Why are there only trivial convergent sequences in the StoneČech compactification of $\mathbb{N}$?

$\begingroup$ I am willing to edit the question for formatting, if someone can please explain what it is asking. What is a KC space? $\endgroup$– Andrej BauerCommented Jul 30, 2013 at 12:08

$\begingroup$ maryam: See also this question at Mathematics(StoneCech compactifications and limits of sequences). @AndrejBauer The definition of KCspace is given in other questions by the same OP as the space where every compact space is closed. But this definition seems to be irrelevant to the question. I guess it asks why there are no nontrivial convergent sequences in $\beta\omega$. (The poster of the only answer given so far understood the question in the same way.) Maybe the OP wants a proof which somehow uses that it is a minimal KCspace? $\endgroup$– Martin SleziakCommented Jul 30, 2013 at 12:14

5$\begingroup$ After the edits, not much remains of the question. $\endgroup$– Andrej BauerCommented Jul 30, 2013 at 13:00

$\begingroup$ I see that somehow I managed to misstype the comment. I wanted to post this link: math.stackexchange.com/questions/35310/… $\endgroup$– Martin SleziakCommented Jul 31, 2013 at 5:49
4 Answers
Let me give a fairly direct proof. Assume $R=(\mathcal{U}_{n})_{n}$ is a sequence of distinct ultrafilters on some set $X$. Since every infinite Hausdorff space has an infinite discrete subspace, there is a subsequence $(\mathcal{V}_{n})_{n}$ of $(\mathcal{U}_{n})_{n}$ such that $\{\mathcal{V}_{n}n\in\mathbb{N}\}$ is a discrete subspace of $\beta X$. In particular, there is a sequence $(A_{n})_{n}$ of sets with $A_{n}\in\mathcal{V}_{m}$ if and only if $m=n$. If we let $B_{n}=A_{n}\setminus(A_{0}\cup...\cup A_{n1})$, then the sequence $(B_{n})_{n}$ is pairwise disjoint and $B_{n}\in\mathcal{V}_{m}$ iff $m=n$. If we let $B=\bigcup_{n}B_{2n}$, then $B\in\mathcal{V}_{m}$ if and only if $m$ is even. In other words, if $\mathcal{B}=\{\mathcal{V}\in\beta XB\in\mathcal{V}\}$, then $\mathcal{B}$ is a clopen set with $\mathcal{V}_{m}\in\mathcal{B}$ iff $m$ is even. Therefore, we conclude that the sequence $(\mathcal{V}_{m})_{m}$ cannot converge to any point, so the sequence $(\mathcal{U}_{n})_{n}$ cannot converge to any point either.

1$\begingroup$ very nice! (of course you mean "injective subsequence") $\endgroup$– YCorCommented Jul 30, 2013 at 21:47

$\begingroup$ (1): What you mean about" If we let $B_{n}=A_{n}\setminus(A_{0}\cup...\cup A_{n1})$, then the sequence $(B_{n})_{n}$ is pairwise disjoint and $B_{n}\in\mathcal{V}_{m}$ iff $m=n$. If we let $B=\bigcup_{n}B_{2n}$, then $B\in\mathcal{V}_{m}$ if and only if $m$ is even." ?(2): What does it mean " clopen set"? $\endgroup$– maryamCommented Aug 25, 2013 at 13:57

