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Calibrating Determinacy Strength in Levels of the Borel Hierarchy
CALIBRATING DETERMINACY STRENGTH IN LEVELS OF THE BOREL HIERARCHY SHERWOOD J. HACHTMAN Abstract. We analyze the set-theoretic strength of determinacy for levels of the Borel 0 hierarchy of the form Σ1+α+3, for α < !1. Well-known results of H. Friedman and D.A. Martin have shown this determinacy to require α+1 iterations of the Power Set Axiom, but we ask what additional ambient set theory is strictly necessary. To this end, we isolate a family of Π1-reflection principles, Π1-RAPα, whose consistency strength corresponds 0 CK exactly to that of Σ1+α+3-Determinacy, for α < !1 . This yields a characterization of the levels of L by or at which winning strategies in these games must be constructed. When α = 0, we have the following concise result: the least θ so that all winning strategies 0 in Σ4 games belong to Lθ+1 is the least so that Lθ j= \P(!) exists + all wellfounded trees are ranked". x1. Introduction. Given a set A ⊆ !! of sequences of natural numbers, consider a game, G(A), where two players, I and II, take turns picking elements of a sequence hx0; x1; x2;::: i of naturals. Player I wins the game if the sequence obtained belongs to A; otherwise, II wins. For a collection Γ of subsets of !!, Γ determinacy, which we abbreviate Γ-DET, is the statement that for every A 2 Γ, one of the players has a winning strategy in G(A). It is a much-studied phenomenon that Γ -DET has mathematical strength: the bigger the pointclass Γ, the stronger the theory required to prove Γ -DET. -
A Proof of Cantor's Theorem
Cantor’s Theorem Joe Roussos 1 Preliminary ideas Two sets have the same number of elements (are equinumerous, or have the same cardinality) iff there is a bijection between the two sets. Mappings: A mapping, or function, is a rule that associates elements of one set with elements of another set. We write this f : X ! Y , f is called the function/mapping, the set X is called the domain, and Y is called the codomain. We specify what the rule is by writing f(x) = y or f : x 7! y. e.g. X = f1; 2; 3g;Y = f2; 4; 6g, the map f(x) = 2x associates each element x 2 X with the element in Y that is double it. A bijection is a mapping that is injective and surjective.1 • Injective (one-to-one): A function is injective if it takes each element of the do- main onto at most one element of the codomain. It never maps more than one element in the domain onto the same element in the codomain. Formally, if f is a function between set X and set Y , then f is injective iff 8a; b 2 X; f(a) = f(b) ! a = b • Surjective (onto): A function is surjective if it maps something onto every element of the codomain. It can map more than one thing onto the same element in the codomain, but it needs to hit everything in the codomain. Formally, if f is a function between set X and set Y , then f is surjective iff 8y 2 Y; 9x 2 X; f(x) = y Figure 1: Injective map. -
The Matroid Theorem We First Review Our Definitions: a Subset System Is A
CMPSCI611: The Matroid Theorem Lecture 5 We first review our definitions: A subset system is a set E together with a set of subsets of E, called I, such that I is closed under inclusion. This means that if X ⊆ Y and Y ∈ I, then X ∈ I. The optimization problem for a subset system (E, I) has as input a positive weight for each element of E. Its output is a set X ∈ I such that X has at least as much total weight as any other set in I. A subset system is a matroid if it satisfies the exchange property: If i and i0 are sets in I and i has fewer elements than i0, then there exists an element e ∈ i0 \ i such that i ∪ {e} ∈ I. 1 The Generic Greedy Algorithm Given any finite subset system (E, I), we find a set in I as follows: • Set X to ∅. • Sort the elements of E by weight, heaviest first. • For each element of E in this order, add it to X iff the result is in I. • Return X. Today we prove: Theorem: For any subset system (E, I), the greedy al- gorithm solves the optimization problem for (E, I) if and only if (E, I) is a matroid. 2 Theorem: For any subset system (E, I), the greedy al- gorithm solves the optimization problem for (E, I) if and only if (E, I) is a matroid. Proof: We will show first that if (E, I) is a matroid, then the greedy algorithm is correct. Assume that (E, I) satisfies the exchange property. -
Primitive Recursive Functions Are Recursively Enumerable
