8
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ That's correct. Having no limit points implies that you are closed. $\endgroup$ – Alex Youcis Oct 2 '13 at 23:48
  • 1
    $\begingroup$ Thanks. Looks up definition of limit point again. $\endgroup$ – Shine On You Crazy Diamond Oct 2 '13 at 23:49
  • 1
    $\begingroup$ Having no limit point also means that the set is discrete. $\endgroup$ – Stefan Hamcke Oct 3 '13 at 0:50
10
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.