Finiteness of a compact subset in $\mathbb R^n$

Question

Let $K$ be a compact subset of $\mathbb R^n$ such that for all $x \in K$, $K\setminus\{x\}$ is also compact. Show that $K$ is finite.

I'm trying to solve it using sequences, but am having difficulty. Could someone help me? Thank you!

Answer 1

We can use the definition of compactness, or properties of compact subsets of $\mathbb{R}^n$.

If $K$ is infinite, it has a limit point $x$. But then $K\setminus\{x\}$ is not closed.

Answer 2

This solution does not use sequences. It is the only one I know. Hope it will help you.

$K \setminus \{x\}$ is a compact set and hence a closed set for any $x \in K$. Therefore the complement $ [K \setminus \{x\}]^c$ must be an open set. It can be seen the following is an open cover for the set $K$:

$$ \bigcup_{x \in K} [K \setminus \{x\}]^C $$

Then there is a finite collection of sets of the form $ [K \setminus \{x\}]^c $ which contain $K$. Each of these sets contain no other elements except $x$ of $K$. Hence if the finite union contains $K$ the set must be finite.