AN ENUMERATION OF THE PRIMITIVE RECURSIVE FUNCTIONS WITHOUT REPETITION SHIH-CHAO LIU (Received April 15,1900) In a theorem and its corollary [1] Friedberg gave an enumeration of all the recursively enumerable sets without repetition and an enumeration of all the partial recursive functions without repetition. This note is to prove a similar theorem for the primitive recursive functions. The proof is only a classical one. We shall show that the theorem is intuitionistically unprovable in the sense of Kleene [2]. For similar reason the theorem by Friedberg is also intuitionistical- ly unprovable, which is not stated in his paper. THEOREM. There is a general recursive function ψ(n, a) such that the sequence ψ(0, a), ψ(l, α), is an enumeration of all the primitive recursive functions of one variable without repetition. PROOF. Let φ(n9 a) be an enumerating function of all the primitive recursive functions of one variable, (See [3].) We define a general recursive function v(a) as follows. v(0) = 0, v(n + 1) = μy, where μy is the least y such that for each j < n + 1, φ(y, a) =[= φ(v(j), a) for some a < n + 1. It is noted that the value v(n + 1) can be found by a constructive method, for obviously there exists some number y such that the primitive recursive function <p(y> a) takes a value greater than all the numbers φ(v(0), 0), φ(y(Ϋ), 0), , φ(v(n\ 0) f or a = 0 Put ψ(n, a) = φ{v{n), a). -
Power Sets Math 12, Veritas Prep
Power Sets Math 12, Veritas Prep. The power set P (X) of a set X is the set of all subsets of X. Defined formally, P (X) = fK j K ⊆ Xg (i.e., the set of all K such that K is a subset of X). Let me give an example. Suppose it's Thursday night, and I want to go to a movie. And suppose that I know that each of the upper-campus Latin teachers|Sullivan, Joyner, and Pagani|are free that night. So I consider asking one of them if they want to join me. Conceivably, then: • I could go to the movie by myself (i.e., I could bring along none of the Latin faculty) • I could go to the movie with Sullivan • I could go to the movie with Joyner • I could go to the movie with Pagani • I could go to the movie with Sullivan and Joyner • I could go to the movie with Sullivan and Pagani • I could go to the movie with Pagani and Joyner • or I could go to the movie with all three of them Put differently, my choices for whom to go to the movie with make up the following set: fg; fSullivang; fJoynerg; fPaganig; fSullivan, Joynerg; fSullivan, Paganig; fPagani, Joynerg; fSullivan, Joyner, Paganig Or if I use my notation for the empty set: ;; fSullivang; fJoynerg; fPaganig; fSullivan, Joynerg; fSullivan, Paganig; fPagani, Joynerg; fSullivan, Joyner, Paganig This set|the set of whom I might bring to the movie|is the power set of the set of Latin faculty. It is the set of all the possible subsets of the set of Latin faculty. -
Polya Enumeration Theorem
Polya Enumeration Theorem Sasha Patotski Cornell University [email protected] December 11, 2015 Sasha Patotski (Cornell University) Polya Enumeration Theorem December 11, 2015 1 / 10 Cosets A left coset of H in G is gH where g 2 G (H is on the right). A right coset of H in G is Hg where g 2 G (H is on the left). Theorem If two left cosets of H in G intersect, then they coincide, and similarly for right cosets. Thus, G is a disjoint union of left cosets of H and also a disjoint union of right cosets of H. Corollary(Lagrange's theorem) If G is a finite group and H is a subgroup of G, then the order of H divides the order of G. In particular, the order of every element of G divides the order of G. Sasha Patotski (Cornell University) Polya Enumeration Theorem December 11, 2015 2 / 10 Applications of Lagrange's Theorem Theorem n! For any integers n ≥ 0 and 0 ≤ m ≤ n, the number m!(n−m)! is an integer. Theorem (ab)! (ab)! For any positive integers a; b the ratios (a!)b and (a!)bb! are integers. Theorem For an integer m > 1 let '(m) be the number of invertible numbers modulo m. For m ≥ 3 the number '(m) is even. Sasha Patotski (Cornell University) Polya Enumeration Theorem December 11, 2015 3 / 10 Polya's Enumeration Theorem Theorem Suppose that a finite group G acts on a finite set X . Then the number of colorings of X in n colors inequivalent under the action of G is 1 X N(n) = nc(g) jGj g2G where c(g) is the number of cycles of g as a permutation of X . -
Cardinality of Sets
