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.
Basic facts about ultrafilters and convergence of a sequence along an ultrafilter
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0Since you ask for a popular form, I think all this is in Jänich's *[Topology](http://books.google.com/books?id=vg1npIhAOQoC)* 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. – 2011-07-14
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0heres a little treatment of ultrafilters http://terrytao.wordpress.com/2007/06/25/ultrafilters-nonstandard-analysis-and-epsilon-management/ – 2011-07-14
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0@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. – 2011-07-14
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0@yoyo: thanks a lot for the reference. – 2011-07-14
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3Here are two links I found useful: David MacIver's [Filters in Topology](http://www.efnet-math.org/~david/mathematics/filters.pdf) and Pete L. Clark's [notes on convergence](http://math.uga.edu/~pete/convergence.pdf) – 2011-07-21
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0@liman I've added a few more references to my post and also a link to my notes. (This is still not the final version and I hope I'll get later to finishing them, but you might find there some things that are related to your question and might be interesting for you.) – 2012-06-21
1 Answers
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 $x=\mathcal F-\lim f$.
- [D] Jacques Dixmier. General Topology. Springer-Verlag, New York, 1984. Undergraduate Texts in Mathematics
- [HS] Hindman, Strauss: Algebra in the Stone-Čech compactification, Walter de Gruyter, Berlin-New York, 1998.
- [F] D. H. Fremlin. Measure theory, Volume 2: Broad Foundations. Torres Fremlin, Essex, 2001.
- [T] Stevo Todorcevic. Topics in Topology. Springer-Verlag, Berlin{Heidelberg, 1997. Lecture Notes in Mathematics 1652.
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 $a\in D$.
More details about this can be found again in my notes here.
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1Thanks 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? – 2011-07-14
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1@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 http://math.stackexchange.com/questions/35450/applications-of-ultrafilters/35461#35461 ) or the existence of Banach limits http://planetmath.org/encyclopedia/ConstructionOfBanachLimitUsingLimitAlongAnUltrafilter.html 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. – 2011-07-14
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0The relationship to the usual limit of sequence is more clear this way. I also like the way the notion of F-limit in my answer generalizes both convergence of nets and convergence of filters in topological spaces. – 2011-07-14
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1BTW this is what Hindman and Strauss about this notion: "As we shall see, the notion is as versatile as the notion of nets, and has two significant advantages: (1) in a compact space a p-limit always converges and (2 ) it provides a "uniform" way of taking limits, as opposed to randomly chosing from among possible limit points of a net." http://books.google.com/books?id=KYXgdiegKDsC&pg=PA63&dq=%22versatile+as+the+notion+of+nets,+and%22&hl=en&ei=KUYfTqazO8bo-gaC1-mfAw&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCgQ6AEwAA#v=onepage&q=%22versatile%20as%20the%20notion%20of%20nets%2C%20and%22 – 2011-07-14
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0Again, thanks a lot for the clarification. I'll have to think a bit before I come to the point to appreciate and understand this way of looking at it fully, but it sure looks nice and your proof definitely *is* very elegant! – 2011-07-14
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0Somehow I overlooked your comment about Stone-Čech compactification and addressed only the first part. Of course, you are perfactly right, one way to look at this notion is to work with $\beta M$ (where $M$ has discrete topology). – 2011-07-14
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1@Theo: Ultralimits can also give nice standard translations of proofs using techniques from non-standard analysis; an example can be seen in John J. Buoni, Albert Klein, Brian M. Scott, & Bhushan J. Wadhwa, *On Power Compactness in a Banach Space*, Indiana Univ. Math. J., Vol. 32, No. 2 (1983), pp. 177-185. – 2011-07-14
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1@Martin, many thanks to you and to all! It would also be nice to see something illustrating the existence of (free?) ultrafilters assigning a limit to every bounded sequence, if possible. – 2011-07-14
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0@liman: I've added some information concerning the existence of ultrafilters and limits of bounded real sequences. Zorn's lemma implies that every system with finite intersection property is contained in some ultrafilter. You can find many notes and books providing the proof of this fact. Ultralimits of bounded real sequences can be regarded as a special cas of the above result on compact spaces. – 2011-07-14
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0@Martin: Thank you so much! I have got a lot of information, now I need to work hard :) – 2011-07-14
