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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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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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X is a finite metric space and A is a proper subset of X. Take A = {$x_1,x_2,....x_n$}. If A is finite then X\A is finite.Let $x_i\in A$ and one can choose r = min {$d(x_i,x_j)$ | $j \not= i$}. Observe that $B(x_i,r)=${x}$\subset A $. This means every subset of X is open implies A is open. Also compliment of A i.e. X\A is also open as X\A is finite. Thus A is closed and open. ie. $\overline A =A $. Hence $\overline A =X $. this implies that A=X as $\overline X =X $. Thus only dense subset of X is itself.