Characterization of Cantor Spaces

Matthew Shaw (ref. Pugh Analysis Textbook) November 2019

1 Introduction

The standard Cantor Middle Thirds Set is compact, perfect, nonempty, and totally disconnected (and since the construction takes place in R with its standard metric, the Cantor Set is also metrizable). Totally Disconnected: Every connected component is a singleton set, and equiv- alently, every point has arbitrarily small clopen neighborhoods. Perfect: A space is called perfect if it has no isolated points. There are two main theorems of this presentation: The Cantor Surjection The- orem, that every nonempty compact metric space is the image of a continuous function with the Cantor set as its domain, and a complete characterization of Cantor spaces, in particular that every compact, nonempty, perfect, and totally disconnected metric space is homeomorphic to the standard Cantor middle thirds set. Notation: The set of words of countably infinite length in a two character alphabet is denoted by Ω, the standard Cantor Middle-Thirds set is denoted by C, and the set of words of length n (where n ∈ N) in a two character alphabet is denoted ω(n). Note that the size of ω(n) is always 2n. ω with no associated number will represent a word of countably infinite length (i.e. an element of Ω) and the notation ω|n for n ∈ N means the word ω truncated to only its first n characters (this is an element of ω(n)). t denotes disjoint union of sets. If α and β are words of finite length, then the notation αβ means the word made up of the characters of α followed by the characters of β, which is also termed concatenation.

2 Construction and Size

We can use an address system to locate and name points in the Cantor set. At each stage, use 0 to denote the left interval and 2 to denote the right interval. 0 1 for example, at the first stage of construction, we use C = [0, 3 ] to denote the 2 2 left interval and C = [ 3 , 1] to denote the right interval. Inside each interval we use the addressing the same way, so at stage 2 we inherit the first character of the name from whichever interval we use in stage 1, for example the furthest left 00 1 interval at stage 2 is denoted C = [0, 9 ], the first 0 in the name coming from

1 it being a subset of the left interval in stage 1 and the second 0 coming from it being the left interval of that set in the next stage of construction. Then at ω1ω2...,ωn some stage n of the construction, Cn is the disjoint union of the sets C , with each ωi ∈ {0, 2}. We can extend from ﬁnite cases to inﬁnite address strings ω1, ω2,... , which corresponds to the intersection of nested intervals

Cω1 ⊇ Cω1ω2 ⊇ Cω1ω2ω3 ⊇ ...

This is an intersection of nested compact sets, so it is nonempty. Since the diameter of the sets tends to zero, the intersection contains exactly one point p(ω), where ω = ω1, ω2,... denotes the infinite word in the alphabet {0, 2}. In particular, \ {p(ω)} = Cω|n n∈N Note that if two points p and p0 have different addresses ω and ω0, then they are different points in the Cantor set because at some digit their addresses disagree, and since the subintervals at that stage are disjoint, they are contained in different disjoint sets and so are different points. Noting also that each point in the Cantor set has an address, we have a bijective correspondence between Ω, the set of words of countably infinite length in the alphabet {0, 2}, and the Cantor set, so the sets have the same cardinality. Clearly |Ω| = |P (N)|, so the Cantor set is uncountable in size (the construction of this bijection is the same as the construction of the Cantor set via ternary expansion without 1).

3 Topological Properties

The Cantor set is perfect, compact, nonempty, and totally disconnected. Nonempty is obvious. Compact follows from Heine-Borel as it is a closed and bounded subset of R. To see why C is perfect and totally disconnected, first note that C contains every endpoint of any interval in any Cn stage of its construction, denote this set of endpoints by E. Now let x ∈ C, let ε > 0, and 1 n 1 let n ∈ N such that 3n < ε. Note that x is in one of the 2 intervals of size 3n of Cn, call this interval I. Note that E ∩ I ⊆ (x − ε, x + ε) contains infinitely many elements, since that interval is contained in Cn and so will have middle thirds removed infinitely many more times, so contributing infinitely many more endpoints inside the interval I. Thus no x is isolated and C is perfect. Note that I is closed in Cn and since Cn \ I is made up of finitely many closed intervals, it is also closed in Cn, so both of these are clopen in Cn. Then I ∩ C and (Cn \ I) ∩ C are both clopen in C with C ∩ I ⊆ (x − ε, x + ε), so every open neighborhood about any x ∈ C contains a clopen neighborhood of x, and so C is totally disconnected.

2 4 Cantor Surjection Theorem

The statement of the theorem is as follows: Given any nonempty compact metric space M, there exists a continuous surjection from C onto M. The proof is broken up into multiple lemmas. Definition: Given a compact metric space M, we call a compact nonempty subset of M a piece of M. Lemma 1: If M is a compact, nonempty metric space and ε > 0, then M can be covered by finitely many pieces, each with diameter no more than ε. ε Proof: Make a covering of M by open balls with radius 2 . By compactness, take a finite subcover. Then the closure of each ball in the subcover constitutes a covering of M by finitely many pieces of diameter no more than ε. This covering is made up of pieces rather than just subsets as closed subsets of compact spaces are compact.

