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Let $C=\{0,1\}^\mathbb{N}$ be the Cantor space. Is $C$ homeomorphic to $\{0,1\}^m$ when $m>\aleph_0$?

Ben
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  • Why would it be? What are your thoughts on the problem? – Daniel Hast Feb 22 '21 at 18:38
  • @DanielHast. Well, I am tempted to say no. But on the other hand, I gather that there is only one Cantor space understood as a nonempty, compact, perfect, 0-dimensional metric space. I am not sure that ${0,1}^m$ for $m>\aleph_0$ is a metric space. – Ben Feb 22 '21 at 18:46
  • It's not metric, net even first countable, when $m > \aleph_0$. For different cardinals $m$ we get non-homeomorphic spaces ${0,1}^m$; a whole class worth of Cantor cubes. – Henno Brandsma Feb 22 '21 at 19:04

2 Answers2

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For any infinite cardinal $\kappa$ we have that $w(\{0,1\}^\kappa) = \kappa$, where the weight $w(X)$ of a space is the minimal size of a base for $X$.

So in particular $\{0,1\}^\kappa \not\simeq \{0,1\}^\lambda$ when $\kappa \neq \lambda$ are infinite cardinals. So we have a lot ( a class of) of topologically distinct Cantor cubes. The only metrisable one is $\{0,1\}^{\aleph_0}$ as a compact metrisable space has countable weight.

I write more on these cubes and dense subsets of them here.

Henno Brandsma
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If $\kappa>\aleph_0$, then $X=\{0,1\}^\kappa$ is not second countable. Let $\mathscr{U}=\{U_n:n\in\omega\}$ be any countable family of open sets in $X$. For each $n\in\omega$ there is a finite $F_n\subseteq\kappa$ and a function $s_n:F_n\to\{0,1\}$ such that $B_n=\{x\in X:x\upharpoonright F_n=s_n\}\subseteq U_n\}$.

Let $C=\bigcup_{n\in\omega}F_n$; $C$ is countable, so there is an $\alpha\in\kappa\setminus C$. Let $$V=\{x\in X:x_\alpha=0\}\,.$$ For each $n\in\omega$ there is a $y^{(n)}\in B_n\subseteq U_n$ such that $y_\alpha^{(n)}=1$, so $y^{(n)}\in U_n\setminus V$. Thus, $V$ is an open set that contains no member of $\mathscr{U}$, and $\mathscr{U}$ is therefore not a base for $X$.

Brian M. Scott
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