# If each subset of a set $X$ is compact then $X$ is finite

$$\textbf{Problem :}$$ Let $$X \subseteq \mathbb{R}^n$$ such that every subset $$A \subseteq X$$ is compact. Prove that $$X$$ is finite.

$$\textbf{Proof}$$

Supose that $$X$$ is infinite, so exists some $$A\subseteq X$$ infinite and by assumption is compact. Then exists some $$a \in A^{'}$$, and because $$A$$ is closed we have : $$a \in A$$.

Define $$B=A-\{ a \} \subseteq X$$, is obviusly that $$B$$ is infinite and by assumption compact in particular closed. Because $$a\in A^{'}$$ , $$a$$ is the limit of a sequence $$x_k \in A-\{ a \}$$ so $$a \in \overline{B}=B$$, a contradiction.

Is good?Another way to solve?, Thanks!

Consider $$X$$ as a space. Then $$X$$ is a Hausdorff space so every compact subset of $$X$$ is closed in the space $$X.$$
So every subset of $$X$$ is closed in the space $$X,$$ but then every subset of $$X$$ is also open in the space $$X.$$
So if $$X$$ is not empty then $$C=\{\{x\}:x\in X\}$$ is an irreducible open cover of $$X:$$ No proper subset of $$C$$ is a cover of $$X.$$ But $$X$$ is compact (because it is a subset of $$X$$) so $$C$$ must be finite.
• This will not work for non-Hausdorff spaces.E.g. if the only open sets in a space $Y$ are $Y$ and $\emptyset$ then every subset of $Y$ is compact. – DanielWainfleet May 10 '19 at 20:48