Let $C \subseteq\mathbb{R}^{n}$ be a compact set. Let $I$ be the set or the isolated points of $C$, prove that $I$ is finite.

Question

I'm thinking of using the fact that $C$ is compact to find a contradiction, because for any open covering of $C$ there should be a finite subcover. Know, if $C$ has an infinite number of isolated points you will need to use an infinite number of open balls (centered at every isolated point) to get a finite subcover. That is absurd. Is my proof going in the right direction?

Answer 1

Again, this is false. Take $$ C = \prod_{i=1}^{n} [0,1]_i \subset \mathbb{R}^n.$$ Then, $C$ is a perfect set in $\mathbb{R}^n$, so every point of $C$ is a limit point. Ergo, there does not exist a point such that it can be surrounded by a neighborhood not containing another point of $C$. In general, any compact connected subset of $\mathbb{R}^n$ will not have an isolated point. There are many more examples, however.

Answer 2

In $\mathbb{R}$, let $X = \{0\} \bigcup \{1, \frac{1}{2}, \frac{1}{3}, \ldots \}$. Then $X$ is compact and has infinitely many isolated points.