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