# Compactness implies limit point compact

In the proof of theorem 28.1 of Munkres that says compactness implies the limit point compact.

Proof: Let X be a compact. Given a set A of X, we wish to prove that if A is infinite, then A has a limit point. We prove the contraposition-if A has no limit point, then A must be finite. So suppose A has no limit point. Then A contains all its limit points, so that A is closed. For each a in A we can choose a neighborhood U_{a} of a that intersect A in a alone. ...

My question is why we can choose such a neighborhood that does not intersect A-{a}?In other word, does such a neighborhood exists for each a in A?

## 2 Answers

If $$x\in X\setminus A$$, then $$X\setminus A$$ is an open nbhd of $$x$$ disjoint from $$A$$. If $$a\in A$$, then $$a$$ is not a limit point of $$A$$, so by the definition of limit point $$a$$ has an open nbhd $$U_a$$ that contains no other point of $$A$$. In other words, $$U_a\cap A=\{a\}$$.

$$x$$ is a limit point of $$A$$, by definition if

for all (open) neighbourhoods $$U$$ of $$x$$, $$U \cap (A\setminus\{x\}) \neq \emptyset$$ or every (open) neighbourhood $$U$$ of $$x$$ contains a point in $$A$$ that is not equal to $$x$$.

So negating this

There is an (open) neighbourhood $$U_x$$ of $$x$$ such that the only possible point of intersection with $$A$$ is $$x$$, which says that $$A \cap U_x = \emptyset$$ or $$A \cap U_x = \{x\}$$, depending on whether $$x \in A$$ or not. In short, we conclude $$A \cap U_x \subseteq \{x\}$$. No need to observe that $$A$$ would be closed, we can then just apply compactness directly to the cover $$\{U_x: x \in X\}$$ and get a similar contradiction with the infiniteness of $$A$$.

In fact this allows us to generalise a bit:

Call $$x \in X$$ a strong limit point of an infinite set $$A$$ if for all (open) neighbourhoods $$U$$ of $$x$$ we have $$|U \cap A|=|A|$$. ($$|\cdot|$$ denoting set cardinality..) Then if $$X$$ is compact we have that every infinite $$A$$ has a "strong limit point" in $$X$$. In this general formulation this property of $$X$$ is even equivalent to compactness which limit point compactness is not.