# Stone-Cech compatification and ultrafilter [closed]

I have been studding about compatification of a topological space $X$. But I have low understanding about the Stone-Cech compatification, specially construction of the Stone-Cech compatification on natural numbers.(I mean $X = \beta\omega$). I have some difficulty with construction of $\beta\omega$ and $\beta\omega - \omega$.

1) : Can you give me brief and to the point explain about $\beta\omega$ and it`s members?

2) : What kind of ultrafilter does put on $\beta\omega$?

3) How are the members of $\beta\omega - \omega$? Which relationship is there between a ultrafilter and members of $\beta\omega - \omega$?

Thanks.

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## closed as off-topic by Joseph Van Name, Ramiro de la Vega, Bill Johnson, Daniel Moskovich, Chris GodsilAug 25 '13 at 14:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Joseph Van Name, Ramiro de la Vega, Bill Johnson, Daniel Moskovich, Chris Godsil
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One should probably consult the book The Stone-Cech Compactification by Russell Walker, The Theory of Ultrafilters by Comfort and Negrepontis, or Rings of Continuous Functions by Gillman and Jerison for such information. –  Joseph Van Name Aug 25 '13 at 12:03
The wikipedia article [Stone–Čech_compactification] (en.wikipedia.org/wiki/Stone–Cech_compactification) contains all the information you need. –  J.-E. Pin Aug 25 '13 at 16:24
We used to say, on the old MO, that "MO is not for requests for people to write an encyclopaedia article". This seems to apply here. If you want to learn the basics of the Stone-Cech compactification, then you should consult a book, or Wikipedia, or a book mentioned on Wikipedia –  Yemon Choi Aug 25 '13 at 16:27
If you can be more specific, perhaps you could make a suitable question for math.SE from this. Asking not so specific question seems to be more suitable for chat. Over there at math.SE we have General Topology chatroom which is rather inactive these days. However, you need at least 20 reputation points to talk in chat. I am not sure whether you need 20 points on math.SE or 20 points on any site would suffice. –  Martin Sleziak Nov 18 at 13:40
You can also have a look at these notes of mine which include proofs of some facts about ultrafiter construction of $\beta\mathbb N$. I would say the proofs there are quite detailed. –  Martin Sleziak Nov 18 at 13:43

## 1 Answer

You are asking what kind of ultrafilter belongs to $\beta\omega$, and the answer is all ultrafilters. You can think of $\beta\omega$ as the set of all ultrafilters on $\omega$, where the topology is defined by letting $\bar{A}=\{p\in\beta\omega\big|A\in p\}$ be a basic open set for every $A\subseteq\omega$. In this way you obtain a compact Hausdorff space with many nice features. Now, in order for $\beta\omega$ to be a compactification of $\omega$, you need a dense embedding of the latter into the former. This is achieved by letting $\iota:\omega\longrightarrow\beta\omega$ be given by $\iota(n)=\{A\subseteq\omega\big|n\in A\}$, i.e. we identify each $n\in\omega$ with the principal ultrafilter on $n$.

Now, as for $\beta\omega\setminus\omega$, it consists of all nonprincipal ultrafilters, since, as I said before, $\omega$ is identified with the set of principal ultrafilters. This is, ultrafilters $p$ belong to $\beta\omega\setminus\omega$ iff they extend the Fréchet filter $\{A\subseteq\omega\big||\omega\setminus A|<\omega\}$ iff $\cap p=\varnothing$. I hope this helps you get some intuition on $\beta\omega$ and $\beta\omega\setminus\omega$.

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please give me more explain. why can we say the set $\omega$ corresponds to the set of principal ultrafilters, and the set $\beta\omega - \omega$ to the set of free ultrafiltrates? –  maryam Aug 25 '13 at 14:08
(1): MAre all point of $\beta\omega$ isolated? why?(2): is it true that $\omega$ is open in $\beta\omega$ , so $\beta\omega - \omega$ is compact? –  maryam Aug 25 '13 at 17:29
Hi. Sorry for the late reply, but I hadn't logged into MathOverflow in a while (writing dissertation, hehe). Given the way the topology in $\beta\omega$ is defined, for each $n\in\omega$ the set $\{p\in\beta\omega\big|\{n\}\in p\}$ is a basic open set... but this basic open set is just the singleton $\{\iota(n)\}$. So every point of (what we chose to identify with) $\omega$ is open in $\beta\omega$. Since $\omega$ is just the union of those singletons, then it is also the case that $\omega$ is open in $\beta\omega$. Thus, $\omega^*=\beta\omega\setminus\omega$ is compact. –  David FernandezBreton Sep 24 '13 at 16:07