has an I proof ≤. This proof is a partial order, and it is well known every partial order can be lengthened to a total order ≤0. The reader may check that the lengthening of an I proof is in its proof closure and is therefore an I proof. Thus ≤0 is an I proof. We proved in Proposition 7.4.5 that every downward closed subset of a proof is autonomous. Hence every ≤0 downward closed subset B of A is I autonomous, and therefore in T . Hence ≤0 is as needed, thus proving (ii).

Now for the converse. We assume T satisﬁes (i) and (ii), and we must show T is the set of autonomous sets of a preceding set proof system on S. Equivalently, we must show T is the set of I autonomous sets of a set I of rules of inference on S.

We need to deﬁne I from T . The idea is to put into I all the rules which have to be there if each A in T is autonomous, but to put in no other rules. Then we show the set of all rules which “have to” be there in fact has T as its set of autonomous sets, thus completing the proof.

To see what rules have to be there if each A in T is autonomous, take A in T . If A is I autonomous then A has an I proof ≤. Take f ∈ A. Since ≤ is an I proof we know {< f} ⇒I f. Now, {< f} is of course a subset of A − f, so by monotonicity of the proof system, C ⇒I f for any subset C of S containing A − f. We use all rules

C ⇒I f obtained in this way as our I.

103 More formally, I is the set of all rules of the form C ⇒ f such that there is some

A ∈ T with f ∈ A and C ⊆ S containing A − f.

We must show with this deﬁnition of I, every A in T is I autonomous and there are no other I autonomous sets.

First, let us show every A in T is I autonomous. By hypothesis, T satisﬁes (ii), which implies A can be ordered as ≤0 such that every ≤0 downward closed subset B of A is in T . It is enough to show ≤0 is an I proof. Take f ∈ A. We must show

0 0 0 {< f} ⇒I f. Well, the set B = {≤ f} is a ≤ downward closed subset of A and so is I autonomous by deﬁnition of ≤0. We see that {<0 f} = B − f and by deﬁnition

0 0 of I, we have B − f ⇒I f, hence {< f} ⇒I f as desired, therefore ≤ is an I proof, which means A is I autonomous.

Second, let us show every I autonomous set is in T . Since T is closed under arbitrary union by hypothesis, it is enough to show that for every f ∈ A there is B ⊆ A containing f such that B is in T . So take an I autonomous set A and f ∈ A and we are looking for such a B. Since A is I autonomous there is an I proof ≤ with domain

A. Since ≤ is an I proof it follows that there is an I rule of inference C ⇒I f such that C is contained in A − f.

By deﬁnition of I, since C ⇒I f in I, it follows that there is a subset B of A containing f that is in T . This completes the proof.

104 7.4.5 Axioms

Axioms are the elements of a proof system you are allowed to infer from nothing at all.

Deﬁnition 7.4.7. Let (S,I) be a preceding set proof system. An I axiom is an element x of S such that ∅ ⇒I x.

The following lemma is trivial, but we include it here to familiarize the reader with deﬁnitions so far presented.

Lemma 7.4.8. Let (S,I) be a preceding set proof system and x ∈ S. Then the following are equivalent.

(i) x is an axiom.

(ii) The single element partial order containing just x is a proof.

(iii) {x} is autonomous.

Proof. If x is an axiom then it is immediate from the deﬁnition of proof that this single element partial order is a proof, therefore (i) implies (ii). If the one point partial order on {x} is a proof then its domain {x} is autonomous by the deﬁnition of autonomous, so (ii) implies (iii). To see that (iii) implies (i), note if {x} is autonomous then it is the domain of a proof ≤, and {< x} is then ∅. Since ≤P is a proof, the rule {< x} ⇒ x is in I, hence the rule ∅ ⇒ x is in I, showing x is an axiom.

105 7.4.6 The Information in the Set of Proofs

The set of preceding sets carries strictly more information than the set of proofs, or equivalently, strictly more information than the set of autonomous sets. In other words, though the preceding sets uniquely determine the proofs, the proofs do not uniquely determine the preceding sets. We can ﬁnd two distinct preceding set proof systems on S with the same set of proofs.

For example, let S = {1, 2, 3}. We give two distinct preceding set proof systems on

S with the same proofs. For the ﬁrst proof system, take the rules of inference ∅ ⇒ 1,

1 ⇒ 2, and 2 ⇒ 3.

