Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The Stone–Čech Compactification of $\mathbb{N}$ as a discrete space has been extensively studied and can be represented using ultrafilters.

Consider $X=(\mathbb{Z},\mathcal{T})$, where $\mathcal{T}$ is the Fürstenberg topology generated by arithmetic sequences. Equipped with this exotic topology, $X$ is a topological ring, metrizable, and totally disconnected.

Since $X$ is metrizable, it is Tychonoff and the map from $X$ to its image in $\beta X$ (its compactification) is a homeomorphism.

Has $\beta X$ been studied? Is there a straightforward description analogous to the compactification of $\mathbb{N}$ with the discrete topology?

share|improve this question
The Fürstenberg topology is just the profinite topology, by the way. It's not so exotic. (That means that one compactification is the profinite completion $\hat{\mathbb{Z}}$. I don't know if this is the Stone-Čech compactification.) –  Qiaochu Yuan Mar 27 '13 at 4:25
The dense embedding $\mathbb{Z} \to \hat{\mathbb{Z}}$ induces a surjective continuous map $\beta \mathbb{Z} \to \hat{\mathbb{Z}}$. Is it also injective? –  Martin Brandenburg Mar 27 '13 at 14:28
@Francois, see Joseph van Name's cardinality argument –  Yemon Choi Mar 27 '13 at 16:40
@Jackson: I think that you are identifying the Furstenberg/profinite compactification with the Stone-Cech and they are not the same –  Yemon Choi Mar 27 '13 at 16:41
I'm just joining the conversation here and I must say that I am very confused since it appears that the notation $\beta \mathbb Z$ is being used by some to denote the Stone-Cech compactification of $\mathbb Z$ with the discrete topology, and by others to denote the Stone-Cech compactification with respect to the topology $\mathcal T$. Perhaps writing $\beta (\mathbb Z, \mathcal T)$ to denote the latter would help? –  Jesse Peterson Mar 27 '13 at 18:33
show 7 more comments

2 Answers

up vote 9 down vote accepted

We can also describe $\beta\mathbb{Z}$ in terms of ultrafilters on Boolean algebras. I claim that $\beta\mathbb{Z}$ is the space of ultrafilters on the Boolean algebra of clopen sets in $\mathbb{Z}$ when $\mathbb{Z}$ has the Fürstenberg topology.

Recall that a space $X$ is zero-dimensional if it has a basis of clopen sets, and recall that a zero set on a space $X$ is a set of the form $f^{-1}(0)$ for some continuous $f:X\rightarrow\mathbb{R}$. A completely regular space $X$ is said to be strongly zero-dimensional if the Stone-Cech compactification $\beta X$ is zero-dimensional. It can be shown that a completely regular space $X$ is strongly zero-dimensional if and only if whenever $Z_{1},Z_{2}\subseteq X$ are disjoint zero sets, there is a clopen set $C\subseteq X$ with $Z_{1}\subseteq C,Z_{2}\subseteq C^{c}$ [1 p. 85]. In other words, a completely regular space is strongly zero-dimensional iff every pair of zero sets is separated by a clopen set. If $X$ is zero-dimensional, then let $\mathfrak{B}(X)$ denote the Boolean algebra of clopen subsets of $X$ and let $\zeta X$ be the space of ultrafilters on $\mathfrak{B}(X)$. Then $\zeta X$ is in a sense the maximal zero-dimensional compactification of $X$ which is called the Banaschewski compactification. If $X$ is strongly zero-dimensional, then the Banaschewski compactification $\zeta X$ is precisely the Stone-Cech compactification. In [1. p. 86] it states that zero-dimensionality and strong zero-dimensionality are equivalent in Lindelof spaces. Therefore since $\mathbb{Z}$ is zero-dimensional and Lindelof, the space $\mathbb{Z}$ is strongly zero-dimensional. We conclude that $\beta\mathbb{Z}=\zeta\mathbb{Z}$ is the space of ultrafilters on $\mathfrak{B}(\mathbb{Z})$.

In order to clear up some confusion about the space $\mathbb{Z}$ and its Stone-Cech compactification, I will outline some basic facts about $\mathbb{Z}$ and $\beta\mathbb{Z}$.

I claim that the space $\mathbb{Z}$ has an infinite partition into clopen sets. It is not too hard to give an explicit example of such a partition. For a more slick proof, assume that $\mathbb{Z}$ has no partition into countably many clopen sets. If $\mathcal{U}$ is an open cover of $\mathbb{Z}$, then there is a clopen cover $\{C_{n}|n\in\mathbb{N}\}$ that refines $\mathcal{U}$. If we set $D_{n}=C_{n}\setminus(C_{0}\cup...\cup C_{n-1})$ for all $n$, then $\{D_{n}|n\in\mathbb{N}\}$ is a partition of $\mathbb{Z}$ into finitely many clopen sets that refines $\mathcal{U}$, so $\mathbb{Z}$ is compact. This is a contradiction. Therefore $\mathbb{Z}$ has a partition into countably many clopen sets.

In particular, there is a continuous surjective function $f:\mathbb{Z}\rightarrow\mathbb{N}$ where $\mathbb{N}$ has the discrete topology. Therefore the map $f$ extends to a continuous surjective function $\bar{f}:\beta\mathbb{Z}\rightarrow\beta\mathbb{N}$. Since $|\beta\mathbb{N}|=2^{\mathbb{c}}$, we conclude that $|\beta\mathbb{Z}|=2^{\mathbb{c}}$ as well. We conclude that the Stone-Cech compactification $\beta\mathbb{Z}$ is much larger than the pro-finite completion of $\mathbb{Z}$.

