uncountable Polish spaces contain Cantor spaceUncountablePolishSpacesContainCantorSpace2009-02-09 22:18:362009-02-09 22:25:31TheoremCantor spacePolish spaceCantor space% almost certainly you want these
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Cantor space is an example of a compact and uncountable Polish space. In fact, every uncountable Polish space contains Cantor space, as stated by the following theorem.
\begin{theorem*}
Let $X$ be an uncountable Polish space. Then, it contains a subset $S$ which is homeomorphic to Cantor space.
\end{theorem*}
For example, the set $\mathbb{R}$ of real numbers contains the \PMlinkname{Cantor middle thirds set}{CantorSet}. Note that, being homeomorphic to Cantor space, $S$ must be a compact and hence closed subset of $X$.
The result is trivial in the case of Baire space $\mathcal{N}$, in which case we may take $S$ to be the set of all $s\in\mathcal{N}$ satisfying $s_n\in\{1,2\}$ for all $n$.
Then, for any uncountable Polish space $X$ there exists a continuous and one-to-one function $f\colon\mathcal{N}\to X$ (see \PMlinkname{here}{InjectiveImagesOfBaireSpace}). Then $f$ gives a continuous bijection from $S$ to $f(S)$. The \PMlinkname{inverse function theorem}{InverseFunctionTheoremTopologicalSpaces} implies that $f$ is a homeomorphism between $S$ and $f(S)$ and, therefore, $f(S)$ is homeomorphic to Cantor space.