Using that lemma, divide the space M into finitely many pieces of diameter 1 and denote this collection of pieces by Σ1. Since this is finite in size, choose n1 n1 ∈ N large enough that 2 ≥ |Σ1|. Let w1 : ω(n1) → Σ1 be any onto function (since |ω(n1)| ≥ |Σ1|, such a function must exist). Then we say w1 labels Σ1 and for L ∈ Σ1 if w1(α) = L, we call α the label of L. Then since each piece L is compact, we can repeat the process, breaking each into finitely many pieces 1 of diameter no more than 2 and for any L ∈ Σ1, we call the collection of such pieces of L Σ2(L). Then the family [ Σ2 = Σ2(L)

L∈Σ1 is a covering of M by ﬁnitely many smaller pieces. Choose n2 ∈ N large enough n2 that 2 ≥ max(|Σ2(L)| : L ∈ Σ1). Then label Σ2 with words αβ ∈ ω(n1 + n2) such that if L = w1(α) then αβ labels all the pieces S ∈ Σ2(L) as β varies with α ﬁxed. This is an onto labeling w2 : ω(n1 + n2) → Σ2 which is coherent with w1 in the sense that matching leading digits in a labeling means the points are close, i.e. contained in the same stage 1 piece. Then proceeding in this fashion inductively, we construct a sequence of divisions

(Σk)k∈N of M and surjections wk : ωk(n1 + n2 + ··· + nk) → Σk for k ∈ N such that the following hold:

1 The maximum diameter of the pieces L ∈ Σk tends to zero as k tends to inﬁnity (speciﬁcally we can do the construction by making pieces of 1 diameter no more than n at stage n of the construction).

2 Σk+1 reﬁnes Σk in the sense that ∀S ∈ Σk+1, ∃L ∈ Σk such that S ⊆ L (i.e. subpieces aren’t broken up across multiple pieces).

3 If L ∈ Σk and Σk+1(L) is the set {S ∈ Σk+1 : S ⊆ L} then [ L ⊆ S

S∈Σk+1(L)

3 (i.e. each piece is entirely covered by its subpieces).

4 The labelings wk are coherent, i.e. if wk(α) = L ∈ Σk, then the association β → wk+1(αβ) labels every S ∈ Σk+1(L) as β varies in ω(nk+1) (with α fixed). Cantor Surjection Theorem: From here the surjection we wish to construct is the natural one given that we know how to label pieces of any nonempty compact space with words of countably infinite length in two letters and we already have a bijective correspondence between such words and the standard Cantor set. Formally, for any p = p(ω) ∈ C, use the labeling provided by \ p(ω) = Cω|n n∈N (technically this isn’t correct as the right side is a set and the left side is a point, but the set on the right is exactly the singleton set containing p, so this notation shouldn’t be too confusing). Then construct the labeling of M with words in a two character alphabet as outlined previously. Then for p ∈ C, consider the sequence of pieces of Lk(p) ∈ Σk of M, where Lk(p) = wk(ω|(n1 +n2 +···+nk)), which is the collection of pieces at stage k of the labeling of M, labeled by the first (n1 + n2 + ··· + nk) characters in the name associated with p. Then the sequence (Lk(p)k∈N) is a nested sequence of pieces of M with diameters tending to zero. Then \ σ(p) = Lk(p) k∈N defines a single point of M (technically the singleton set containing only that point of M) as a nested intersection of compact sets is nonempty and if the diameter tends to zero, the intersection contains only one point. This σ : C → M is the surjection we seek, and it remains to be shown that σ is continuous. To see that σ is continuous, let ε > 0 and let x ∈ C. Note that if two points x, y ∈ C are close, then that corresponds to the first many characters in their associated words matching. Then make δ > 0 small enough that the associated words in the interval match up to the first n1 + n2 + ··· + nk characters with 1 k ∈ N large enough that ε > k so that the points in M are within the same piece of diameter less than ε. To explicitly see why σ is surjective, note that since M is covered by pieces labeled by words in two letters at every stage k, each p ∈ M is the intersection of at least one nested sequence of pieces labeled in this way, corresponding to an element of Ω, and so since we have a bijection between Ω and C, this surjection from Ω to M is equivalent to having a surjection from C to M.

