# Basic facts about ultrafilters and convergence of a sequence along an ultrafilter

Could you help, please. I need the information about the ultrafilters, namely, any ideas how one can see that they exist and a proof of the fact that for any ultrafilter every sequence on a compact topological space has a limit. I hope these basic facts can be collected somewhere in a popular form, I would be grateful for a reference.

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Since you ask for a popular form, I think all this is in Jänich's Topology in the chapter in which he proves Tychonoff's theorem. I don't think you can get it much cheaper than the way he does it. – t.b. Jul 14 '11 at 18:46
heres a little treatment of ultrafilters terrytao.wordpress.com/2007/06/25/… – yoyo Jul 14 '11 at 18:50
@Theo: If I understand liman's question correctly, he is not asking about convergence of filter on a topological space (which is addressed in Janich) but about a slightly different (but closely related) situation - he works with a filter on $\mathbb N$ and studies the convergence of a sequence along this filter. – Martin Sleziak Jul 14 '11 at 19:07
@yoyo: thanks a lot for the reference. – liman Jul 14 '11 at 20:17
Here are two links I found useful: David MacIver's Filters in Topology and Pete L. Clark's notes on convergence – kahen Jul 21 '11 at 10:58

I hope that we both have in mind the same notion of convergence of a sequence along an ultrafilter on $\mathbb N$.

I know that you asked for a reference. Instead of giving a reference, I have copied a part of my LaTeXed notes I made for myself sometimes ago instead. In these notes I work with functions and ultrafilters on arbitrary set. (For $M=\mathbb N$ you'll get sequences.) I wrote these notes basically for the reason that I found the results useful, I did not know any reference at that time, and wanted to keep proof for myself somewhere. I later found book [HS], where this is shown, and I mention it in these notes. But I do not think that this book would be good for you if this is for the first time you encountered this notion. (In my opinion, if someone has the maturity required to read this book, he should be able to figure out the proof of the result about ultrafilters and compact spaces by himself.) Added later: In the meantime I found a few more references and I've added them to this post.

For other references: Some facts about the limit along an ultrafilter are collected in the paper M.A.Alekseev, L.Yu. Glebskii, E.I.Gordon. On approximations of groups, group actions and Hopf algebras. You can also find something in the book Komjath, Totik: Problems and Theorems in Classical Set Theory, but they work only with real sequences. The proof for the case of real sequences is also given at planetmath. If the case of bounded real sequences is sufficient for you, you can find many resources.

You can find many references for the existence of free ultrafilters. E.g. already mentioned books by Komjath and Totik, or Janich. If you do not have access to these books, you could try to Google for filter ultrafilter zorn.

Snippet from my notes (this links to version which is still unfinished):

The following result can be found in [HS,Theorem 3.48]. [HS,Theorem 3.52] shows that this is a characterization of compact spaces. Further references: This blog post and [D, Theorem 4.3.5], [T, p.64, Claim 14.1], [F, 2A3Se(i)].

Proposition: Let $\mathcal F$ be an ultrafilter on $M$, $X$ be a compact space and $f:M\to X$ be a map. Then $\mathcal F-\lim f$ exists.

Recall that $\mathcal F-\lim f=x$ means that $f^{-1}(U)\in\mathcal F$ holds for each neighborhood $U$ of $x$. This is equivalent to the claim that the filterbase $f[\mathcal F]$ converges to $x$ in $X$.

We give a direct proof and also transformation to a known result from general topology. (Namely the result that in a compact space every ultrafilter has a limit - in the usual topological sense, see here or in the book suggested by Theo.)

Proof 1. Suppose that no point $x\in X$ is an $\mathcal F$-limit of $f$. Hence for every $x$ there is a neighborhood $U_x$ such that $f^{-1}[U_x]\notin\mathcal F$. By compactness, there is an finite subcover of $\{U_x; x\in X\}$.

