# Problem of the Week #22 - October 29th, 2012

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#### Chris L T521

##### Well-known member
Staff member
Here's this week's problem.

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Problem: Let $X$ be a topological space, and suppose that $X$ is compact. Show that compactness implies limit point compactness, but not conversely (i.e. one does not have limit point compactness imply compactness). Under what condition is the converse true?

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Remember to read the POTW submission guidelines to find out how to submit your answers!

#### Chris L T521

##### Well-known member
Staff member
No one answered this week's question. Here's my solution:

Proof: Let $X$ be a compact (topological) space. We seek to show that if $A\subseteq X$ is infinite, then $A$ has a limit point. We proceed by proving the contrapositive: If $A$ has no limit point, then $A$ is finite.

Suppose that $A$ has no limit points. Then $A$ contains all it's limits points, implying that $A$ is closed. Furthermore, for each $a\in A$, there is a neighborhood $U_a$ of $a$ such that $U_a\cap A = \{a\}$. The space $X$ is covered by the open sets $X\backslash A$ and the open sets $U_a$; since $X$ is compact, it can be covered by finitely many of these open sets. Since $X\backslash A$ doesn't intersect $A$, and each set $U_a$ contains only one point of $A$, it follows that the set $A$ must be finite. Q.E.D.

It turns out that limit point compactness implies compactness only when $X$ is a metrizable space.

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