Let (X, τ ) be a topological space. Show that ${\{x\}}= \bigcap_{G \in τ }G $

Question

Let (X, τ ) be a topological space. Suppose that for any x ∈ X one has that {x} is a closed set. Show that:

It is known that $\bar{\{x\}}=\{x\}$ by theorem

$${\{x\}}= \bigcap_{\{x\} \subset F}F $$ with F closed.

Then $$X-\{x\}=X- \bigcap_{\{x\} \subset F}F $$

by DeMorgan Law

$$X-\{x\}=\bigcup_{\{x\} \subset F}(X-F) $$

The problem is that I arrive at the union and not at the intersection.

Answer 1

Hint:

For every $y\neq x$ the set $X-\{y\}$ is an open set with $\{x\}\subseteq X-\{y\}$.

Answer 2

$(X, \tau) $ is a $T_1$ space iff $\{x\}$is closed for all $x\in X$ iff for any two distinct points, there exists open set containing one point but not containg other.

To show equality between two sets, we have to show one is subset of others.

One side set inclusion is trivial.

We only need to show $$\bigcap_{x\in G\in\tau } G \subseteq \{x\}$$

Choose, $y\in \bigcap_{x\in G\in\tau } G$

Claim :$y=x$ .

If not, then by the $T_1$ axiom there exists $G_o \in \tau $ [ in fact we can take $G_o=X \setminus \{y\} $ ]such that $x\in G_0$ but $y\notin G_o$.

Hence, $y\notin G_o \implies y\notin \bigcap_{x\in G\in\tau } G$ (contradiction) $

Hence, $y=x$