Space such that Intersection of Open Sets containing Point is Singleton may not be Hausdorff

Theorem

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

Let $x \in S$ be arbitrary.

Let $T$ be such that the intersection of all open sets containing $x$ is $\set x$:

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

Then it is not necessarily the case that $T$ is a Hausdorff space.

Let $T$ be the finite complement topology on the real numbers $\R$, for example.

The open sets of $T$ are subsets of $\R$ of the form $U$ such that $\R \setminus U$ is finite, together with $\O$.

Let $x \in \R$ be arbitrary.

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

Let $y \in \R$ such that $y \ne x$.

Note that the set $\set y$ is finite.

Thus $\R \setminus \set y$ is an open set of $T$ which contains $x$ but not $y$.

Hence $y \notin K$.

As $y$ is arbitrary, it follows that the only element of $\R$ which is in $K$ is $x$ itself.

That is:

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

$\blacksquare$

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