## Uncountability of real numbers — Topological proof

Perhaps the fastest proof of uncountability of $\mathbb{R}$ is using the diagonalization trick. However, there’s another proof that I learnt from Topology by Munkres (Theorem 27.7 in page 176) which is nice in its own way.

It is enough to show that the closed interval $[0, 1]$ is uncountable. Note that $[0, 1]$ has the following three properties:

(Definition: If $X$ is a topological space, then $x$ is called an isolated point of $X$ if the singleton $\{x\}$ is an open set in $X$). And now the general theorem follows:

Theorem. If $X$ is a non-empty compact Hausdorff topological space having no isolated points, then $X$ is uncountable.

Proof. (Following Munkres). The proof proceeds in two steps. The first step will use Hausdorff condition and the non-existence of isolated points. The second step will use compactness (via the equivalent formulation stating that if any family $\mathcal{C}$ of closed sets in $X$ satisfies the finite intersection property, then $\bigcap_{C\in\mathcal{C}} C\neq\emptyset$).

Step 1: We show that if $x\in X$ and $U$ is any non-empty open set in $X$ then there exists a subset $V\subset U$ such that $x\notin\overline{V}$. First we choose $y\neq x$ such that $y\in U$ (this is possible: if $x\notin U$, then let $y$ be any element of $U$; if $x\in U$ then $\{x\}\subsetneq U$ because $\{x\}$ is not open but $U$ is open, hence we can pick $y\in U$ different from $x$). Next using Hausdorff condition, we can find disjoint neighbourhoods $W_{1}$ and $W_{2}$ of $x$ and $y$ respectively. It follows that $V=W_{2}\cap U$ satisfies $x\notin\overline{V}$ (Why? If $x\in\overline{V}$, then every neighbourhood of $x$ would intersect $V$. But $W_{1}$ is a neighbourhood of $x$, and $W_{1}\cap W_{2}=\emptyset$, and so $W_{1}\cap V=\emptyset$, contradiction).

Step 2. We will show that any function $f:\mathbb{N}\to X$ is not surjective. This will establish uncountability of $X$. Let $x_{n}=f(n)$. Applying Step 1 to $x=x_{1}$ and $U=X$, we get a subset $V_{1}\subset U$ such that $x_{1}\notin\overline{V}_{1}$. We inductively define $V_{n}$ as follows. If $V_{k}$ is already determined, we apply Step 1 to $x=x_{k+1}$ and $U=V_{k}$ to get a subset $V_{k+1}\subset V_{k}$ such that $x_{k+1}\notin V_{k+1}$. So we have a nested sequence of non-empty closed sets: $\overline{V}_{1}\supset\overline{V}_{2}\supset\cdots$
Since $X$ is compact, the intersection $\bigcap_{n\in\mathbb{N}} \overline{V}_{n}$ is not empty. If $x\in\bigcap_{n\in\mathbb{N}} \overline{V}_{n}$, then $x\neq x_{n}$ for each $x_{n}$ (otherwise, $x_{n}=x\in \overline{V_{n}}$ which is a contradiction). Thus, $f$ is not surjective.