2013-07-30 11:17:40 8 Comments

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

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## 4 comments

## @Joseph Van Name 2013-07-30 19:49:29

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 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 $B_{n}=A_{n}\setminus(A_{0}\cup...\cup A_{n-1})$, 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 $\mathcal{B}=\{\mathcal{V}\in\beta X|B\in\mathcal{V}\}$, 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.

## @YCor 2013-07-30 21:47:24

very nice! (of course you mean "

injectivesubsequence")## @maryam 2013-08-25 13:57:08

(1): What you mean about" If we let $B_{n}=A_{n}\setminus(A_{0}\cup...\cup A_{n-1})$, 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"?

## @Joseph Van Name 2013-08-25 17:47:50

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.

## @Joseph Van Name 2013-07-30 15:17:15

This result follows from much stronger results from general topology. These results can be found in [1].

$\phantom{Here is a secret message.}$

As a result of the above corollary, since every discrete space of non-measurable cardinality is realcompact, if $A$ is a discrete space of non-measurable 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 non-measurable 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 non-trivial 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

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]

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 non-trivial 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 Stone-Cech Compactification. Berlin: Springer-Verlag, 1974.

## @maryam 2013-07-31 05:33:45

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?

## @Joseph Van Name 2013-07-31 16:18:35

Every compact F-space is a minimal-KC space with no non-trivial convergent sequence. Under Stone duality compact zero-dimensional F-spaces 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 σ-complete Boolean algebra satisfies the countable separation property, and the countable separation property is closed under taking quotients.

## @YCor 2013-07-30 13:24:23

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 : 583-585, 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 Koppelberg-Tits 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 Koppelberg-Tits argument is given in Proposition 4.4 here, but some better references are hopefully available.

## @maryam 2013-07-30 19:15:45

what is your references? I have read Engelking, but it is not clear to me.

## @Joseph Van Name 2013-07-30 20:36:35

@maryam. You should read about Stone-duality 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 Stone-Cech 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.

## @André Henriques 2013-07-30 11:35:06

The Stone-Čech 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.

## @Ramiro de la Vega 2013-07-30 11:39:56

I guess you are showing there are no convergent sequences of

principalultrafilters.## @André Henriques 2013-07-30 11:40:28

You're right. I'll edit my answer.

## @André Henriques 2013-07-30 12:25:40

Hmm... I guess I don't know how to do this.