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Following a question I asked yesterday Is this the correct definiton of $T_1$ space?

I was left with a claim:

$(X, \tau)$ is $T_1$ iff $\forall A \subseteq X, A = \bigcap\{U \subseteq X: U \in \tau, A \subseteq U\}$

I can borrow two additional results: singletons are closed, finite union of singletons is closed (by def), but I don't see that they are useful at all

I am not sure how to prove this but let's try:

$(\Rightarrow)$

Suppose $(X, \tau)$ is a $T_1$ space, then for all $x,y \in X$, $x \neq y$, there exists open sets $U, V \in \tau$ such that $x \in U, y \not\in U$ and $y \in V, x \not\in V$

Let $A \subseteq X$, we wish to show that $A$ is the intersection of all open sets containing it.

Pick $x \in A$, then there exists $ U \in \tau$, $ x \in U$ such that $y \not\in U$ for all $x \neq y$. Then $x \in A$ and $x \in U$...what does this imply? Stuck! Help!

$(\Leftarrow)$ Let $x \in \bigcap\{U \subseteq X: U \in \tau, A \subseteq U\}$, I want to show that $x \in A$

Suppose $A$ is open, then trivially $x\in\bigcap\{U \subseteq X: U \in \tau, A \subseteq U\} \subset A$.

Otherwise suppose $A$ is closed. By contradiction assume that $x \not\in A$, then $x \in X \backslash A$ is open, but $A \cap X \backslash A = \varnothing$, then $x \not \in \bigcap\{U \subseteq X: U \in \tau, A \subseteq U\}$ a contradition.

This shows that all sets are intersection of open sets....Wrong!

(Not using $T_1$ properties at all..)

Can someone use his or her magic touch to revive this proof from the dead?

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  • $\begingroup$ There are a few issues here... $\endgroup$ – Berci Jun 14 '16 at 22:57
  • $\begingroup$ A set that is not open doesn't need to be closed. $\endgroup$ – Matt Samuel Jun 15 '16 at 0:24
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Suppose $X$ is T$_{1}$. Let $A\subseteq X$. Let $x\in X\setminus A$. For each $a\in A$ there is an open set $U_a$ with $a\in U_a$ and $x\notin U_a$. Then $U_x:=\bigcup_{a\in A} U_a$ is open containing $A$, missing $x$. Then $A=\bigcap _{x\in X\setminus A} U_x$, so that $A$ is the intersection of all open sets containing it.

Suppose $X$ is not T$_1$. There exist $x,y\in X$ such that every open set containing $x$ also contains $y$. Then $\{x\}$ is a subset of $X$ which is not the intersection of all open sets containing it.

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