There are opens $U,V$ such that $a\in U-V, b\in V-U$ $\iff $ every finite subset is closed Definition: the $T_1$ axiom says that given $2$ distinc points $a,b\in X$, there will exist opens $U$ and $V$ such that $a\in U-V$ and $b\in V-U$.
I need to show that a topological space has the $T_1$ property $\iff$ every finite subset is closed.
I'm trying to do the following: I need to prove that every finite subset is closed, that is, the complementar is open. But I don't even know if the complementar is finite or infinite. So... Maybe I need to suppose that there exists a finite subset which is open. I don't know also how this would help. I guess that, since it's a finite set of points, I can take balls as in the $T_1$ axiom and maybe take their intersection and arrive at something.
Could somebody help me?
 A: You can show that result as follows:
Let $(X, \tau)$ be a topological space. The following conditions are equivalent


*

*$X$ is a $T_{1}$ space. 

*For each $x \in X$ the set $\left\{x\right\}$ is closed in $X$.

*All finite $A\subset X$ is closed in $X$.

*$\tau_{cf} \subset \tau$ $\qquad$ (where $\tau_{cf}$ is the topology of finite complements)


Demonstration
1) implies 2) Let $x \in X$, consider the set $X-\left\{x\right\}$.
Let $w \in X-\left\{x\right\}$ $\Rightarrow$ $w\neq x$, by hypothesis exist $U, V \in \tau$ such that $x \in U-V$, $w \in V-U$.
Clearly $V\subset X-\left\{x\right\}$ since supposing otherwise would lead to $x \in V$, which is not true.
In short we have given $w \in X-\left\{x\right\}$ there exists $V \in \tau$ such that $w \in V\subset X-\left\{x\right\}$, then $w \in (X-\left\{x\right\})^{\circ}$ therefore $X-\left\{x\right\} \subset (X-\left\{x\right\})^{\circ}$ so $X-\left\{x\right\} \in \tau$, so $\left\{x\right\}$ is closed in $X$.
2) implies 3) Let $A\subset X$ be a finite set.
Suppose $A:= \left\{a_{1},\ldots , a_{n}\right\}$ with $a_{i} \in X$, for each $i=1,\ldots , n$.
By hypothesis $\left\{a_{i}\right\}$ is closed in $X$, then $A=\bigcup_{i=1}^{n}\left\{a_{i}\right\}$ which is closed in $X$ being finite union of closed sets in $X$.
3) implies 4) Let $B \in \tau_{cf}$ $\Rightarrow$ $X-B$ is a finite set, by hypothesis $X-B$ is closed in $X$, so $B \in \tau$, therefore $\tau_{cf} \subset \tau$
4) implies 1) Let $x, y \in X$ such that $x\neq y$, then the sets $X-\left\{x\right\}, X-\left\{y\right\} \in \tau_{cf}$, by hypothesis $X-\left\{x\right\}, X-\left\{y\right\} \in \tau$
Consider $U:=X-\left\{y\right\}$ and $V:=X-\left\{x\right\}$, clearly $x \in U-V$, and  $y\in V-U$ thus we have that $X$ is a $T_{1}$ space.
A: To see any set of the form $\{a\}$ is closed, where $a \in X$: let $b \in X\setminus \{a\}$. The assumption gets us open sets $U$, $V$ such that $a \in U \setminus V$ and $b \in V \setminus U$. So $b \in V \subseteq X \setminus \{a\}$ (the last inclusion is just a restatemenr of $a \notin V$).
So every $b \in X\setminus\{a\}$ is an interior point of it, so $X\setminus \{a\}$ is open and so $\{a\}$ is closed. Now any finite set is just a finite union of such sets so also closed....
A: First let $X$ be $T_1$ we show that every finite subset is closed. As Marco suggested it suffices to prove that every singleton set is closed. Now assume this is not the case, say there exists $\{a\}\subseteq X$ which is not closed, this means $X-\{a\}$ is not open, thus by definition there exists $b\in X-\{a\}$ such that $X-\{a\}$ does not contain any open neighbourhood of $b$. This means every open neighbourhood of $b$ intersects $X-(X-\{a\})=\{a\}$, which contradicts the assumption $X$ is $T_1$ (because from the definition of $T_1$ we can find at least one open neighbourhood of $b$ which does not contain $a$). 
The converse is straight forward: If every finite subset is closed, this means every singleton set is closed. Thus for distinct $a,b\in X$, choose $U=X-\{b\}$ and $V=X-\{a\}$, we are done.
