# Does a perfect compact metric space have a closed subset homeomorphic to a countable product of 2 point sets?

I recently found a proof here that a perfect complete metric space has a closed subset homeomorphic to $\{0,1\}^\mathbb{N}$. In a metric space, I know that compactness is equivalent to being complete and totally bounded.

If the conditions on $X$ are changed to being a perfect, compact metric space instead, does the result still hold? Is there proof that a perfect compact metric space contains a closed subset homeomorphic to $\{0,1\}^\mathbb{N}$ with the usual product topology? Thanks.

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Completeness and compactness are not equivalent in metric spaces: a metric space is compact iff it’s complete and totally bounded. – Brian M. Scott Sep 29 '11 at 1:54
@Brian, shoot, you're right. I've changed my question now. – Steven Lin Sep 29 '11 at 2:12
What makes you think that strengthening completeness to compactness makes this any easier than the "messy" proof you linked to in the previous version of your question (Prop 3.2.8 in this document)? If anything it should make it more difficult, at first glance, since you deprive yourself of some wiggle room... I'm not aware of a significantly different proof from the one that is given by Katok. Think about what you want to prove and what you will have proved as soon as you have the result. – t.b. Sep 29 '11 at 2:15
@t.b. I'm sorry, I completely changed the question, and I don't think it will make it any easier. My misunderstanding was pointed out by Brian Scott, so I wanted to accept the proof by Katok, and then I want to see if the same holds if we only know the metric space is perfect and compact, not perfect and complete. – Steven Lin Sep 29 '11 at 2:46
I'm so confused now. Compact metric spaces are automatically complete. It's the other direction in which the implication fails (as indicated by Brian M. Scott above). – user83827 Sep 29 '11 at 3:06

Added: I wrote the answer below for the original question. Strengthening the hypothesis by making $X$ compact doesn’t make matters any easier; you just get an extra reason for knowing that the sets $F_\sigma$ (see below) are non-empty.

The idea underlying that argument is actually pretty straightforward; it’s just the details that are a little messy. The construction basically just mimics the construction of the middle-thirds Cantor set. $X_0$ and $X_1$ are analogous to the intervals $[0,1/3]$ and $[2/3,1]$. $X_{0,0}, X_{0,1}, X_{1,0}$, and $X_{1,1}$ are analogous to the intervals $[1,1/9],[2/9,1/3],[2/3,7/9]$, and $[8/9,1]$. And so on. The details just ensure that the intersection of each of the resulting infinite nests of closed sets is a singleton and that the relative topology on $C$ is the right one.

I don’t offhand know of any essentially different argument, though it can be presented a little more cleanly. For instance, one could begin by proving the following technical

Lemma: Let $X$ be a perfect metric space. If $B$ is an open ball of diameter $d$, there are points $x,y\in B$ and a positive $r$ such that \begin{align*}&\operatorname{cl}B(x,r)\cap \operatorname{cl}B(y,r)=\varnothing,\\ &\operatorname{cl}B(x,r)\cup \operatorname{cl}B(y,r)\subseteq B,\text{ and}\\ &\operatorname{diam cl}B(x,r),\operatorname{diam cl}B(y,r)\le \frac{d}2. \end{align*}

This is where most of the really fiddly detail is. Once you have this, the proof of the main result is relatively straightforward, though the notation can be simplified only so far. Let $x\in X$ be arbitrary, and apply the lemma with $B = B(x,1)$ to get $x_0$, $x_1$, and $r_0=r_1>0$ such that

\begin{align*}&\operatorname{cl}B(x_0,r_0)\cap \operatorname{cl}B(x_1,r_1)=\varnothing,\\ &\operatorname{cl}B(x_0,r_0)\cup \operatorname{cl}B(x_1,r_1)\subseteq B(x,1),\text{ and}\\ &\operatorname{diam cl}B(x_0,r_0),\operatorname{diam cl}B(x_1,r_1) \le 1. \end{align*} Given any $\sigma = \langle i_1,\dots,i_n\rangle \in \{0,1\}^n$, apply the lemma with $B = B(x_\sigma,r_\sigma)$ to get $x_{\sigma^\frown 0}$, $x_{\sigma^\frown 1}$, and $r_{\sigma^\frown 0}=r_{\sigma^\frown 1}>0$ such that \begin{align*}&\operatorname{cl}B(x_{\sigma^\frown 0},r_{\sigma^\frown 0})\cap \operatorname{cl}B(x_{\sigma^\frown 1},r_{\sigma^\frown 1})=\varnothing,\tag{1}\\ &\operatorname{cl}B(x_{\sigma^\frown 0},r_{\sigma^\frown 0})\cup \operatorname{cl}B(x_{\sigma^\frown 1},r_{\sigma^\frown 1})\subseteq B(x_\sigma,r_\sigma),\text{ and}\tag{2}\\ &\operatorname{diam cl}B(x_{\sigma^\frown 0},r_{\sigma^\frown 0}),\operatorname{diam cl}B(x_{\sigma^\frown 1},r_{\sigma^\frown 1}) \le \frac12 \operatorname{diam cl}B(x_\sigma,r_\sigma).\tag{3} \end{align*}

(Here $\sigma^\frown i$ is the sequence in $\{0,1\}^{n+1}$ obtained by concatenating $\sigma$ with $i$.)

Now let $\sigma$ be an infinite sequence of $0$’s and $1$’s. For each positive integer $n$ let $\sigma_n$ be the finite sequence consisting of the first $n$ terms of $\sigma$. Let $\mathscr{F}_\sigma = \{\operatorname{cl}B(x_{\sigma_n}, r_{\sigma_n}):n\in\mathbb{Z}^+\}$, and let $F_\sigma = \bigcap\mathscr{F}_\sigma$. Then

• $(2)$ ensures that $\mathscr{F}_\sigma$ is a nest of closed sets and that $F_\sigma\ne\varnothing$;
• $(1)$ ensures that if $\tau$ is a different infinite sequence of $0$’s and $1$’s, $F_\tau\cap F_\sigma = \varnothing$; and
• $(3)$ ensures that $F_\sigma$ contains at most one point, and since we already know that $F_\sigma\ne\varnothing$, it must contain exactly one point, which I’ll call $x_\sigma$.

The map $x_\sigma\mapsto\sigma \in \{0,1\}^{\mathbb{Z}^+}$ is then easily shown to be a homeomorphism, just as in the PDF.

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Thanks a ton for this answer. – Steven Lin Sep 29 '11 at 5:46
+1. I'm curious why there was only one upvote, cool answer. – Ilya Oct 18 '11 at 9:31