# Help in proof of Compactness implies limit point compactness.

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 2 '13 at 23:48
• Thanks. Looks up definition of limit point again. – Shine On You Crazy Diamond Oct 2 '13 at 23:49
• Having no limit point also means that the set is discrete. – Stefan Hamcke Oct 3 '13 at 0:50

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.