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.