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0@Brian Thank you very much for this reference! @Martin, thank you, again, this was really illuminating. – 2011-07-14
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1Here's a link http://dx.doi.org/10.1512/iumj.1983.32.32015 to the paper mentioned by @Brian. (I guess I was not the only one interested enough to look it up.) – 2011-07-14
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0I have got the impression that ultrafilters are points of the Stone-Czech compactification of the natural numbers. Is this correct? – 2011-07-15
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0@liman Yes, ultrafilters on N can be considered as the points of Stone-Cech compactification of the countable discrete space. With this approach, the F-limit of a sequence $x_n$ of points of a compact space is exactly the value of the unique extencion $\overline x:\beta\mathbb{N}\to X$ at the point $F$, i.e. $\overline x(F)=F-\lim x$. – 2011-07-15
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0@Martin: Do I understand correctly that some ultrafilters are translation-invariant ($\mu E=\mu(n+E)$ for any natural $n$ and for any $E\subset\mathbb N$), and some are not? Many thanks for your nice comments! – 2011-07-19
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0@liman: I will quote from Komjath-Totik, p. 342: "There is no translation-invariant ultrafilter on $\omega$ as exactly one of the sets of the odd, resp. even numbers is in any ultrafilter." However, there exist translation invariant filters. (E.g. cofinite sets.) – 2011-07-19
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0@Martin: (if you read this) Each ultrafilter determines a bounded linear functional on $l^\infty$; can one say that these functionals are something like the extreme points of the unit ball, or what? – 2011-08-03
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0Well, I seem to have found the answer myself. If we think of the functionals over $l^\infty$ as measures on the Stone-Cech compactification, the ultrafilters are then the $\delta$-measures, and so on. – 2011-08-03
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0@liman: Yes, it is correct, they are precisely the extreme points of unit ball of $\ell_\infty^*$. And you are right that one way to show this is to work with $C^*(\beta\mathbb N)$ instead. Extreme points of unit ball of $C^*(K)$ for *K* compact are described in Lemma 3.116 in [this book](http://books.google.com/books?id=5BDX2NNsqR4C&pg=PA373&&q=%22Lemma%203.116%22&f=false) or in Lemma 3.42 in [this book](http://books.google.com/books?id=Hcqm4_lW4EkC&printsec=frontcover&dq=fabian+functional&hl=en&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCgQ6AEwAA#v=snippet&q=%22lemma%203.42%22). – 2011-08-03
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0@Martin: this means that the ultrafilters can be defined without the Zorn lemma. I don't see where it could be used implicitly, am I wrong? – 2011-08-03
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0@liman: Existence of *free* ultrafilters cannot be shown in ZF, I do not quite follow how you want to derive them from the above. Krein-Milman theorem needs some form of choice (see here http://mathoverflow.net/questions/15654/extreme-point-compact-convex-set or at wikipedia), I guess the existence of extreme points does not hold in ZF either.\\ Perhaps you could post this as a separate question - I think several math.SE users can are experts in question concerning AC. – 2011-08-04
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0@Martin: I am sorry, what is _ZF_? – 2011-08-04
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0@liman: By ZF I mean Zermelo-Fraenkel set theory (i.e., the usual axioms with the omitting the axiom of choice). – 2011-08-04
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1@Martin: I meant that we can define the ultrafilters formally as extreme points of the unit ball of the set of all functionals on $l^\infty$ with the property that they take constant functions to the same constant. Probably, the Zorn lemma will appear when we prove that this definition is compatible with the standard one. – 2011-08-04
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0@liman: I do believe that Zorn lemma (or Axiom of Choice) will appear in the proof that such functionals exist. At least some weak form of AC is needed. A model of ZF, where no free ultrafilters exist is, as far as I know, due to [Pincus and Soloway](http://projecteuclid.org/euclid.jsl/1183739943). This [paper](http://dx.doi.org/10.1016/S0019-3577(98)80039-6) might be of interset too. – 2011-08-04
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0As I've said earlier, several people here have good knowledge about ZF and AC, so you might get a better answer if you posted this as a question. In fact, some such questions have already been asked: http://mathoverflow.net/questions/59157/reference-request-independence-of-the-ultrafilter-lemma-from-zf http://mathoverflow.net/questions/15872/non-principal-ultrafilters-on – 2011-08-04
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0@Martin: thanks a lot, I will first try to understand what I need and then, if necessary, post a new question. Once again, many thanks! – 2011-08-04
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0@MartinSleziak thank you so much!!! these notes are awesome!!! – 2013-04-07
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0Amazing answer very helpful! – 2013-06-26