It is understood when a preceding set proof system is described in such a way, we tend not to bother to say, for instance, that C ⇒ x for all C ⊆ {1, 2, 3} such that C contains 1. We simply say let 1 ⇒ 2.

With that understanding, the reader may check that the proofs here are exactly the downward closed subsets of {1, 2, 3} with the usual ordering of the integers. For the second proof system, take the same rules of inference as the ﬁrst, and add the rule 3 ⇒ 1. The preceding sets in these two proof systems are diﬀerent, but simple checking shows that the set of proofs is the same in each case.

The reason that adding the rule yields no new proofs is that there is no chance to use the rule; inferring 1 from 3 gives nothing when we must prove 1 before ever proving 3.

This suggests the only information lost in passing from the rules of inference to the set of proofs is the superﬂuous rules. We make this precise in the following lemma for ﬁnite proof systems, whose simple proof is not included.

106 Lemma 7.4.9. Let (S,I) be a ﬁnite preceding set proof system. Then there is a minimum (not just minimal) set J under inclusion of rules of inference on S such that the set of I proofs and J proofs are equal. In fact, J is the set of all rules T ⇒ x such that {< x} ⊆ T for some I proof (P, ≤).

While a fact that is similar in spirit is true for proof systems in general, it is more complicated and not central to what we do, so we do not state it here.

Since, in passing to the set of proofs, the only information we lose is (roughly speaking) that of which superfluous rules had been included, it is plausible, and in fact turns out to be true, that little of significance is lost and both clarity and efficiency are gained by “forgetting” the superfluous rules and only “remembering” the proofs, or equivalently, the autonomous sets. The definition of autonomous system in the following section accomplishes precisely this.

7.5 Autonomous Systems

7.5.1 Introduction

The following deﬁnition is basic to all that follows.

Deﬁnition 7.5.1. An autonomous system, or ausys, is a set S together with a set T of subsets of S, called autonomous sets, satisfying the following two conditions:

(i) T is closed under arbitrary union.

107 (ii) (The Order Property) For every autonomous set A there is a total order ≤0 of

A such that every ≤0 downward closed set is autonomous.

Note that since the union of autonomous sets is autonomous, there is a largest autonomous subset S0 of S. Since all the structure is contained within S0, we assume unless otherwise noted that S = S0.

We call this maximum set S0 the autonomous part of S. It is convenient to use this terminology more generally.

Deﬁnition 7.5.2. Let S be a subset of an ausys. The autonomous part of S, denoted by aut(S), is the largest autonomous subset of S under inclusion.

7.5.2 Deﬁnition of Proof Revisited

Proposition 7.4.5 characterizes proofs in a preceding set proof system in terms of autonomous sets. For autonomous systems, in which “autonomous” is the basic notion, this lemma becomes the deﬁnition of proof.

Deﬁnition 7.5.3. Let (S,T ) be an ausys. A T proof is a partial order (P, ≤P ) of a subset P of S such that every downward closed set is autonomous.

108 7.6 The Canonical Orders

7.6.1 Canonical Order Deﬁnition and Basics

This sections formalizes the notion of “y needs x”, or, “in order to prove y, one must

ﬁrst prove x”, in an autonomous system. However, need is often context dependent.

The element y may need x if only elements from the autonomous set A are allowed, while not in general needing x. Taking this context dependence into account yields the following deﬁnition, which can be thought of as, “y needs x if all tools used are from A”, or “y needs x in A”.

Definition 7.6.1. Let A be an autonomous set. The canonical order ≤A is defined by letting x ≤A y iff x ≤ y for all proofs ≤ with domain A, where x, y ∈ A.

It is clear from the deﬁnition that ≤A is in fact a partial order on each autonomous set A.

For the following lemma, we need the notion of the infimum under lengthening. The reader may take it as an exercise to show that given a set S, the set of partial orders on S is itself a partial order under lengthening. We think of ≤1 as less than ≤2 if ≤2 lengthens ≤1. It is also a simple exercise to show that the infimum of every nonempty set of partial orders on S exists. In fact, given partial orders ≤i on S for i in an index set I, the infimum ≤ is the unique partial order on S such that for all x, y in S, x ≤ y iff x ≤i y for all i in I. We refer to this as the inf under lengthening. We note the supremum of a set of partial orders on the set S does not always exist as there may in fact be no upper

109 bound at all under lengthening. For instance, if x <1 y and y <2 x then ≤1 and ≤2 have no sup under lengthening. However, if an upper bound exists then we may take the inﬁmum of all such upper bounds to obtain a supremum under lengthening.

