# A finite topological space is metrizable iff it is discrete

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.

• In the first part, how do you justify that the singletons are open ? This is the key. In the second part, you don't even talk about a metric, so I don't see how you are proving that the space is metrizable. Sep 23, 2019 at 19:11
• @CaptainLama please look at my recent edit.
– user643073
Sep 23, 2019 at 19:15
• @hardmath the proposition I used was used in the proof of what I wanted to show.
– user643073
Sep 23, 2019 at 19:20
• @hardmath Yes, I agree. For instance, the set $\{$ a,b $\}$ with the trivial topology is not metriziable, as the topology is not discrete.
– user643073
Sep 23, 2019 at 19:25
• I just changed it to remove any confusion. Sep 23, 2019 at 22:43

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).

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.

• Please look at my recent edit.
– user643073
Sep 23, 2019 at 19:15
• Why $R>0$? If $X$ is finite, then there is no problem, but what if $X$ is infinite? Jul 9 at 23:30
• @ParcoMacelli The entire question starts with "Let $X$ be a finite topological space..." Otherwise, discrete implies metrizable (same proof!) but not the other way round ($\mathbb R$ with usual topology is metrizable but not discrete).
– user700480
Jul 9 at 23:38
• Oh, you are right; I misread the title. I was thinking of a proof for "discrete $\Rightarrow$ metrisable" without referring to the discrete metric. If $X$ is an infinite discrete space, is it also true that $R>0$? Jul 10 at 10:45
• @ParcoMacelli That question sadly does not make much sense to me. You are proving discrete $\implies$ metrizable. How can you even use the distance $d$ or calculate the infimum (in that case) if you don't even know if there is metric on it?
– user700480
Jul 10 at 12:13