# Prove: $F$ is countable and closed $\implies F$ has an isolated point.

Let $F$ be a closed countable non-empty subset of $\Bbb R$. Prove that $F$ has an isolated point.

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Could anyone tell me what is the Negation of the statement? –  La Belle Noiseuse Feb 8 '13 at 6:57
Here is a proof. (A perfect set is a closed set with no isolated points, so, requirement, you want to prove a perfect set of reals cannot be countable.) –  Omar Antolín-Camarena Feb 8 '13 at 7:15
@CityOfGod If $F$ has no isolated points then $F$ is not closed or $F$ is not countable. –  JSchlather Feb 8 '13 at 7:45
@JacobSchlather Thank you! –  La Belle Noiseuse Feb 8 '13 at 8:16

Hint: Every complete metric space without isolated points is uncountable.

See, Elementary Real and Complex Analysis, P 87.

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If X is a countable complete metric space with no isolated points, then each singleton {x} in X is nowhere dense, and so X is of first category in itself. See wikipedia

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A perfect set is a closed set without isolated points. This answer shows that a (non-empty) perfect subset of $\Bbb R$ has $2^\omega=\mathfrak c$ points. In particular, it must be uncountable. Thus, a countable, closed subset of $\Bbb R$ cannot be perfect and must have an isolated point.

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