A Characterization of Hausdorff Spaces?

Question

Problem. Let $(X,\tau)$ be a Hausdorff space and let $x\in X$. Define, $$\mathcal{N}_x:=\{U\in\tau:x\in U\}$$Then show that, $$\displaystyle\bigcap_{U\in\mathcal{N}_x}U=\{x\}$$Is the converse true?

My Attempt

Let $(X,\tau)$ be an Hausdorff space. Now observe that since $x\in U$ for all $U\in \mathcal{N}_x$ we must have,$$\displaystyle\bigcap_{U\in\mathcal{N}_x}U\supseteq\{x\}\tag{1}$$Suppose if possible that, $$\displaystyle\bigcap_{U\in\mathcal{N}_x}U\supset\{x\}$$So there is an $y\in \displaystyle\bigcap_{U\in\mathcal{N}_x}U$ such that $y\ne x$. Now since $X$ is Hausdorff, there exists $U\in \mathcal{N}_x$ and $V\in\mathcal{N}_y$ such that $U\cap V=\emptyset$. This implies that, $$V\cap\left(\displaystyle\bigcap_{U\in\mathcal{N}_x}U\right)=\emptyset\tag{2}$$But this is a contradiction since we have assumed that $y\in V$ and $y\in \displaystyle\bigcap_{U\in\mathcal{N}_x}U$. This contradiction arose due to our assumption that $$\displaystyle\bigcap_{U\in\mathcal{N}_x}U\supset\{x\}$$Hence we are forced to conclude that, $$\displaystyle\bigcap_{U\in\mathcal{N}_x}U\supseteq\{x\}$$ and we are done.

Main Question

However, I don't know how to prove (or disprove) the converse. To be more precise I don't know how to prove (or disprove) the following,

Let $(X,\tau)$ be a topological space and let $x\in X$. Define, $$\mathcal{N}_x:=\{U\in\tau:x\in U\}$$for all $x\in X$. Suppose, $$\displaystyle\bigcap_{U\in\mathcal{N}_x}U=\{x\}$$for all $x\in X$. Prove that $(X,\tau)$ is Hausdorff.

Can anyone help?

Answer 1

As others have already pointed out, $\bigcap \mathcal{N}_x =\{x\}$ is true for all $x \in X$ iff $x$ is $T_1$.

But $\bigcap \{\overline{U}: U \in \mathcal{N}_x\} = \{x\} $ for all $x$ is true iff $X$ is Hausdorff: suppose the equality holds for all $x$, and let $x \neq y$ be two points of $X$. Then $y \notin \{x\} = \bigcap \{\overline{U}: U \in \mathcal{N}_x\}$ so there exists some $U \in \mathcal{N}_x$ such that $y \notin \overline{U}$. But then $U$ and $X\setminus \overline{U}$ are both open disjoint and contain $x$ resp. $y$, so $X$ is Hausdorff.

Conversely if $X$ is Hausdorff and $x \in X$ we clearly have $\{x\} \subseteq \bigcap \{\overline{U}: U \in \mathcal{N}_x\}$ and when $y \neq x$, we have disjoint open sets $U \in \mathcal{N}_x$ and $V \in \mathcal{N}_y$. As $V$ is open and misses $U$, $y \notin \overline{U}$ and so $y \notin \bigcap \{\overline{U}: U \in \mathcal{N}_x\}$ and we have equality.

When we use the term neighbourhood of $x$ for any set $N$such that there is an open set $O$ with $x \in O \subseteq N$, as is commonly done, we can formulate this property as:

$X$ is Hausdorff if every point (singleton, really) is the intersection of its closed neighbourhoods.

Answer 2

The converse is not true. The property is in fact a characterization of $T_1$.

$⋂\mathcal{N}_x = \{x\}$ is equivalent to $(∀y ≠ x)(∃U ∈ \mathcal{N}_x): y ∉ U$, which (holding for every $x$) is a definition of $T_1$. $T_1$ is also equivalent to every singleton being closed, or equivalently to every finite set being closed.

On the other hand, the collection of all co-finite sets (and ∅) forms a topology, so on every set there is the coarsest $T_1$ topology – a natural candidate for a $T_1$ topology that is not $T_2$. And indeed, on every infinite set, this topology is not $T_2$.

Answer 3

The converse if false. Consider $\mathbb{R}$ with the cofinite topology (that is, the closed sets are the finite sets).

Fix $x$ and consider $y$ distinct from $x$. Then the set $\mathbb{R} \setminus \{y\}$ is an open neighbourhood of $x$ distinct from $y$, so we have the property in the hypothesis.

However its clear this topology is not Hausdorff as for open sets $U$ and $V$, $U \cap V $ is open, and so has finite complement (therefore it cannot be empty).