1$\begingroup$ Maryam. Read about ultrafilters and Stone duality from the books that I mentioned in other comments and these basic questions will have a clear answer. $\endgroup$ Commented Aug 25, 2013 at 17:47
This result follows from much stronger results from general topology. These results can be found in [1].
$\mathbf{Theorem}$ Each nondiscrete closed subset of $\beta X\setminus\upsilon X$ contains a copy of $\beta\mathbb{N}$ (and in particular, its cardinality is at least $2^{2^{\aleph_{0}}}$).
$\phantom{Here is a secret message.}$
$\mathbf{Corollary}$ If $X$ is locally compact and realcompact, then every infinite closed set in $\beta X\setminus X$ contains a copy of $\beta\mathbb{N}$ (thus, its cardinality is at least $2^{2^{\aleph_{0}}}$).
As a result of the above corollary, since every discrete space of nonmeasurable cardinality is realcompact, if $A$ is a discrete space of nonmeasurable cardinality, then every closed set in $\beta A\setminus A$ contains a copy of $\beta\mathbb{N}$.
Now assume that $A$ is a set of nonmeasurable cardinality and $(x_{n})_{n}$ is a convergent sequence in $\beta A$ that converges to some point $x\in\beta A$. Take a subsequence $(y_{n})_{n}$ such that $\{y_{n}n\in\mathbb{N}\}\subseteq A$ or $\{y_{n}n\in\mathbb{N}\}\cap A=\emptyset$. If $\{y_{n}n\in\mathbb{N}\}\subseteq A$, and $\{y_{n}n\in\mathbb{N}\}$ takes infinitely many values, then take a subsequence $(z_{n})_{n}$ where each $z_{n}$ is distinct. Let $f:A\rightarrow[0,1]$ be a function where $z_{n}=0$ whenever $n$ is even and $z_{n}=1$ whenever $n$ is odd. Then $f$ extends to a continuous $\overline{f}:A\rightarrow[0,1]$. In particular, $\overline{f}(z_{n})\rightarrow f(x)$. This is a contradiction since $\overline{f}(z_{n})$ oscillates between $0$ and $1$ endlessly. Therefore $\{y_{n}n\in\mathbb{N}\}$ can only take finitely many values, so $(y_{n})_{n}$ is eventually constant.
On the other hand, if $(y_{n})_{n}\subseteq\beta A\setminus A$, then since $\beta A\setminus A$ is closed, we have $x\in\beta A\setminus A$ as well. Therefore $\{y_{n}n\in\mathbb{N}\}\cup\{x\}$ is a closed subset of $A$ of cardinality less than $2^{2^{\aleph_{0}}}$. Therefore, by the above corollary, the set $\{y_{n}n\in\mathbb{N}\}\cup\{x\}$ is finite, so the sequence $(y_{n})_{n}$ must be eventually constant. QED
$\textbf{Added later}$ Using the notion of an $F$space, we have other results that prove that $\beta\mathbb{N}$ has no nontrivial convergent sequence. If $X$ is a completely regular space, then a subset $Z\subseteq X$ is said to be a cozero set if there is a function $f:X\rightarrow[0,1]$ with $Z=f^{1}(0,1]$. A subset $Z$ of a completely regular space $X$ is said to be $C^{*}$embedded if every bounded continuous map $f:Z\rightarrow\mathbb{R}$ extends to a bounded continuous function $g:X\rightarrow\mathbb{R}$. A completely regular space is said to be an $F$space if every cozero set is $C^{*}$embedded. There are many characterizations of $F$spaces and these characterizations can be found in [1]. The following result lists some of these characterizations
$\mathbf{Theorem}$. Let $X$ be a completely regular space. Then the following are equivalent.
$X$ is an $F$space.
If $U,V$ are disjoint cozero sets in $X$, then there is a continuous function $f:X\rightarrow[0,1]$ with $U\subseteq f^{1}[\{0\}]$ and $V\subseteq f^{1}[\{1\}]$.
$\beta X$ is an $F$space.
From the above result, we conclude that if $D$ is a discrete space, then $\beta D$ is an $F$space. The following result can be found in [2]
$\mathbf{Theorem}$ Every countable subspace of an $F$space is $C^{*}$embedded and hence, every infinite compact $F$space contains a copy of $\beta\mathbb{N}$.
From the above result, we conclude that if $X$ is an $F$space, then every sequence in $X$ is eventually constant. If $(x_{n})_{n}$ is not eventually constant, then take a subsequence $(y_{n})_{n}$ where each $y_{n}$ is distinct and where the set $Y:=\{y_{n}n\in\mathbb{N}\}$ is discrete. Let $f:Y\rightarrow [0,1]$ be the map where $y_{n}=0$ whenever $n$ is even and $y_{n}=1$ whenever $n$ is odd. Then $f$ extends to continuous map $g:X\rightarrow[0,1]$. However, the sequence $g(y_{n})$ cannot converge to anything, so $(x_{n})_{n}$ is not convergent. In particular, since $\beta D$ is an $F$space for discrete $D$, there are no nontrivial convergent sequences in $\beta D$.
[1] Gillman, Leonard, and Meyer Jerison. Rings of Continuous Functions,. Princeton, NJ: Van Nostrand, 1960.
[2] Walker, Russell C. The StoneCech Compactification. Berlin: SpringerVerlag, 1974.