Let us denote the sets from this subcover by $U_1,\dots,U_n$. For each $i=1,\dots,n$ we have $f^{-1}[{U_i}]\notin\mathcal F$. Since $\mathcal F$ is ultrafilter, this is equivalent to $f^{-1}[{X\setminus U_i}]\in\mathcal F$.

Now $\bigcap_{i=1}^n (X\setminus U_i)=\emptyset$, since $U_1,\dots,U_n$ is a cover and this implies $\bigcap_{i=1}^n f^{-1}[{X\setminus U_i}]= f^{-1}[{\bigcap_{i=1}^n {X\setminus U_i}}]=\emptyset$. Consequently $\emptyset\in\mathcal F$, a contradiction.

Proof 2. Is is easy to observe that the filter given by the filterbase $f[\mathcal F]$ is an ultrafilter on $X$. Indeed, if $A\subseteq X$, then $f^{-1}[A] \cup f^{-1}[X\setminus A]=M$, hence one the sets $f^{-1}[A]$, $f^{-1}[X\setminus A]$ belongs to $\mathcal F$ and thus one of the sets $A$, $X\setminus A$ is in $f[\mathcal F]$. Since $X$ is compact and $f[\mathcal F]$ is an ultrafilter, there is a limit $x$ of $f[\mathcal F]$ in $X$. Then $x=\mathcal F-\lim f$.

Since this question appeared twice in comments, it might be good to add this information to the answer. (For some people this can be useful approach - depending on your background. Or it can work other way too - if you already know something about $\mathcal F$-limits, this might help you when you learn about Stone-Čech compactification.)

The $\mathcal F$-limit is related to Stone-Čech compactification in a very natural way. Let us work with Stone-Čech compactification $\beta M$ of $M$ endowed with the discrete topology. One of possibilities how to construct $\beta M$ is to define $\beta M$ to be the set of all ultrafilters on $M$ and endow it with the topology generated by the sets $A^*=\{\mathcal F\in\beta M; A\in\mathcal F\}$, where $A\subseteq M$. It can be shown that the map which assigns to a point $m\in M$ the corresponding principal ultrafilter is an embedding an that this topological space fulfills all conditions from the definition of Stone-Čech compactification.

Now for any function $f:M \to X$ where $X$ is compact we have unique extension to $\overline f: \beta M \to X$. The $\mathcal F$-limits can be understood as the values of this extension: For any ultrafilter $\mathcal F\in\beta M$ we have $$\overline f(\mathcal F)=\mathcal F-\lim f.$$

Let me add something about another thing that was addressed in the comments below. The $\mathcal F$-limit defined in the way described in this post generalizes both the notion of limit of a net and limit of a filter in the way it is usually defined in general topology. (This is Bourbaki's approach - they define this rather general notion first and various notions of limits are special cases. I do not claim that this approach would be good for students who see the nets or filters for the first time. But for someone, who is already familiar with both of them, it might be interesting to know about a unifying approach.)

Namely if $X$ is a topological space if we take the identity map $id_X \colon X\to X$, then $\mathcal F-\lim id_X$ is the same thing as the same thing as the usual definition of a limit of a filter. For net on a directed set $(D,\le)$ we can take section filter generated by the set $D_a=\{d\in D; d\ge a\}$, where $a\in D$.

Thanks for the information and the answer. I usually think in terms of nets, so I fail to see any substantial difference in the two concepts (am I being sloppy and miss something crucial?), but the argument becomes quite elegant in this language. The way I see it: you extend the map $f: M \to X$ to the Stone-Cech compactification of $M$, using its universal property, no? – t.b. Jul 14 '11 at 19:24
@Theo: I think the most important difference is language. Ultralimits can be used to show the existence of finitely additive measure on N extending asymptotic density (Hrbacek-Jech, Theorem 2.9; or this post math.stackexchange.com/questions/35450/… ) or the existence of Banach limits planetmath.org/encyclopedia/… We should be able to work out both constructions using an ultrfilter on $[0,1]$ or $\mathbb R$; but this way it seems to be more elegant and brings more insight. – Martin Sleziak Jul 14 '11 at 19:34