topological condition for a set to be uncountable

Theorem.

A nonempty compact Hausdorff space with no isolated points is 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 $\alpha$.

Figure 1: The induction step: separating points in $U_{\alpha }$.

Now define a function $f:2^{\omega }\to X$ as follows. If $\alpha$ iseventually zero, put $f(\alpha )=x_{\alpha }$. Otherwise, consider the sequence$(x_{({\alpha }_0)},x_{({\alpha }_0,{\alpha }_1)},x_{({\alpha }_0,{\alpha }_1,{\alpha }_2)},\mathrm{\dots })$ 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 $({\alpha }_0,\mathrm{\dots },{\alpha }_n)$ of $\alpha$, the point $f(\alpha )$ is in $U_{({\alpha }_0,\mathrm{\dots },{\alpha }_n)}$.

Figure 2: Defining $f(\alpha )$ as a limit point of the $x_{({\alpha }_0,\mathrm{\dots },{\alpha }_n)}$.

Suppose $\alpha$ and $\beta$ are distinct sequences in $2^{\omega }$. Let $n$ be the first position where ${\alpha }_n\ne {\beta }_n$. Then $f(\alpha )\in U_{({\alpha }_0,\mathrm{\dots },{\alpha }_n)}$ and $f(\beta )\in U_{({\beta }_0,\mathrm{\dots },{\beta }_n)}$, and by construction $U_{({\alpha }_0,\mathrm{\dots },{\alpha }_n)}$ and $U_{({\beta }_0,\mathrm{\dots },{\beta }_n)}$ are disjoint. Hence $f(\alpha )\ne f(\beta )$, 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 $\mathrm{[}\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{]}$ is uncountable.

Proof.

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