$\begingroup$ In topological KC  space , every compact set is closed.a topological space( X,τ ) is called minimal KC if ( X,τ )is KC and there is no topology σ ⊂ τ such that ( X, σ ) is KC. so, X = βω is KC  minimal.is there any example except it that is minimal KC but does not have non trivial sequence? $\endgroup$– maryamCommented Jul 31, 2013 at 5:33

$\begingroup$ Every compact Fspace is a minimalKC space with no nontrivial convergent sequence. Under Stone duality compact zerodimensional Fspaces correspond to the Boolean algebras with the countable separation property. We say that a Boolean algebra B satisfies the countable separation property if whenever R,S⊆B are subsets with r∧s=0 whenever r∈R,s∈S, then there is some b∈B with r≤b,s≤b′ for r∈R,s∈S. Every σcomplete Boolean algebra satisfies the countable separation property, and the countable separation property is closed under taking quotients. $\endgroup$ Commented Jul 31, 2013 at 16:18
Here's an answer for sequences of arbitrary ultrafilters (I replace $\mathbf{N}$ by an arbitrary set $X$ since it holds in general). Let $\mathcal{U}_n$ be a sequence of ultrafilters converging to an ultrafilter $\mathcal{U}$. Define $A_n\subset 2^X$ as the set of subsets $B\subset X$ such that the sequence $(\mathbf{1}_{\mathcal{U}_k}(B))_{k\ge n}$ is constant. Then $A_n$ is a Boolean subalgebra of $2^X$, there are inclusions $A_n\subset A_{n+1}$ for all $n$ and $\bigcup A_n=2^X$. A result of Koppelberg and Tits ("Une propriété des produits directs infinis de groupes finis isomorphes". CR Math. Acad. Sci. Paris, Sér. A 279 : 583585, 1974) then implies that $A_n=2^X$ for some $n$. This means that $\mathcal{U}_k=\mathcal{U}$ for all $k\ge n$.
The KoppelbergTits result is now stated as "2^X" has cofinality $\neq\omega$ as a Boolean algebra. The above reference is in French; a simple generalization of the KoppelbergTits argument is given in Proposition 4.4 here, but some better references are hopefully available.

$\begingroup$ what is your references? I have read Engelking, but it is not clear to me. $\endgroup$– maryamCommented Jul 30, 2013 at 19:15

$\begingroup$ @maryam. You should read about Stoneduality and ultrafilters. The book A Course in Universal Algebra Chapter IV Section 4 may be a good place to start for the relation between ultrafilters on Boolean algebras and compact spaces. It turns out that most of the basic textbooks in general topology do not point out the fact that the StoneCech compactification of a discrete space is the set of ultrafilters on that space. More generally, any compactification of any completely regular space can be represented in terms of ultrafilters. $\endgroup$ Commented Jul 30, 2013 at 20:36
The StoneČech compactification is compatible with finite disjoint unions: $\beta(A\sqcup B)=\beta A\sqcup \beta B$.
If $X=(x_n)_{n\in\mathbb N}$ were a convergent sequence with limit $x\in\beta\mathbb N$, then for any partition of $\mathbb N=A\sqcup B$ such that both $A\cap X$ and $B\cap X$ are infinite, we would have $x\in \beta A$ and $x\in \beta B$, a contradiction.

$\begingroup$ I guess you are showing there are no convergent sequences of principal ultrafilters. $\endgroup$ Commented Jul 30, 2013 at 11:39

$\begingroup$ You're right. I'll edit my answer. $\endgroup$ Commented Jul 30, 2013 at 11:40

$\begingroup$ Hmm... I guess I don't know how to do this. $\endgroup$ Commented Jul 30, 2013 at 12:25