A finite topological space is metrizable iff it is discrete

Question

What I want to show: Let X be a finite topological space. X is metrizable if and only if the topology is discrete.

proposition that I used in the proof: Let X be a finite metric space. Let $X=$ $\{$ $x_1,....,x_n$ $\}$. Let $x \in X$. Show that for $r=min\{$ $d(x_i,x_j)$ $:$ $i\neq j$ $\}$, we have $B(x,r)$ $\subset$ $\{$ $x$ $\}$

My proof:

Let $a\in B(x,r)$. That means $d(a,x) < r = $ $min \{$ $d(x_i,x_j)$ $:$ $i\neq j$ $\}$ implies that $d(a,x)=0$. Because if $d(a,x)>0$ then $a\neq x$ and so $r\leq d(a,x)$ which contradicts our assumption that $a \in B(x,r)$. Hence $a=x \in $ $\{$ $x$ $\}$.

So proof of: Let X be a finite topological space. X is metrizable if and only if the topology is discrete:

Suppose $X$ is metrizable. That means there exists a metric d on X so that $\tau$ $=$ $\{$ All Open subsets of $(X,d)$ $\}$. Let $A \in P(X)$. As $X$ is finite, so is A. Hence, A is the union of singletons, all of which are open . As the arbitrary union of open sets in a metric space are open, it follows that A is likewise open in $(X,d)$. Hence $A\in \tau$; this shows that $\tau$ is the discrete topology.

Suppose $\tau$ is the discrete topology. This means that $\tau$ $=$ $P(X)$. Let the metrix on X be the discrete metric. Therefore for any subset A of X; A is finite. This means that it is open in X. Is the proof correct? I would love feedback.

Answer 1

In any metric space $(X,d)$ a singleton set $\{x\}$ is closed (if $y \neq x$, then $B(y, d(x,y))$ is an open ball around $y$ that misses $\{x\}$) and so all finite sets are closed as finite unions of singletons. And if $X$ is finite this means that all subsets of $X$ are closed and hence open too. So $X$ is discrete.

That a discrete $X$ is always metrisable is classical (the discrete metric will witness that).

Answer 2

For the proof "metrizable $\Rightarrow$ discrete", it wouldn't harm to add why singletons are open. (Let $x\in X$ and set $R=\min_{y\in X, y\ne x}d(x,y)$. Then the open ball $B(x,R)$ contains only $x$.) (UPDATE: You seem to have provided the proof of that.)

For the opposite, I don't see you having proven that it is metrizable. However, this is not hard: define $d(x,y)=\begin{cases}1&x\ne y\\0&x=y\\ \end{cases}$, then prove that it induces the discrete topology.