Help in proof of Compactness implies limit point compactness.

Question

Munkres' Topology says

Theorem 28.1. Compactness implies limit point compactness, but not conversely.

Proof. Let $X$ be a compact space. Given a subset $A$ of $X$, we wish to prove that if $A$ is infinite, then $A$ has a limit point. We prove the contrapositive - if $A$ has no limit point, then $A$ must be finite. Suppose $A$ has no limit point. Then $A$ contains all its limit points, so that $A$ is closed. ....

That's confusing. Do they mean vacuously $A$ must be closed since it has no limit points even to contain?

That's correct. Having no limit points implies that you are closed. — Alex Youcis, Oct 02 '13 at 23:48
Wow, I was just confused by the exact same point in Munkres 28.1 and found this. How useful! Thanks :) — clay, Mar 10 '20 at 05:02

score 9 · Accepted Answer · answered Oct 03 '13 at 02:32

If a set has no limit points, then it is closed, as it automatically contains all its limit points. By the same reasoning, each subset of $A$ is closed, as a limit point of the subset would also be a limit point of $A$. Therefore, it is discrete. Conversely, being closed and discrete implies that a set has no limit points.
As a closed subset of $X$, $A$ is compact. But a discrete compact set must be finite.

Help in proof of Compactness implies limit point compactness.

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