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Prove that if $(X,d)$ is a compact metric space, and $K$ is an infinite set in $(X,d)$, then if $K$ has no limit point, $K$ is a closed set.

Idea : Just like most topology proofs, the way I want to approach this problem is to show that $X - K$ is open however I am unsure how to do this. I actually have an idea that does not involve open sets which I show below, but it would be nice if I could show $X - K$ is open. The help would be appreciated!

Suppose $K$ has no limit points. If $K'$ is the set of limit points of $K$ then $K' = \emptyset$. $K$ is closed if every limit point of $K$ is a point of $K$. Then $K$ is closed if $K' \subset K$. Since $K' = \emptyset$, $K\ \subset K$ since $\emptyset \subset K \forall K$. Then $K$ is closed.

Nick Freeman
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    I think this is correct way to approach this problem. Showing that the complement is open just requires that every point which is not a limit point of $K$ or a point of $K$ has a ball contained in $K^c$. – Bruce Zheng Oct 16 '14 at 16:57
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    You could also use the fact that sequential closedness and closedness are equivalent in a compact metric space. If there are no limit points, any sequence of points in $K$ converging to a point in $K$ is eventually constant. – Joonas Ilmavirta Oct 16 '14 at 17:07

3 Answers3

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What can we deduce from $K$ not having any limit points?

Well, that means if we take $x \in X - K$, then we can find some $\epsilon > 0$ such that $B(x, \epsilon) \cap K = \emptyset$ (otherwise, $x$ would be a limit point of $K$).

But $B(x, \epsilon) \cap K = \emptyset \implies B(x, \epsilon) \subseteq X - K$, and so for each $x \in X - K$, we found an open ball around $x$ entirely contained in $X - K$, which shows $X - K$ is open, and thus $K$ is closed.

layman
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Suppose $x \notin K$. Suppose $B(x,{1 \over n})$ intersects $K$ for all $n$, then $x$ is a limit point of $K$, which is a contradiction. Hence for some $n_x$ we have $K \cap B(x, {1 \over n_x}) = \emptyset$. Hence $K^c$ is open.

copper.hat
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You can show that every point of $X-K$ is an interior point. Since for $x \in X-K$, there exist an open set $O \ni x$, such that $O \cap K= \phi$, otherwise $x$ would be the limit point for $K$. and thus for an arbitrary $x \in X-K$, we hav $x \in O \subset X-K$. Hence $X-K$ is open.

WikiSandhu
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