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Is it possible to describe every finite set in a metric space as closed, even though empty set is open, and in discrete metric the finite subset is open? For every subset of a metric space, its closure being the set of all $a \in M$ such that the intersection of the subset and $B[a;e]$ is finite for any $e>0$.

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    $\begingroup$ I don't understand your question. Always remember that a set may be both closed and open, or it may be neither. $\endgroup$
    – Eric Auld
    Dec 9 '13 at 3:06
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    $\begingroup$ Yes, it is fine. A closed set can be open, nothing forbids that. $\endgroup$ Dec 9 '13 at 3:07
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    $\begingroup$ Yes, every finite subset of a metric space is closed. This is completely independent of whether any of them are also open: a set can be both. $\endgroup$ Dec 9 '13 at 3:07
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    $\begingroup$ @Bob : yes, every finite set in a metric space is closed. It is not just "possible to describe" it as closed, it actually is closed. You should avoid weasel words when writing, talking, or asking about mathematics. However, sometimes a closed set is also open, as all the previous comments note. $\endgroup$ Dec 9 '13 at 3:10
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    $\begingroup$ @Bob : I'm not sure what you mean by "strictly closed". Either a set is closed or it isn't. By the way, using the discrete metric, all sets are open, not just finite sets. $\endgroup$ Dec 9 '13 at 3:13
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Is it possible to describe every finite set in a metric space as closed

Yes: every finite set in a metric space is closed. (As said in comments).

For every subset of a metric space, its closure being the set of all $a\in M$ such that the intersection of the subset and $B[a;\epsilon]$ is finite for any $\epsilon >0$.

This is not what the closure of a set means. Its definition is similar to the above, but the intersection with $B[a;\epsilon]$ must be nonempty, not finite.

In fact, the property you stated describes the set of all isolated points of the given subset.

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