Jtdylktuy

up vote 0 down vote favorite 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. real-analysis general-topology compactness share | cite | improve this quest