# A Step in a Proof that Compactness implies Limit Point Compactness

I have a question about the last part of this proof (bold letters).

Why from the empty intersection between $X-A$ and $A$ and that each set $U_a$ contains only one point of $A$, can we conclude that the set $A$ must be finite?

Compactness implies limit point compactness.

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.

Furthermore, for each $a\in A$ we can choose a neighborhood $U_a$ of $a$ such that $U_a$ intersects A in the point $a$ alone. The space $X$ is covered by the open set $X - A$ and the open sets $U_a$; being compact, it can be covered by finitely many of these sets.

Since $X - A$ does not intersect $A$, and each set $U_a$ contains only one point of $A$, the set $A$ must be finite.

• $\left\{ X - A , \{ U_a\}_{a \in A} \right\}$ is a cover of $X.$ Since $X$ is compact, there exists a finite subcover of the above cover, say $\left\{X - A, \{U_i\}_{i \le n} \right\}$. Now, since $X - A$ does not intersect $A,$ we must have, $A \subseteq \bigcup U_i,$ but the latter contains only a finite number of the elements of $A,$ by construction. – stochasticboy321 Dec 7 '17 at 3:54
• If $A$ was NOT finite, then it could not be covered by finitely many $U_a$, since each $U_a$ contains only one element of $A$ (and $X\setminus A$ contains zero elements of $A$) – David M. Dec 7 '17 at 3:56
• A word on the title: $(X-A)\cap A = \emptyset$ is always true, so your question looks much sillier than it is. – spaceisdarkgreen Dec 7 '17 at 3:59
• @stochasticboy321 clear and good explanation, thank you. – Isa Dec 8 '17 at 0:22

## 1 Answer

The relevant assertions for the last line of the proof are the following. First, for all $a\in A$, $a\in U_a$ and $|U_a|=1$. Second, there are only finitely many sets in the family, $\{X-A\}\cup\{U_a:a\in A\}$.

From the first assertion, we deduce $U_a=\{a\}$ for all $a\in A$. Thus, for all $a,a'\in A$, $U_a=U_{a'}$ if and only if $a=a'$. This means that there are no repeats when we write $\{U_a:a\in A\}$. More formally, the function $A\to \{U_a:a\in A\}$ given by $a\mapsto U_a$ is bijective.

Now, as a consequence of the second assertion, the family $\{U_a:a\in A\}$ has only finitely many sets in it. So, because $A$ is in a bijective correspondence with $\{U_a:a\in A\}$, it follows that $A$ must be a finite set.

One will notice that we actually do not need the observation that $X-A$ and $A$ are disjoint, because $\{U_a:a\in A\}$ covers $A$ by definition.