5 Peano Curves and Moore-Kline

Peano Curves exist: That is, there exists a continuous surjection τ : [0, 1] → B2, where B2 denotes

4 the unit disk in R2. Proof: First construct the continuous surjection σ : C → B2. Define a gap interval (a, b) to be any interval with endpoints in C which has empty intersection with C. Then define  σ(x) if x ∈ C  τ(x) = (1 − t)σ(a) + tσ(b) if x = (1 − t)a + tb ∈ (a, b)  and (a,b) is a gap interval This function amounts to taking the continuous surjection from C and inter- polating to points in the interval outside of C by line segments, and is so also continuous with B2 as its image. Definition: We say a metric space M is a Cantor space if it is compact, nonempty, perfect, and totally disconnected. Moore-Kline Theorem: Every Cantor space M is homeomorphic to the standard Cantor Middle-Thirds set C. To prove this, we first need a definition and a lemma. Definition: A Cantor piece is a nonempty clopen subset S of a Cantor space M. Observations: Every Cantor piece is a Cantor space in its own right and any Cantor space can be divided into two disjoint Cantor pieces (total disconnect- edness tells us such nonempty clopen neighborhood exist about each point). Cantor Partition Lemma: Given a Cantor space M and ε > 0, ∃N ∈ N such that ∀d ≥ N, there is a partition of M into exactly d Cantor pieces of diameter no more than ε. Proof: We already know we can make a covering by finitely many pieces, but we also require the pieces be disjoint. Begin by covering M with finitely many clopen subsets U1,U2,...,Um, each with diameter no more than ε. To make m the sets disjoint, define the new family {Vi}i=1 as follows: V1 = U1,V2 = U2 \ U1,...,Vm = Um \ (U1 ∪ U2 ∪ · · · ∪ Um−1), and if any Vi is empty, dis- N card it from the collection to make a new collection {Xi}i=1 of N ≤ m disjoint pieces. If d = N, we’re done at this point. If d > N, inductively divide XN into 2, then 3, and eventually d − N + 1 disjoint Cantor pieces so that

XN = Y1 t Y2 t · · · t Yd−N+1 and so M = X1 t X2 t · · · t XN−1 t Y1 t · · · t Yd−N+1 is a partition of M.

Proof of the Moore-Kline Theorem: This proof follows largely the same ideas as the Cantor Surjection Theorem. Given a Cantor space M, use the previous lemma with ε = 1 to construct a partition Σ1 of M into exactly a dyadic number of nonempty pieces, that is, into n1 d1 = 2 pieces for some n1 ∈ N. Then we have a bijection w1 : ω(n1) → Σ1 as the sets have the same cardinality. Then partition each L ∈ Σ1 into N(L)

5 1 Cantor pieces of diameter no more than 2 and choose a dyadic

n2 d2 = 2 ≥ max({N(L): L ∈ Σ1}) and use the lemma to partition each L ∈ Σ1 into exactly d2 nonempty pieces. Then again deﬁning [ Σ2 = (L)

L∈Σ1 we make a partition of M into exactly d1d2 pieces, coherently labeled by ω(n1 + n2). Specifically, we have a bijection for each L ∈ Σ1 wL : ω(n2) → Σ2(L), and defining w2 : ω(n1 + n2) → Σ2 by w2(αβ) = S ∈ Σ2 if and only if w1(α) = L and wL(β) = S. The w2 defined in this way is a bijection. Proceeding in this way by induction, we refine our partitions of M to partitions made up of pieces with strictly decreasing diameters (which tend to zero) and we extend our bijective labelings. The homeomorphism we construct is exactly the surjection constructed in the Cantor Surjection Theorem, for p ∈ C, \ σ(p) = Lk(p) k∈N where Lk(p) ∈ Σk has (unique) label ω(p)|(n1 + n2 + ··· + nk). Since distinct 0 points p, p ∈ C have distinct labels and the Σk divisions are partitions of M 0 0 (they are disjoint), ω(p) 6= ω(p ) =⇒ ∃k ∈ N where Lk(p) 6= Lk(p ), so σ(p) 6= σ(p0), which means σ is an injective function. We already know from the Cantor Surjection Theorem’s construction that σ is a continuous surjection, so σ : C → M is a continuous bijection, so it’s a homeomorphism. Therefore any Cantor space M is homeomorphic to the standard Cantor Middle-Thirds set C.

6 Bonus Material

C is nowhere dense in R as it can contain no open interval because its width is less than the width of any interval in R. However, one can construct a fat Cantor set by removing middle n! portions at stage n rather than simply middle thirds. This leaves a set with positive width, but the fat Cantor set is still a Cantor space. There are a few obvious corollaries to the Moore-Kline Theorem included here: Corollary 1: Every Cantor space is homeomorphic to every other Cantor space. Proof: Homeomorphism is an equivalence relation on topological spaces, so since they are all homeomorphic to the standard Cantor Middle-Thirds set, by transitivity, they are all homeomorphic to each other. Corollary 2: The fat Cantor set is homeomorphic to C despite C having width 0 and the fat Cantor set having positive width. Corollary (mine): Any compact metric space has cardinality no larger than the cardinality of R.

6 Proof: There is an onto function from C to any such space, so no such space can have larger cardinality than C, which has the same cardinality as R.