2
$\begingroup$

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?

  • 0
    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.2012-06-13
  • 0
    This is true if the convergence spaces are topological by the argument you just gave. The general case though...2012-06-13
  • 0
    Sorry... I seem to have mentally confused "convergence space" with "sequential space" for some reason...2012-06-13

1 Answers 1

1

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.

  • 0
    Ah, of course! Thanks a lot.2012-06-13