# Show that for a finite metric space A, every subset is open

Let A be a finite metric space .I want to prove that every subset of A is open. I let the set B, be any subset of A. Since A is finite,then I know that A/B is also finite.I'm stuck here how can this help me reach to a proof? I beg your help

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## 4 Answers

Every finite metric space is equivalent to a discrete space.

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That's not a proof, is it? You just reworded the question and asserted it as a theorem. – nik Apr 17 '12 at 15:26
Well, I can't quite agree; mathematics is all about transformation of problem statements. In this case I transformed the problem into the direct application of a well-known theorem. – akkkk Apr 17 '12 at 15:42

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But your theorem is exactly what is being asked, the definition of a discrete space is "every subset is open"... If you asked "Show that theorem X is true", and someone answered "It's obvious by theorem X", how would you feel? – nik Apr 17 '12 at 15:43
Viewing problems in a more general light can sometimes help. In this case, an abstract question (about open and closed sets) is asked, and I clarified it by the more intuitive understanding of discrete spaces. And apparently the question owner was helped. – akkkk Apr 17 '12 at 16:31

Hint: If $(A,d)$ is a finite metric space and $x \in A$ and we let $$\delta=\min_{y \in A \setminus \{x\}}d(x,y)$$ then what is in $B(x,\delta)$?

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Massive hint: In a metric space, finite point sets are closed. So suppose that you have a subset $B$ of $A$. Then $A \setminus B$ is a finite point set so.....

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A space is discrete iff every singleton set is open. If M is a finite metric space and $x\in M$. Let $\epsilon$ be the minimum distance from x to other points of M, the $B_{\epsilon}(x)$ contains x only So $\{x\}$ is open for every x.So M is discrete.

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