Intersection of Open Sets of Hausdorff Space containing Point is Singleton

Theorem

Let $T = \struct {S, \tau}$ be a Hausdorff space.

Let $x \in S$ be arbitrary.

Then the intersection of all open sets containing $x$ is $\set x$:

$\ds \bigcap_{\substack {H \mathop \in \tau \\ x \mathop \in H} } = \set x$

Let $x \in S$.

Let $K = \ds \bigcap_{\substack {H \mathop \in \tau \\ x \mathop \in H} }$, that is, the intersection of all open sets containing $x$.

Aiming for a contradiction, suppose there exists $y \in S$ such that $y \in K$ but $y \ne x$.

By definition of Hausdorff space, there exist disjoint open sets $U, V \in \tau$ containing $x$ and $y$ respectively.

We have that $y \in V$.

But we also have that $y \in K$.

$K \subseteq U$

and so:

$y \in U$

So $y \in U$ and $y \in V$ and so $U \cap V \ne \O$.

This directly contradicts our assertion that $U, V \in \tau$ are disjoint.

In turn, this contradicts our supposition that $T$ is a Hausdorff space.

Hence by Proof by Contradiction there can be no such $y$ such that $y \in K$.

The result follows.

$\blacksquare$

1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $4$: The Hausdorff condition: Exercise $4.3: 4 \ \text {(a)}$