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?

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    $\begingroup$ 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.) $\endgroup$ – Qiaochu Yuan Mar 27 '13 at 4:25
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    $\begingroup$ Also, isn't the profinite completion going to be a group (which Stone-Cech isn't)? $\endgroup$ – Yemon Choi Mar 27 '13 at 4:35
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    $\begingroup$ 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? $\endgroup$ – Martin Brandenburg Mar 27 '13 at 14:28
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    $\begingroup$ @Francois, see Joseph van Name's cardinality argument $\endgroup$ – Yemon Choi Mar 27 '13 at 16:40
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    $\begingroup$ 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? $\endgroup$ – Jesse Peterson Mar 27 '13 at 18:33

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)

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    $\begingroup$ 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. $\endgroup$ – François G. Dorais Mar 27 '13 at 14:44
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    $\begingroup$ ... 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 $p$. $\endgroup$ – François G. Dorais Mar 27 '13 at 14:50
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    $\begingroup$ That's great! But can we please stop calling it the Furstenberg topology? $\endgroup$ – HJRW Mar 27 '13 at 15:03
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    $\begingroup$ @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 $\endgroup$ – Yemon Choi Mar 27 '13 at 16:21
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    $\begingroup$ 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. $\endgroup$ – Joseph Van Name Mar 27 '13 at 16:28

Topologically speaking $\mathbb{Z}$ with the topology mentioned above is just (homeomorphic to) the space of rational numbers. The space $\beta\mathbb{Q}$ has been studied a lot (not as much as $\beta\mathbb{N}$), for example by Eric van Douwen in Remote Points


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 ...

  • $\begingroup$ $|\beta\mathbb{Z}|=2^{2^{\aleph_{0}}}$ and clearly $|\hat{\mathbb{Z}}|$ has cardinality continuum. $\endgroup$ – Joseph Van Name Mar 27 '13 at 16:34
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    $\begingroup$ 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) $\endgroup$ – Yemon Choi Mar 27 '13 at 16:39
  • $\begingroup$ (OK, you do use compactness of range - wish I could edit comments! - but my first point still stands) $\endgroup$ – Yemon Choi Mar 27 '13 at 16:47
  • $\begingroup$ 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. $\endgroup$ – Joseph Van Name Mar 27 '13 at 16:52
  • $\begingroup$ 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. $\endgroup$ – Martin Brandenburg Mar 27 '13 at 17:07

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