It's known that every countable, metrizeable space with no isolated points is homeomorphic to the rationals with the standard topology.

Suppose you wanted to reformulate the above without referencing metrizability directly. Since a countable space is clearly separable, you can use Urysohn's characterization of separable metric spaces to replace metrizeable with 2nd countable, regular (I take regular to mean regular and Hausdorff). You could even get away with 1st countable, regular since, for a countable space, 1st countable implies 2nd countable.

My question is, what happens if we try to relax metrizability to regularity? Is the space automatically 1st countable (and hence homeomorphic to the rationals)? To summarize:

Is there a countable space $X$ which is regular Hausdorff with no isolated points, but not 1st countable?

The Arens-Fort space (example 26 from Counterexamples in Topology) shows there is a countable, regular Hausdorff (in fact $T_5$) space which is not 1st countable. Unfortunately, the space has lots of isolated points.


There is such a space. For any $\mathscr{U}\in\beta\omega\setminus\omega$ we can start with the subspace $\omega \cup \{\mathscr{U}\}$ of $\beta\omega$ and ‘fatten up’ each isolated point to a copy of the rationals.

Let $\mathscr{U}$ be a free ultrafilter on $\omega$. Let $p$ be a point not in $\omega\times \mathbb{Q}$, and let $X = (\omega\times \mathbb{Q})\cup \{p\}$. We topologize $X$ as follows. For each $q\in\mathbb{Q}$ let $\mathscr{B}(q)$ be the set of clopen nbhds of $q$ in the usual topology on $\mathbb{Q}$. For $\langle n,q \rangle \in \omega\times \mathbb{Q}$ let $\mathscr{B}(n,q) = \{\{n\}\times B:B \in \mathscr{B}(q)\}$ be a local base at $\langle n,q\rangle$. Finally, take $\mathscr{B}(p) = \{U\times \mathbb{Q}:U\in\mathscr{U}\}$ as a local base at $p$. Let $$\mathscr{B} = \mathscr{B}(p)\cup \bigcup_{\langle n,q\rangle\in \omega\times \mathbb{Q}}\mathscr{B}(n,q)\;;$$ then $\mathscr{B}$ is a base for a topology $\mathscr{T}$ on $X$, and it’s easy to check that $\langle X,\mathscr{T}\rangle$ has no isolated points, is not first countable at $p$, and is regular. Indeed, the members of $\mathscr{B}$ are clopen in $\langle X,\mathscr{T}\rangle$, so $\langle X,\mathscr{T}\rangle$ is zero-dimensional and hence completely regular.

The same idea of ‘fattening up’ isolated points to isolated copies of $\mathbb{Q}$ can be applied to the Arens-Fort space. Start with $Y={\langle 0,0\rangle}\cup (\mathbb{Z}^+\times\mathbb{Z}^+$), where each point of $\mathbb{Z}^+\times\mathbb{Z}^+$ is isolated, and a set $V$ containing $\langle 0,0\rangle$ is open iff $\{m\in\mathbb{Z}^+:V\setminus(\{m\}\times\mathbb{Z}^+)\text{ is infinite}\}$ is finite (i.e., $V$ contains all but finitely many points of all but finitely many ‘columns’ of $\mathbb{Z}^+\times\mathbb{Z}^+$).

To get the desired space $X$, first replace each $\langle m,n\rangle \in \mathbb{Z}^+\times\mathbb{Z}^+$ by a copy, $Q(m,n)$, of $\mathbb{Q}$ with its usual topology. If $V$ is an open nbhd of $\langle 0,0\rangle$ in $Y$, let $$V^* = \{\langle 0,0\rangle\}\cup \bigcup_{\langle m,n\rangle\in V\setminus\{\langle 0,0\rangle\}} Q(m,n),$$ and take the family of such sets $V^*$ as a local base at $\langle 0,0\rangle$. The resulting space is countable, has no isolated points, is not first countable at $\langle 0,0\rangle$, and has a clopen base.

  • $\begingroup$ Thanks a lot, very natural approach! I'd thought of "doubling" each isolated point (which wrecks regularity of course), but this did not occur to me. $\endgroup$ – Mike F Oct 12 '11 at 23:57

Another naturally occurring example is the following: by the Hewitt-Marczewksi theorem, the set $I^I$, where $I$ is the unit interval, in the product topology, is a separable space, so has a countable dense subset $D$. As the product $I^I$ is completely regular, so is $D$, and so it's normal (being Lindelöf) and thus hereditarily normal, even. But at no point can $D$ have a local base of size $< \mathfrak{c}$ (very non-isolated), as is shown in Engelking's book, IIRC (no copy at hand).

As an aside: I have sometimes wondered: are all such countable dense subsets homeomorphic?

  • $\begingroup$ Very cool! Unless I'm much mistaken two explicit examples of countable dense subsets of $I^I$ would be (i) all rational valued functions $I \to I$ which are piecewise constant with respect to some finite, rational partition of $I$ and (ii) all polynomials with rational coefficients mapping $I \to I$. If all the countable dense subsets of $I^I$ were homeomorphic, it would have the (maybe?) surprising consequence that these two particular $D$ are homeomorphic in the topology of pointwise convergence. $\endgroup$ – Mike F Oct 14 '11 at 18:27

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