Cardinality of Sets MAT231 Transition to Higher Mathematics Fall 2014 MAT231 (Transition to Higher Math) Cardinality of Sets Fall 2014 1 / 15 Outline 1 Sets with Equal Cardinality 2 Countable and Uncountable Sets MAT231 (Transition to Higher Math) Cardinality of Sets Fall 2014 2 / 15 Sets with Equal Cardinality Definition Two sets A and B have the same cardinality, written jAj = jBj, if there exists a bijective function f : A ! B. If no such bijective function exists, then the sets have unequal cardinalities, that is, jAj 6= jBj. Another way to say this is that jAj = jBj if there is a one-to-one correspondence between the elements of A and the elements of B. For example, to show that the set A = f1; 2; 3; 4g and the set B = {♠; ~; }; |g have the same cardinality it is sufficient to construct a bijective function between them. 1 2 3 4 ♠ ~ } | MAT231 (Transition to Higher Math) Cardinality of Sets Fall 2014 3 / 15 Sets with Equal Cardinality Consider the following: This definition does not involve the number of elements in the sets. It works equally well for finite and infinite sets. Any bijection between the sets is sufficient. MAT231 (Transition to Higher Math) Cardinality of Sets Fall 2014 4 / 15 The set Z contains all the numbers in N as well as numbers not in N. So maybe Z is larger than N... On the other hand, both sets are infinite, so maybe Z is the same size as N... This is just the sort of ambiguity we want to avoid, so we appeal to the definition of \same cardinality." The answer to our question boils down to \Can we find a bijection between N and Z?" Does jNj = jZj? True or false: Z is larger than N. -
Enumeration of Finite Automata 1 Z(A) = 1
INFOI~MATION AND CONTROL 10, 499-508 (1967) Enumeration of Finite Automata 1 FRANK HARARY AND ED PALMER Department of Mathematics, University of Michigan, Ann Arbor, Michigan Harary ( 1960, 1964), in a survey of 27 unsolved problems in graphical enumeration, asked for the number of different finite automata. Re- cently, Harrison (1965) solved this problem, but without considering automata with initial and final states. With the aid of the Power Group Enumeration Theorem (Harary and Palmer, 1965, 1966) the entire problem can be handled routinely. The method involves a confrontation of several different operations on permutation groups. To set the stage, we enumerate ordered pairs of functions with respect to the product of two power groups. Finite automata are then concisely defined as certain ordered pah's of functions. We review the enumeration of automata in the natural setting of the power group, and then extend this result to enumerate automata with initial and terminal states. I. ENUMERATION THEOREM For completeness we require a number of definitions, which are now given. Let A be a permutation group of order m = ]A I and degree d acting on the set X = Ix1, x~, -.. , xa}. The cycle index of A, denoted Z(A), is defined as follows. Let jk(a) be the number of cycles of length k in the disjoint cycle decomposition of any permutation a in A. Let al, a2, ... , aa be variables. Then the cycle index, which is a poly- nomial in the variables a~, is given by d Z(A) = 1_ ~ H ~,~(°~ . (1) ~$ a EA k=l We sometimes write Z(A; al, as, .. -
The Axiom of Choice and Its Implications
THE AXIOM OF CHOICE AND ITS IMPLICATIONS KEVIN BARNUM Abstract. In this paper we will look at the Axiom of Choice and some of the various implications it has. These implications include a number of equivalent statements, and also some less accepted ideas. The proofs discussed will give us an idea of why the Axiom of Choice is so powerful, but also so controversial. Contents 1. Introduction 1 2. The Axiom of Choice and Its Equivalents 1 2.1. The Axiom of Choice and its Well-known Equivalents 1 2.2. Some Other Less Well-known Equivalents of the Axiom of Choice 3 3. Applications of the Axiom of Choice 5 3.1. Equivalence Between The Axiom of Choice and the Claim that Every Vector Space has a Basis 5 3.2. Some More Applications of the Axiom of Choice 6 4. Controversial Results 10 Acknowledgments 11 References 11 1. Introduction The Axiom of Choice states that for any family of nonempty disjoint sets, there exists a set that consists of exactly one element from each element of the family. It seems strange at first that such an innocuous sounding idea can be so powerful and controversial, but it certainly is both. To understand why, we will start by looking at some statements that are equivalent to the axiom of choice. Many of these equivalences are very useful, and we devote much time to one, namely, that every vector space has a basis. We go on from there to see a few more applications of the Axiom of Choice and its equivalents, and finish by looking at some of the reasons why the Axiom of Choice is so controversial. -
17 Axiom of Choice
