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Let $I$ be a directed set, $\mathcal U$ a free ultrafilter on $I$, i.e. an ultrafilter which contains the Fréchet filter (or equivalently $\bigcap \mathcal U = \emptyset$) and $(x_i)_{i\in I}$ a bounded $\mathbb R$-valued net.

I want to show that the ultrafilter limit $\displaystyle\lim_\mathcal U$ satisfies $$\lim_{\mathcal U}x_i\leq \limsup_{i\in I}x_i.$$

This is not true for a non-free Ultrafilter of course and I am not quite sure if it must holf for free ultrafilters. The reason why I think it should hold is because I have seen an approach which defines ultraproducts of Banach spaces with the help of a Limit functional with the above property. To construct this functional however, no ultrafilters are used. On the other hand, I know that ultraproducts are usually defined via ultrafilters which makes me think that ultrafilter limits should satisfy this aswell.

What I have found so far is that it would suffice to know that $\{i\in I: i\geq i_0\}\in\mathcal U$ for each $i_0\in I$ but I don't know if that leads to anything. (Edit: This isn't true as the example $I = \mathbb N\cup\{\infty\}$ shows, here we have $\{i\in I: i\geq \infty\} = \{\infty\}\notin \mathcal U$. Is there another way to show this or is it simply not true?)

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  • $\begingroup$ This does not necessarily hold. We could take $I = \mathbb{R}$ (with the usual order), and $\mathcal{U}$ an ultrafilter containing all the sets $(-\infty, x)$ for $x\in \mathbb{R}$. You need a property that could be informally stated as "$\mathcal{U} \to +\infty$". Which formally is "for all $i_0 \in I$, we have $\{ i \in I : i \geqslant i_0\} \in \mathcal{U}$". [Which means $\mathcal{U}$ can't be free if $I$ has a largest element.] $\endgroup$ – Daniel Fischer Jan 4 '17 at 13:38
  • $\begingroup$ @DanielFischer That makes sense, thank you. I need to see if I can just switch to ultrafilters with that property instead of free ones then. If $I$ doesn't have a largest element, then $\mathcal U$ with that property is free, right? $\endgroup$ – Tim B. Jan 4 '17 at 13:58
  • $\begingroup$ Yes. If an ultrafilter is non-free, it contains a singleton $\{i_0\}$. Since $i_0$ is not the largest element, there is an $i_1$ with $i_1 \not\leqslant i_0$, and an $i_2$ with $i_0 \leqslant i_2$ and $i_1 \leqslant i_2$. We can't have $i_2 \leqslant i_0$ (otherwise $i_1 \leqslant i_2 \leqslant i_0$), so $\{ i \in I : i \geqslant i_2\}$ doesn't belong to the non-free ultrafilter. // It looks very much like the idea is to do something like subnets or, essentially the same, filters finer than the one generated by $\bigl\{ \{ x_i : i \geqslant i_0\} : i_0 \in I\bigr\}$. $\endgroup$ – Daniel Fischer Jan 4 '17 at 14:06
  • $\begingroup$ Thank you, that helped a lot. @DanielFischer $\endgroup$ – Tim B. Jan 4 '17 at 14:10

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