[1] The Stone-Cech Compactification, Russell Walker (1970)

share|improve this answer
Excellent! Incidentally, I think this shows that $\beta\mathbb{Z}$ is $\widehat{\mathbb{Z}}$ since (if my computations are correct) the clopen sets in the Fürstenberg topology are precisely the finite unions of arithmetic progressions. Thus, an ultrafilter $p$ on the clopen algebra is such that there is exactly one arithmetic progression $a_p(N) + \mathbb{Z}N \in p$ for each modulus $N$. These must be coherent and thus determine a point $\hat{p} \in \widehat{\mathbb{Z}}$. The map $p \mapsto \hat{p}$ is a homeomorphism. –  François G. Dorais Mar 27 '13 at 14:44
... because the arithmetic progressions are dense in the clopen algebra. So the clopen sets that contain the arithmetic progressions $q + \mathbb{Z}N$ where $q \equiv \hat{p} \pmod{N}$ form an ultrafilter, which has to be $p$. –  François G. Dorais Mar 27 '13 at 14:50
That's great! But can we please stop calling it the Furstenberg topology? –  HJRW Mar 27 '13 at 15:03
@Francois, as I commented above, I don't see how beta Z can be the profinite completion if the profinite competion is a group. See the discussion I linked to in my comment –  Yemon Choi Mar 27 '13 at 16:21
I think there is an infinite partition of $\mathbb{Z}$ into clopen sets. If not, then $\mathbb{Z}$ would be pseudocompact by my old question mathoverflow.net/questions/103543/…. Therefore $\mathbb{Z}$ would be compact since it is realcompact and pseudocompact. This is a contradiction. Therefore $\mathbb{Z}$ has an infinite partition into clopen sets. –  Joseph Van Name Mar 27 '13 at 16:28
show 11 more comments

Edit. The comments and the other answers reveal that my proof has some gap. But I won't delete it, instead I've rewritten it as an attempt to prove $\beta \mathbb{Z} = \hat{\mathbb{Z}}$. I hope that the failure of this naive proof motivates to read the more sophisticated answers.

When $\mathbb{Z}$ is equipped with the Fürstenberg topology, do we have $\beta \mathbb{Z} = \hat{\mathbb{Z}}$?

The Fürstenberg topology is the subspace topology induced by the profinite completion $\hat{\mathbb{Z}} = \lim_{n>0} \mathbb{Z}/n$. The embedding $\mathbb{Z} \to \hat{\mathbb{Z}}$ is dense, hence for every compact Hausdorff space $X$ we get an injective map $\hom(\hat{\mathbb{Z}},X) \to \hom(\mathbb{Z},X)$. The question is whether it is surjective, because this would mean that $ \hat{\mathbb{Z}}$ satisfies the defining universal property of $\beta \mathbb{Z}$.

Let $f : \mathbb{Z} \to X$ be a continuous map. This means that for every $a \in \mathbb{Z}$, every open subset $U \subseteq X$ containing $f(a)$ already contains $f(a+n \mathbb{Z})$ for some $n>0$. Let $a=(a_1,a_2,\dotsc) \in \hat{\mathbb{Z}}$, i.e. $a_n \equiv a_m \bmod n$ for $n|m$. Since $X$ is compact, the net $(f(a_n))_{n>0}$ (using divisibility for the indices) has a convergent subnet, say $(f(a_{n(i)}))_{i \in I} \longrightarrow x$.

Actually any two subnets have the same limit: Assume that $(f(a_{m(j)}))_{j \in J} \longrightarrow y$. Choose open neighborhoods $U,V$ of $x,y$, it is enough to prove $U \cap V \neq \emptyset$ since $X$ is Hausdorff. For large $i$ we have that $f(a_{n(i)}) \in U$, and for large $j$ we have $f(a_{m(j)}) \in V$. Choose $b>0$ with $f(a_{n(i)} + b \mathbb{Z}) \subseteq U$ and $f(a_{m(j)} + b \mathbb{Z}) \subseteq V$. We may assume $n(i),m(j)|b$. For $p=n(i) m(j)$ we have $a_p \equiv a_{n(i)} \bmod n(i)$, hence $a_p \equiv a_{n(i)} \bmod b$. Similarily we get $a_p \equiv a_{m(j)} \bmod b$. Hence $f(a_p) \in U \cap V$.

Hence $\tilde{f}(a) := $(the limit of some subnet of $f(a_n)$) gives a well-defined map $\hat{\mathbb{Z}} \to X$. Clearly it agrees with $f$ on constant sequences. But now the problem seems to be that $\tilde{f}$ is not continuous ...

share|improve this answer
$|\beta\mathbb{Z}|=2^{2^{\aleph_{0}}}$ and clearly $|\hat{\mathbb{Z}}|$ has cardinality continuum. –  Joseph Van Name Mar 27 '13 at 16:34
Martin, doesn't your argument only work if when "let $f: Z \to X$ be continuous" you are equipping Z with the subspace topology from its profinite completion, hence you're just showing that a continuous function on a dense subset had a unique continuous extension? (You don't use compactness of range) –  Yemon Choi Mar 27 '13 at 16:39
(OK, you do use compactness of range - wish I could edit comments! - but my first point still stands) –  Yemon Choi Mar 27 '13 at 16:47
The problem is that subnets are terrible to work with. Subnets should be avoided whenever possible. I have never found a legitimate use for subnets. –  Joseph Van Name Mar 27 '13 at 16:52
I think you all refer to $\beta \mathbb{Z}$ when $\mathbb{Z}$ is equipped with the discrete topology. But of course here it is equipped with the Fürstenberg topology, as in the question. –  Martin Brandenburg Mar 27 '13 at 17:07
show 4 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.