It is a non-trivial result that there is a countable connected Hausdorff space.

Let ${\cal T}$ be a set of connected Hausdorff topologies on $\omega$ such that whenever $\tau_1\neq\tau_2\in {\cal T}$ we have $(\omega,\tau_1)\not\cong (\omega,\tau_2)$. What is the largest cardinality that ${\cal T}$ can have?


It seems that the number of such topologies is $2^{\mathfrak c}$. Such (huge) number of connected Hausdorff topologies can be constructed by a suitable modification of the Bing's construction of a connected Hausdorff space.

First we introduce the necessary definitions. A subset $U$ of the real line is called regular open if $U$ coincides with the interior of the closure of $U$ in $\mathbb R$.

A filter $\mathcal F$ on $\mathbb R$ will be called

$\bullet$ (regular) open if $\mathcal F$ has a base consisting of (regular) open subsets of $\mathbb R$;

$\bullet$ vanishing if each neighborhood of zero in $\mathbb R$ belongs to $\mathcal F$.

It can be shown that the set of regular open vanishing filters on $\mathbb R$ has cardinality $2^{\mathfrak c}$.

For any open vanishing filter $\mathcal F$ we define a countable connected Hausdorff space $X_{\mathcal F}$ as follows. On the countable set $X=\{(x,y)\in\mathbb Q\times\mathbb Q:y\ge 0\}$ consider the topology $\tau_{\mathcal F}$ in which a neighborhood base at a point $(x,y)\in X$ consists of the sets $$\Delta_U=\{(x,y)\}\cup\{(t,0)\in X:t\in (x-\sqrt{3}y+U)\cup(x+\sqrt{3}y+U)\}$$where $U\in\mathcal F$.

It can be shown that the topological space $X_{\mathcal F}$ is Hausdorff and the closures of any non-empty open sets in $X_{\mathcal F}$ have a common point, which implies that $X_{\mathcal F}$ is connected.

It can be shown that for distinct regular open vanishing filters $\mathcal U,\mathcal V$ the topologies $\tau_{\mathcal U}$, $\tau_{\mathcal V}$ on the set $X$ are distinct, which implies that on the countable set $X$ there exist $2^{\mathfrak c}$ Hausdorff connected topologies. Since the cardinality of bijections of the set $X$ is continuum, the number of pairwise non-homeomorphic connected Hausdorff topologies on $X$ remains equal to $2^{\mathfrak c}$.

  • $\begingroup$ Amazing! - I would already have been surprised if the number of such topologies were "only" $2^{\aleph_0}$! $\endgroup$ – Dominic van der Zypen Sep 13 '17 at 6:57
  • $\begingroup$ It is very nice, but, unfortunately, does not provide invariants which may distinguish such spaces. $\endgroup$ – Fedor Petrov Oct 31 '17 at 19:56

That there are $2^{\mathfrak{c}} = 2^{2^{\aleph_0}}$ homeomorphism types of countably infinite connected Hausdorff spaces is already proved in

Kannan, V.; Rajagopalan, M. Regularity and dispersion in countable spaces. Duke Math. J. 39 (1972), 729–734.

There are certainly no more than $2^{2^{\aleph_0}}$ homeomorphism types of countably infinite topological spaces, so "the largest cardinality" is $2^{\mathfrak{c}}$.

In fact the authors prove a bit more. I quote from F. Burton Jones' Math Review, taken from MathSciNet:

A regular Hausdorff space (with more than one point) cannot be both connected and countable. A separation axiom weaker than regularity (but stronger than Hausdorff) is Uryson's: If $p$ and $q$ are distinct points, there exist open sets $U$ and $V$ such that $p \in U$, $q \in V$ and $\overline{U} \cap \overline{V} = \varnothing$. Countable connected Uryson spaces have been known for some time, but only recently P. Roy has given an example [same J. 33 (1966), 331–333; MR0196701] of a countable connected Uryson space with a dispersion point, i.e., a point whose complement is totally disconnected. Roy's space is regular at only one point, and he raised the question whether such a space can be regular almost everywhere. The authors supply such an example. In fact, they show that there exist $2^{\mathfrak{c}}$ mutually non-homeomorphic, countable, connected Uryson spaces with dispersion points, each of which is regular at each point of a dense subset. In order to carry through their construction, the authors characterize the countable, totally disconnected, Hausdorff spaces that admit a regular dispersion point, i.e., those that may be made connected by the addition of a single point at which the space becomes regular.

With regard to Fedor Petrov's comment:

It is very nice, but, unfortunately, does not provide invariants which may distinguish such spaces.

Indeed it seems difficult in practice to sort countably connected Hausdorff spaces into homeomorphism types! In Section 3 of this paper with Lebowitz-Lockard and Pollack we associate to every countably infinite integral domain $R$ with zero Jacobson radical -- hence in particular to every countably infinite principal ideal domain (PID) with infinitely many prime ideals -- a countably infinite, connected Hausdorff "Golomb space" $G(R)$. The number of isomorphism types of countably infinite PIDs with infinitely many prime ideals is easily seen to be $\mathfrak{c}$ (which is also the number of isomorphism types of countably infinite rings). So one may hope that we are producing $\mathfrak{c}$ homeomorphism types of countably infinite connected Hausdorff spaces via this algebraic construction...but it is of course conceivable that $G(R)$ and $G(S)$ are homeomorphic when $R$ and $S$ are not isomorphic. Does this ever happen? We don't know. In our first draft of the paper we could prove nothing about this, but later we were able to show that if $\mathbb{F}_1$ and $\mathbb{F}_2$ are finite fields and $G(\mathbb{F}_1[t])$ and $G(\mathbb{F}_2[t])$ are homeomorphic then $\# \mathbb{F}_1 = \# \mathbb{F}_2$, which shows that these spaces fall into at least $\aleph_0$ homeomorphism types.

In order to show this we adapt work of T. Banakh, J. Mioduszewski and S. Turek. Many of the key ideas first appeared in a MathOverflow answer of Taras Banakh.

  • $\begingroup$ Very interesting! Does not this work with complements of Jacobson radicals in more general domains? $\endgroup$ – მამუკა ჯიბლაძე Jul 8 '18 at 5:40
  • 1
    $\begingroup$ @მამუკაჯიბლაძე: I suspect not. At least the proof of (i) $\implies$ (ii) in Theorem 5 does not immediately carry over. In particular there will be distinct elements $x$ and $y$ not in the Jacobson radical, but such that $x-y$ is in the Jacobson radical, and this look problematic. $\endgroup$ – Pete L. Clark Jul 8 '18 at 18:33

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