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.

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