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If $K \subset E_1 \cup E_2$, where $K$ is compact and $E_1, E_2$ are disjoint open subsets of a topological space, is $K \cap E_1$ compact? Is that always the case if $E_1, E_2$ are not disjoint?

I've seen other threads that have this, so I was wondering why the solution I thought of is incorrect:

Let $U_{\alpha}$ be an open covering of $K$. Then because $K$ is compact, there is a finite subcovering $U_1, U_2, \ldots, U_N \in U_{\alpha}$ that cover $K$. But then $U_1, \ldots, U_N, E_1$ is a finite collection of open sets that covers $K$ and $E_1$, so it covers $K \cap E_1$, and so $K \cap E_1$ must be compact.

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  • $\begingroup$ If $E_1$ and $E_2$ are not disjoint, you might as well take $E_2=X$ (the whole space); so you're just asking, if $K$ is compact and $E_1$ is open, is $K\cap E_1$ compact? $\endgroup$ – bof Nov 14 '18 at 8:05
  • $\begingroup$ If $E_1,E_2$ don't have to be disjoint, here is a counterexample: $K=[0,1]$, $E_1=(0,1)$, $E_2=\mathbb R$. $E_1$ and $E_2$ are open subsets of the space $\mathbb R$, $K$ is a compact subset of $E_1\cup E_2$, but $K\cap E_1=E_1$ is not compact. $\endgroup$ – bof Nov 14 '18 at 8:08
  • $\begingroup$ If your argument were correct (which it is not), it would prove that any subset of a compact set is compact. $\endgroup$ – bof Nov 14 '18 at 8:09
  • $\begingroup$ Yes, I realize the conclusion of this "proof" is incorrect, but I was wondering where the flaw was that led to this false conclusion. $\endgroup$ – user386867 Nov 14 '18 at 8:10
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    $\begingroup$ To prove that $K\cap E_1$ is compact, you have to show that any open cover of $K\cap E_1$ has a finite subcover. Proving that an open cover of $K$ has a finite subcover doesn't do it. $\endgroup$ – bof Nov 14 '18 at 8:11
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Assuming that your space is Hausdorff $K\setminus E_2=K\cap E_1$ and $K\setminus E_2$ is compact. On second thoughts you don't need Hausdorff property!.

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  • $\begingroup$ Where do you think you need Hausdorff? If $E_1$ and $E_2$ are disjoint and $K\subseteq E_1\cup E_2$ then $K\cap E_1=K\setminus E_2$ is just set theory. If $K$ is compact and $E_2$ is open then $K\setminus E_2$ is compact in any topological space. $\endgroup$ – bof Nov 14 '18 at 8:15
  • $\begingroup$ @bof You are right. $\endgroup$ – Kavi Rama Murthy Nov 14 '18 at 8:18
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Your proof should start with a cover $\mathcal{U}$ (WLOG by open sets of $X$) of $K \cap E_1$, and produce a finite subcover of that:

Add the one set $E_2$ to $\mathcal{U}$ and we have an open cover of $K$ (any set in $K$ lies in $E_1$ or $E_2$ and the first ones are covered by $\mathcal{U}$ the other by $E_2$..) and so $\mathcal{U} \cup \{E_2\}$ has a finite subcover $\mathcal{U}'$ and then $\mathcal{U}'\setminus \{E_2\}$ is still finite (smaller than the finite $\mathcal{U}'$) and a subcover of $\mathcal{U}$. So $K \cap E_1$ is compact.

Alternatively note that $K\cap E_1 = K \cap (X\setminus E_2)$ and thus is a closed subset of the compact $K$ and hence compact too.

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