Equivalent properties of T1 spaces

Question

Let (X,T) be a topological space:

How does every finite subset of X is closed imply that for every subset $A \subseteq X, A = \bigcap \{ U \subseteq X : U$ is open and $A \subseteq U\}$

I know that both these imply that (X, T ) is T1, but I'm not sure how to show the above equivalence.

Answer 1

Suppose that each finite subset of $X$ is closed, and let $A\subseteq X$. For each $x\in X\setminus A$ the singleton $\{x\}$ is closed, so $U_x=X\setminus\{x\}$ is open. Clearly $A\subseteq U_x$ for each $x\in X\setminus A$, and in fact it’s not hard to see that

$$A=\bigcap_{x\in X\setminus A}U_x\;.$$

Let $\mathscr{U}=\{U\subseteq X:A\subseteq U\text{ and }U\text{ is open}\}$, and let $\mathscr{U}_0=\{U_x:x\in X\setminus A\}$. Clearly $\mathscr{U}_0\subseteq\mathscr{U}$, so

$$\bigcap\mathscr{U}\subseteq\bigcap\mathscr{U}_0=A\;.$$

On the other hand, $A\subseteq U$ for each $U\in\mathscr{U}$, so

$$A\subseteq\bigcap\mathscr{U}\;,$$

and hence

$$\bigcap\mathscr{U}=A\;.\tag{1}$$

Now assume that $(1)$ holds for all $A\subseteq X$, and let $x\in X$. For each $y\in X\setminus\{x\}$ let $\mathscr{U}_y$ be the family of open nbhds of $y$. By hypothesis $\bigcap\mathscr{U}_y=\{y\}$, so there is a $U_y\in\mathscr{U}_y$ such that $x\notin U_y$. It’s straightforward to verify that

$$X\setminus\{x\}=\bigcup_{y\in X\setminus\{x\}}U_y\;,$$

which is an open set, and hence that $\{x\}$ is closed. Finally, the union of finitely many closed sets is closed, so every finite subset of $X$ must be closed.

Answer 2

I've never heard of the second definition and it's actually pretty cool. Given $y\notin A$, you can construct an open set $U_y$ containing $A$ that does not contain $y$ (namely, the complement of $\{y\}$.) The intersection of all $U_y$ for $y\notin A$ must be $A$.

Answer 3

Suppose every finite subset of $X$ is closed. What that means is that every cofinite set is open - in particular, for each $x\in X$ the set $B_x=X\setminus\{x\}$ is open.\

So fix a set $A$, and let's look at the intersection $I$ of all opens containing $A$. Clearly this contains $A$; and for $x\not\in A$, $B_x$ is an open containing $A$ but not $x$ - do you see why this implies that $x\not\in I$?