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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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4 Answers 4

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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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  • $\begingroup$ 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 $\endgroup$
    – Abramo
    Nov 10, 2011 at 10:37
  • $\begingroup$ @Theoneandonly How is this set compact? Every point is itself an open set, so I can define an infinite cover of X that doesn't have a finite subcover. $\endgroup$
    – Mark
    Oct 1, 2020 at 17:48
  • $\begingroup$ @Mark: $0\in X$, and every open set that contains $0$ contains all but finitely many of the other points of $X$. $\endgroup$ Oct 1, 2020 at 17:50
  • $\begingroup$ @BrianM.Scott When you say that $X$ inherits the topology from $\mathbb{R}$, you mean that open sets are then of the form $U \cap X$, where $U$ is open in $\mathbb{R}$. That would mean that all one point sets are open, right? Ahhh....except for the point $0$! Thanks! $\endgroup$
    – Mark
    Oct 1, 2020 at 17:57
  • $\begingroup$ @Mark: Yes, the open sets in $X$ are the sets of the form $U\cap X$ for $U$ open in $\Bbb R$, but it is not true that all of the singletons of $X$ are open: the sets $\{2^{-n}\}$ are open in $X$, but $\{0\}$ is not open in $X$. $\endgroup$ Oct 1, 2020 at 17:58
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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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    $\begingroup$ Loved the part about compactification!! $\endgroup$ Nov 10, 2011 at 14:37
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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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Singleton sets are open .there union is a cover for X.so this is the cover which do not have a subcover hence X is not compact

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