MHB Is Compactness Equivalent to Limit Point Compactness in Topological Spaces?

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Compactness in a topological space implies limit point compactness, as every open cover of a compact space has a finite subcover, ensuring that every infinite subset has a limit point. However, limit point compactness does not guarantee compactness, as demonstrated by examples like the open interval (0, 1) in the standard topology, which is limit point compact but not compact. The converse holds true if the space is also Hausdorff, where limit point compactness and compactness coincide. The discussion remains focused on the implications and conditions surrounding these concepts in topology. Understanding these relationships is crucial for deeper insights into topological properties.
Chris L T521
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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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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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