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Is it true that if $A$ is discrete as a subspace of $X$, and $X \;$ is compact, then $A$ is finite?

If this doesn't hold, then does it hold for $X\;$ manifold?

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up vote 14 down vote accepted

It is not true in general. Let $X = \{0\}\cup\{2^{-n}:n\in\mathbb{N}\}$ with the topology inherited from $\mathbb{R}$; then $X$ is compact, and $X\setminus \{0\}$ is an infinite discrete subset of $X$. Of course every closed discrete subset of a compact space is finite, so infinite discrete subsets won’t be closed, but in general they will exist. For instance, the space $X$ just described can be embedded in any infinite compact metric space.

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Thank you very much! For my purpose the fact that "every closed discrete subset of a compact space is finite" is just what I need – Abramo Nov 10 '11 at 10:37

Take the subspace $\{0\}\cup\{\frac1n; n\in\mathbb N\}$ of the real line. It is compact and contains an infinite discrete subspace $\{\frac1n; n\in\mathbb N\}$.

You can construct a similar example as a subspace of unit circle, so this fails for manifolds too. (It suffices to make a quotient of $[0,1]$ by identifying zero and one, which leaves the above subspace unchanged.)

More generally, for arbitrary discrete space you can construct a compactification

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Loved the part about compactification!! – André Caldas Nov 10 '11 at 14:37

Abramodj: I just want to point out that closed discrete subset of a compact space is finite is consequence of a more general fact that a discrete space is compact iff it is finite. Now since closed subset of a compact space is compact, so closed discrete subset of compact space are compact and hence finite.

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Thanks! This is a very important remark – Abramo Nov 10 '11 at 18:00

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