With these notions at hand, the following lemma is immediate from the deﬁnition.

Proposition 7.6.2. Let A be an autonomous set. Then the canonical order ≤A is the inf under lengthening of the set of proofs with domain A.

The canonical orders are very often used even in proofs of facts whose statement makes no mention of them. The following is the most important characterization of these orders. It is used frequently and without comment.

Proposition 7.6.3. Let A be autonomous. Then x ≤A y iﬀ every autonomous subset of A containing y also contains x.

Proof. We prove each direction, but it will be more convenient in each case to prove the contrapositive instead.

First then, we must prove if x 6≤A y then there is an autonomous subset of A containing y and not x. Well, by the deﬁnition of the canonical order, since x 6≤A y there is a proof ≤ with domain A such that x 6≤ y. Consider the set {≤ y}. This set is a ≤ downward closed subset of the proof ≤, and is therefore autonomous. Since x 6≤ y, it follows that x is not in this set. Hence we have an autonomous subset of A containing y and not x.

110 For the second half of the proof, we assume there is an autonomous subset of A containing y and not x and we must prove that x 6≤A y. So let C be such an autonomous subset of A containing y and not x. C is autonomous so there is a proof ≤1 with domain C. And A is autonomous by hypothesis so there is a proof ≤2 with domain A. We take the merge ≤m of ≤1 and ≤2 in this order. The merge has A ∪ C = A as its domain and is a proof since it is in the proof closure of the proofs

≤1 and ≤2. By the deﬁnition of the merge, it follows that x 6≤m y. By the deﬁnition of the canonical order and the fact that ≤m is a proof it follows that x 6≤A y.

7.6.2 Descendability

Suppose x is a nonaxiom in an ausys. Then intuitively, x must “need” something in order to prove it. Formally, we may ask if there is some autonomous set A and y in A such that y

In general, however, this is not the case.

Let S = {x1, x2,...} ∪ {x}. Let each xi be an axiom, and let each inﬁnite subset of

{xi}i≥1 imply x. x is a nonaxiom in this ausys, but suppose there is an autonomous set

A containing x and an element y such that y

The ausys in the previous paragraph is somewhat of an all purpose counterexample,

111 so we refer to it by name as the standard nondescendable ausys, or sna. By “descendable”, we vaguely mean various properties that state that in any given context, there should be a minimal (under inclusion) set from which we are able to prove x. The word “descendability” comes from the fact that one such notion, strong aut descendability, is deﬁned as the intersection of every chain of autonomous sets decreasing under inclusion itself being autonomous, which may be thought of saying if autonomous sets are “descended” then the limiting set is autonomous.

The vagueness of the previous paragraph is intentional, as there are many and varied descendability hypotheses. The descendability axioms are ﬁniteness conditions on ausyses much as Noetherian ring, well order, compact topological space, et cetera, are all ﬁniteness conditions on their respective structures.

While many ausys theorems are true in general, many involving canonical orders require descendability hypotheses. For instance, if x is a nonaxiom then x “should” need some y in some set A. Typically, any theorem that says certain structural information “should” follow from knowledge of the canonical orders is very likely to require an appropriate descendability hypothesis. Descendability hypotheses are also required for a number of theorems making no reference to canonical orders, such as the unique lexicographic decomposition theorem for ﬁnite ausyses, and forbidden minor/subdot theorems. The fact that descendability hypotheses are required for the truth of these theorems suggests the canonical orders are sitting in the background, and further suggests the importance of these orders for the study of ausyses in general.

With that said, we now prove some basic facts regarding descendability and canonical

112 orders. We note the deﬁnition of strong aut descendability is particularly important as the forbidden minor theorems we later prove are proved under this assumption.

Deﬁnition 7.6.4. An ausys is strong aut descendable if the intersection of every chain of autonomous sets under inclusion is also autonomous.

Deﬁnition 7.6.5. An ausys is weak aut descendable if for all autonomous sets A and for all x in A, there is a minimal autonomous subset B of A containing x.