# topological condition for a set to be uncountable

###### Proof.

Let $X$ be a nonempty compact Hausdorff space with no isolated points. To each finite $0,1$-sequence $\alpha$ associate a point $x_{\alpha}$ and an open neighbourhood $U_{\alpha}$ as follows. First, since $X$ is nonempty, let $x_{0}$ be a point of $X$. Second, since $x_{0}$ is not isolated, let $x_{1}$ be another point of $X$. The fact that $X$ is Hausdorff implies that $x_{0}$ and $x_{1}$ can be separated by open sets. So let $U_{0}$ and $U_{1}$ be disjoint open neighborhoods of $x_{0}$ and $x_{1}$ respectively.

Now suppose for induction that $x_{\alpha}$ and a neighbourhood $U_{\alpha}$ of $x_{\alpha}$ have been constructed for all $\alpha$ of length less than $n$. A $0,1$-sequence of length $n$ has the form $(\alpha,0)$ or $(\alpha,1)$ for some $\alpha$ of length $n-1$. Define $x_{(\alpha,0)}=x_{\alpha}$. Since $x_{(\alpha,0)}$ is not isolated, there is a point in $U_{\alpha}$ besides $x_{(\alpha,0)}$; call that point $x_{(\alpha,1)}$. Now apply the Hausdorff property to find disjoint open neighbourhoods $U_{(\alpha,0)}$ and $U_{(\alpha,1)}$ of $x_{(\alpha,0)}$ and $x_{(\alpha,1)}$ respectively. The neighbourhoods $U_{(\alpha,0)}$ and $U_{(\alpha,1)}$ can be chosen to be proper subsets of $U_{\alpha}$. Proceed by induction to find an $x_{\alpha}\in U_{\alpha}$ for each finite $0,1$-sequence $\alpha$.

Now define a function $f\colon 2^{\omega}\to X$ as follows. If $\alpha$ is eventually zero, put $f(\alpha)=x_{\alpha}$. Otherwise, consider the sequence $(x_{(\alpha_{0})},x_{(\alpha_{0},\alpha_{1})},x_{(\alpha_{0},\alpha_{1},\alpha% _{2})},\dots)$ of points in $X$. Since $X$ is compact and Hausdorff, it is closed and limit point compact, so the sequence has a limit point in $X$. Let $f(\alpha)$ be such a limit point. Observe that for each finite prefix $(\alpha_{0},\dots,\alpha_{n})$ of $\alpha$, the point $f(\alpha)$ is in $U_{(\alpha_{0},\dots,\alpha_{n})}$.

Suppose $\alpha$ and $\beta$ are distinct sequences in $2^{\omega}$. Let $n$ be the first position where $\alpha_{n}\neq\beta_{n}$. Then $f(\alpha)\in U_{(\alpha_{0},\dots,\alpha_{n})}$ and $f(\beta)\in U_{(\beta_{0},\dots,\beta_{n})}$, and by construction $U_{(\alpha_{0},\dots,\alpha_{n})}$ and $U_{(\beta_{0},\dots,\beta_{n})}$ are disjoint. Hence $f(\alpha)\neq f(\beta)$, implying that $f$ is an injective function. Since the set $2^{\omega}$ is uncountable and $f$ is an injective function from $2^{\omega}$ into $X$, $X$ is also uncountable. ∎

###### Corollary.

The set $[0,1]$ is uncountable.

###### Proof.

Being closed and bounded, $[0,1]$ is compact by the Heine-Borel Theorem; because $[0,1]$ is a subspace of the Hausdorff space $\mathbb{R}$, it too is Hausdorff; finally, since $[0,1]$ has no isolated points, the preceding theorem implies that it is uncountable. ∎

Title topological condition for a set to be uncountable TopologicalConditionForASetToBeUncountable 2013-03-22 16:15:15 2013-03-22 16:15:15 mps (409) mps (409) 15 mps (409) Theorem msc 54D10 msc 54A25 msc 54D30