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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?

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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, $U_a\cap A=\{a\}$.

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$x$ 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.

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