Math 361 Axiom of Choice 17 Axiom of Choice De¯nition 17.1. Let be a nonempty set of nonempty sets. Then a choice function for is a function f sucFh that f(S) S for all S . F 2 2 F Example 17.2. Let = (N)r . Then we can de¯ne a choice function f by F P f;g f(S) = the least element of S: Example 17.3. Let = (Z)r . Then we can de¯ne a choice function f by F P f;g f(S) = ²n where n = min z z S and, if n = 0, ² = min z= z z = n; z S . fj j j 2 g 6 f j j j j j 2 g Example 17.4. Let = (Q)r . Then we can de¯ne a choice function f as follows. F P f;g Let g : Q N be an injection. Then ! f(S) = q where g(q) = min g(r) r S . f j 2 g Example 17.5. Let = (R)r . Then it is impossible to explicitly de¯ne a choice function for . F P f;g F Axiom 17.6 (Axiom of Choice (AC)). For every set of nonempty sets, there exists a function f such that f(S) S for all S . F 2 2 F We say that f is a choice function for . F Theorem 17.7 (AC). If A; B are non-empty sets, then the following are equivalent: (a) A B ¹ (b) There exists a surjection g : B A. ! Proof. (a) (b) Suppose that A B. -
Formal Construction of a Set Theory in Coq
Saarland University Faculty of Natural Sciences and Technology I Department of Computer Science Masters Thesis Formal Construction of a Set Theory in Coq submitted by Jonas Kaiser on November 23, 2012 Supervisor Prof. Dr. Gert Smolka Advisor Dr. Chad E. Brown Reviewers Prof. Dr. Gert Smolka Dr. Chad E. Brown Eidesstattliche Erklarung¨ Ich erklare¨ hiermit an Eides Statt, dass ich die vorliegende Arbeit selbststandig¨ verfasst und keine anderen als die angegebenen Quellen und Hilfsmittel verwendet habe. Statement in Lieu of an Oath I hereby confirm that I have written this thesis on my own and that I have not used any other media or materials than the ones referred to in this thesis. Einverstandniserkl¨ arung¨ Ich bin damit einverstanden, dass meine (bestandene) Arbeit in beiden Versionen in die Bibliothek der Informatik aufgenommen und damit vero¨ffentlicht wird. Declaration of Consent I agree to make both versions of my thesis (with a passing grade) accessible to the public by having them added to the library of the Computer Science Department. Saarbrucken,¨ (Datum/Date) (Unterschrift/Signature) iii Acknowledgements First of all I would like to express my sincerest gratitude towards my advisor, Chad Brown, who supported me throughout this work. His extensive knowledge and insights opened my eyes to the beauty of axiomatic set theory and foundational mathematics. We spent many hours discussing the minute details of the various constructions and he taught me the importance of mathematical rigour. Equally important was the support of my supervisor, Prof. Smolka, who first introduced me to the topic and was there whenever a question arose. -
Enumeration of Structure-Sensitive Graphical Subsets: Calculations (Graph Theory/Combinatorics) R
Proc. Nati Acad. Sci. USA Vol. 78, No. 3, pp. 1329-1332, March 1981 Mathematics Enumeration of structure-sensitive graphical subsets: Calculations (graph theory/combinatorics) R. E. MERRIFIELD AND H. E. SIMMONS Central Research and Development Department, E. I. du Pont de Nemours and Company, Experimental Station, Wilmington, Delaware 19898 Contributed by H. E. Simmons, October 6, 1980 ABSTRACT Numerical calculations are presented, for all Table 2. Qualitative properties of subset counts connected graphs on six and fewer vertices, of the -lumbers of hidepndent sets, connected sets;point and line covers, externally Quantity Branching Cyclization stable sets, kernels, and irredundant sets. With the exception of v, xIncreases Decreases the number of kernels, the numbers of such sets are al highly p Increases Increases structure-sensitive quantities. E Variable Increases K Insensitive Insensitive The necessary mathematical machinery has recently been de- Variable Variable veloped (1) for enumeration of the following types of special cr Decreases Increases subsets of the vertices or edges of a general graph: (i) indepen- p Increases Increases dent sets, (ii) connected sets, (iii) point and line covers, (iv;' ex- X Decreases Increases ternally stable sets, (v) kernels, and (vi) irredundant sets. In e Increases Increases order to explore the sensitivity of these quantities to graphical K Insensitive Insensitive structure we have employed this machinery to calculate some Decreases Increases explicit subset counts. (Since no more efficient algorithm is known for irredundant sets, they have been enumerated by the brute-force method of testing every subset for irredundance.) 1. e and E are always odd. The 10 quantities considered in ref.