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If $X,\tau$ is a topological space and K is a closed subspace. If $U$ is a compact set of $X,\tau$ then $U\cap K$ is compact.

Attempted proof:

Since $U$ is compact there is a finite sub covering such that $U\subset \bigcup_\limits{i=1}^{n}B_i$ $B_i\in \tau\forall 1\leqslant i\leqslant n$

$U\cap K\subset (\bigcup_\limits{i=1}^{n}B_i)\cap K=\bigcup_\limits{i=1}^{n}(B_i\cap K)$

$B_i\cap K$ is open for the subspace topology for all $i$.

Hence $\bigcup_\limits{i=1}^{n}(B_i\cap K)$ is a finite sub covering of $U\cap K$ proving $U\cap K$ to be compact in the subspace topology.

Question:

Is my proof right? If not how should I correct it?

Thanks in advance!

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  • $\begingroup$ I think you're misusing the definition of compactness: it says that every open cover has a finite subcover (not just that a finite cover exists). $\endgroup$ – Uncountable Dec 25 '18 at 18:59
  • $\begingroup$ The statement is equivalent to: a closed subspace A of a compact space B is compact. You can prove this by constructing an open cover of B from an open cover of A (which is given by open subsets of B intersected with A) by adding the complement of A, and extracting a finite subcover. $\endgroup$ – Uncountable Dec 25 '18 at 19:04
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The proof is wrong from the start. Every topological space has a finite covering: just take the covering with a single open set, which is the whole space.

Let $(O_\lambda)_{\lambda\in\Lambda}$ be an open cover of $U\cap K$. Add to this cover the set $U\cap K^\complement$ (which is an open subset of $U$). Then you have an open cover of $U$. Since $U$ is compact, it has a finite subcover. The elements of this finite subcover which are distinct from $U\cap K^\complement$ form a finite subcover of $U\cap K$.

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