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If [itex]X \LARGE[/itex] is a metric space and [itex]E \subset X[/itex] is a discrete set then is [itex]E \LARGE[/itex] open or closed or both?

Here's my understanding:

[itex]E \LARGE[/itex] is closed relative to [itex]X \LARGE[/itex].

proof: If [itex]p \subset E[/itex] then by definition [itex]p \LARGE[/itex] is an isolated point of [itex]E \LARGE[/itex], which implies that [itex]p \LARGE[/itex] is not an interior point of [itex]E\LARGE[/itex]. Since a subset of a metric space is open if every point of that subset is an interior point, it follows that [itex]E \large[/itex] is closed.

Now that I write this proof, I believe the problem is that a subset is open IF every point is an interior point, not IF AND ONLY IF. There are other ways for the subset to be open other than all points being interior. Is this my error?

The web tells me a discrete set is open and closed. Proofs I've seen use collections of neighborhoods with radius s.t. the only point contained in the nbhd is the discrete point at its center and since any collection of open sets is open then the discrete set is open.

The Problem:when proving closure by taking complements, wouldn't we need ANY collection of closed sets to be closed when if fact this is only true if the collection is finite?

Thanks for the help,

pob

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# A discrete subset of a metric space is open and closed

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