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Suppose that $x_{i \in I}$ is a generalized sequence on a compact Hausdorff space $X$, indexed by the directed set $I$, and with the property that $\exists \, j \in I$ such that $\forall i \geqslant j$, then $x_i = k$, where $k \in X$. We then pick a non-principal ultrafilter $U$ on the index set $I$ (so $U \subset \mathcal{P}(I)$).

My claim is that we then have $\lim_{U} (x_i) = k$.

First we notice that $X$ being compact and Hausdorff, the ultralimit, $\lim_U (x_i)$, of this sequence always exists uniquely. Also, we can just assume that such an ultrafilter on $I$ exists ($I$ being an infinite index is actually sufficient and necessary for that).

I have been unable so far to prove that this claim is true and I am starting to think I am missing an assumption for it to be true.

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This isn't true. For instance, let $I=\mathbb{N}\cup\{\infty\}$; then any sequence $(x_i)_{i\in I}$ is eventually constant with value $k=x_\infty$ (since you can always just take $j=\infty$). But if, say, there is a $y\in X$ distinct from $k$ such that $x_n=y$ for all $n\in \mathbb{N}$, then $\lim_U(x_i)=y$, not $k$.

For the result to be true, you could require that $U$ contains the set $\{i\in I:i\geq j\}$ for all $j\in I$. Note, however, that there may be no nonprincipal ultrafilter with this property, as the previous example (or more generally, any directed set $I$ which has a greatest element) shows.

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  • $\begingroup$ Thanks! If we assume that there is no greatest element in $I$, is it reasonable to assume there exists such an ultrafilter. I realize now that the proof I was trying to write down revolved exactly around proving that $\{ i \in I; i \geqslant j \} \in U$. For what I want to do with that result, proving the existence of that ultrafilter would be sufficient, but I feel that in general this a very strong assumption. $\endgroup$ – Olivier Melançon Oct 22 '15 at 22:38
  • $\begingroup$ If $I$ has no greatest element, then directedness of $I$ implies that the sets $\{i\in I:i\geq j\}$ generate a proper filter $F$ on $I$ which contains the cofinite filter (this is not obvious, and is a good exercise to try to prove). So any ultrafilter containing $F$ is nonprincipal. $\endgroup$ – Eric Wofsey Oct 22 '15 at 22:42
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The claim is in fact false.

For an explicit counterexample, let $I=\mathbb{Z}$ with the usual ordering, let $X$ be the two-point Hausdorff space $X=\{a, b\}$, let $x_i=a$ if $i<0$ and $x_i=b$ if $i\ge 0$, and let $\mathcal{U}$ be a nonprincipal ultrafilter on $\mathbb{Z}$ such that $\mathbb{Z}_{<0}\in\mathcal{U}$. Then the limit of $(x_i)_{i\in I}$ along $\mathcal{U}$ is $a$, even though it clearly "ought" to be $b$.


From here it's an easy exercise to find the necessary-and-sufficient condition $\mathcal{U}$ needs to satisfy in order to make this statekement work. (Note that as Eric Wofsey's answer shows, this condition will actually contradict non-principality unless $I$ satisfies "there is no largest element.")

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  • $\begingroup$ OK, fine, "explicit" might be a bit of a reach . . . :P $\endgroup$ – Noah Schweber Oct 22 '15 at 22:36

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