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Studying Rudin's Real Analysis and after proving the theorem "Closed subsets of compact sets are compact" I have came up with the following thought: The theorem means that if a set is closed and has a compact superset, then it is compact. This also means that:

A set $K$ is not compact $\implies$ $K$ is not closed or $K$ has no compact superset.

Assume that $K$ is not compact and $K$ is closed. Then it must be that $K$ has no compact superset. I wanted to show that. Trivially, if for every $E$, which is $ E \subseteq K$, $E$ is not compact, then since $K \subseteq K$, $K$ is automatically not compact. While this looks valid, it seemed too trivial to be true to me. Is this really valid? If not, how can one show that when $K$ is not compact, but closed, it has no compact supersets?

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You want to show that $K$ has no compact superset. That means you have to show that if $K \subseteq E$, then $E$ is not compact.

Let if possible, $E$ is compact. As closed subsets of compact sets are compact, then $K$ is compact, which is a contradiction.

Alternatively, if you want to prove this without using the above theorem then:

Let $\bigcup_{\alpha} U_{\alpha}$ be any open cover of $K$. As $K$ is closed in $X$(our topological space) and $K \cap E=K$, therefore $K$ is closed in $E$. Hence, $E \setminus K$ is open in $K$. Now, $$E=\bigcup_{\alpha} U_{\alpha} \cup (E \setminus K)$$ If, $E$ is compact, then this open cover has a finite subcover. That is, there exists, $\alpha_1, \alpha_2, \cdots \alpha_n$ such that $$E=\bigcup_{i=1}^n U_{\alpha_i} \cup (E \setminus K)$$Hence, $$K\subseteq\bigcup_{i=1}^n U_{\alpha_i}$$ Thus, we have shown that an arbitrary open cover of $K$ has a finite subcover. It follows that $K$ is compact, which is a contradiction.

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  • $\begingroup$ The last equality should be inclusion. $\endgroup$ – tomasz Jun 29 '17 at 22:39

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