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I have got the following question which I could not solve:

can a metric space have exactly 36 open sets?

I believe if the metric space is finie, then it has to be discrete and so the number of open sets will be some power of 2. Am I right? Please help!

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Belief is not proof. But your belief in this case is correct. Now go prove it. – Daniel Fischer Jul 23 '14 at 14:22

Hint: show that in any finite metric space, all singletons (sets with a single element) are open. From there, it is easy to show that every subset of a finite metric space is open.

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In every metric space all finite sets are closed. Thus if the metric space itself is finite definitely all sets are open. – Anupam Jul 23 '14 at 14:31
@Anupam that's not what I had in mind, but it certainly works! At any rate, you have your answer then. $36$ is not a power of $2$, so no. – Omnomnomnom Jul 23 '14 at 14:32

Let $(X,d)$ be a finite metric space, pick $x\in X$. We want to show that $\{x\}$ is open which is the same as saying there is a radius $r>0$ such that $\{y\,:\, d(x,y)<r\}=\{x\}$.

How do we show such a radius exists? Well for each $y\in X$ such that $y\neq x$ we know that $d(x,y)>0$. As there are only finitely many such $y$s there is a least $r$ such that $d(x,y)=r>0$. So let $r' = r/2$ then $d(x,y)>r'$ for each $y\neq x$ and so $B(x,r')=\{x\}$ and so the space is discrete, as you thought.

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Let the assumed metric space space be (X,d).If X was infinite then then there would be infinite open sets,thus X is finite.Assume it has exactly 36 open sets,then if A and B are two distinct open subsets of X then their union and intersection must also be open sets,which gives rise to a power set structure to this collection of 36 sets.But power sets cardinalities are powers of 2 so it gives a contradiction.So no.

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