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


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.

  • $\begingroup$ 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. $\endgroup$ – DanielWainfleet May 10 '19 at 20:48

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