Let $(X,d)$ to be a arbitrary metric space. I know that: every finite subspace of a metric space is closed. Then $\{a\}$ ,with $a\in X$, is closed $\Rightarrow X \setminus \{a\}$ is open. (1)

But if $X$ is a finite metric space then $X\setminus \{a\}$ is also finite so $X\setminus\{a\}$ is closed. (2)

My question is: from (1) + (2) $\Rightarrow$ if $X$ is a finite metric space $X\setminus \{a\}$ is open and closed in the same time?


The distinction becomes less important in this finite case if, for $d$ the minimum distance between two points, one works with open or closed balls of radius $d/2.$

So every subset is both open and closed.

Edit: OP has asked for "more examples". Let $X$ be the finite metric space consisting of vertices of a square of side-length $1$ in the $x,y$ plane, with the metric from that in the plane. Then for each vertex, the open ball of radius $1/2$ centered there contains only that vertex. So each point is open (it is the only point in an open ball). Then any subset of $X$ is open, since it is a union of open sets. Any set is also closed, since its complement in $X$ is also a union of points of $X.$

This same thing can be done in any finite metric space as already outlined in the first paragraph.

  • $\begingroup$ i need some examples. $\endgroup$ – Ica Sandu Apr 24 '18 at 23:21

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