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Let $X$ be a convergence space and let $K_1, K_2, \ldots, K_n$ be compact subsets of $X$. I'm trying to prove for myself that the union $K$ of the $K_i$ is compact. By definition, $K$ is compact if every ultrafilter on it converges, so given an ultrafilter $\mathcal U$ on $K$, I have to show that it converges. Somehow I need to produce ultrafilters on each of the $K_i$ and relate them to $\mathcal U$ but I have no idea how to do this. Any tips?

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I think this is true in any topological space: given an open cover of $K_1\cup\cdots\cup K_n$, it is also an open cover of each $K_i$. For each $i$, find a finite subcover that covers $K_i$; the union of $n$ finite subcovers is a finite subcover, and it will cover the union. – Arturo Magidin Jun 13 '12 at 2:53
This is true if the convergence spaces are topological by the argument you just gave. The general case though... – echoone Jun 13 '12 at 3:16
Sorry... I seem to have mentally confused "convergence space" with "sequential space" for some reason... – Arturo Magidin Jun 13 '12 at 3:18
up vote 1 down vote accepted

If $\mathscr{U}$ is an ultrafilter on $K$, there must be an $i\in\{1,\dots,n\}$ such that $K_i\in\mathscr{U}$. Then the trace of $\mathscr{U}$ on $K_i$, $\{U\cap K_i:U\in\mathscr{U}\}$, is an ultrafilter on $K_i$ that converges iff $\mathscr{U}$ does.

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Ah, of course! Thanks a lot. – echoone Jun 13 '12 at 16:16

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