# Prove that the set consisting of isolated points is finite.

1. Let A be a compact subset in $R^n$. Investigate whether the following assertion is true or not: If A consists of isolated points only then A is finite.

I couldn't demonstrate my answer.We know that when A is a compact subset then it is closed and bounded. I was looking and I found that it resembles this question Is a closed subset of isolated points in a compact set necessarily finite? I am still not fully convinced.

• If $A$ is infinite then it has an accumulation point. – OR. Mar 31 '15 at 18:00

If the points of $A$ are isolated, then each singleton $\{a\}$ is an open set in $A$, for all $a\in A$. Therefore, the collection of these singletons form an open covering of $A$. By compactness of $A$, there exists $a_1,\ldots, a_n \in A$ such that $A = \{a_1\} \cup \cdots \cup \{a_n\}$. Thus $A = \{a_1,\ldots, a_n\}$.
For each $a \in A$, let $U_a$ be an open neighborhood of $a$ that does not otherwise intersect $A$ (i.e. $U_a \cap A = \{a\}$); this is possible because each $a$ is an isolated point.
Then $\{U_a\}$ is an open subcover of $A$, and because $A$ is compact, there is a finite subcover. What does this imply?