# Topology on a finite set with closed singletons is discrete

Could you help me to prove this proposition below?

$X$ is a finite set and $(X,\tau)$ is a topological space. Show that if every subset which have single element is closed set then $\tau$ describes discrete topology.

I started with let $X=\{x_1,x_2,\cdots,x_n\}$ . If every subset with single element is closed set then for $1\leq i \leq n$ then $\{x_i\}$ is closed, at the same time $X \setminus \{x_i\}$ is open set.

for$\quad$ $1\leq i \leq n$ $\quad$ $X \setminus \{x_i\}=\{x_1,x_2,\cdots,x_{i-1},x_{i+1},\cdots,x_n\}$ and we know that $(X,\tau)$ is a topological space so it provides the axioms. It is the point where i stuck. How can i build discrete topology with using the set $\quad$ $X \setminus \{x_i\}=\{x_1,x_2,\cdots,x_{i-1},x_{i+1},\cdots,x_n\}$ by using finite intersection and (finite or infinite) union set operations. thank you in advance :)

In fact it is easier to work with properties of closed sets. If $\{ x_i\}$ and $\{ x_j\}$ are both closed, then so is $\{ x_i\} \cup \{x_j\} = \{x_i, x_j\}$.
Call $\{1,..,n\}=I$. As you wrote, $A_i = X \setminus \{x_i\}=\{x_1,x_2,\cdots,x_{i-1},x_{i+1},\cdots,x_n\}$ is open. If you take the finite intersection of $\bigcap_{i \in I \setminus \{j\} }A_i$, you get an $\{ x_j \}$.