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I know that for the standard metric on R, every singleton is closed. However, I have a claim that on a finite set for any arbitrary metric every singleton is open, how can I prove this?

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    $\begingroup$ Hint: If $X$ is a finite metric space, then the set $\{d(x, y) \in (0, \infty) : x, y \in X, x \neq y\}$ is finite and hence has a (strictly positive) minimum. $\endgroup$ – Theo Bendit 2 days ago
  • $\begingroup$ Great hint, I wil try to sketch a proof and update $\endgroup$ – Beaba 2 days ago
  • $\begingroup$ How do I prove that {0} is not a subset of this set? $\endgroup$ – Beaba 2 days ago
  • $\begingroup$ If $0$ were in this set, then $d(x, y) = 0$ for some $x, y \in X, x \neq y$, which violates one of the axioms. $\endgroup$ – Theo Bendit 2 days ago
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If $F$ is finite, and $x \in F$, then the set $D_x:=\{d(x,y) : y \in F, y \neq x\}$ is also a finite set and does not contain $0$ (as $d(x,y)=0$ iff $x=y$ and $y$ is never equal to $x$) so $r=\min D_x$ is well-defined and $>0$ (set it to $1$ if $D_x$ is empty, which happens for $F=\{x\}$). Then for all $y \neq x \in F$, $d(x,y) \ge r$, so $y \notin B(x,r) \cap F$. It follows that $B(x,r) \cap F = \{x\}$ and so $\{x\}$ is relatively open in $F$.

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