Is there a ZFC example of a Tychonoff space $X$ such that:

  1. $X$ is separable.
  2. $X$ has countable tightness (that is, a subset of $X$ is closed if and only if it contains the closure of each one of its countable subsets)
  3. $X$ has cardinality larger than the continuum?

The existence of a separable space of countable tightness and cardinality larger than the continuum is consistent with ZFC. Take any hereditarily separable space of cardinality larger than the continuum, like Fedorchuk's compact $S$-space. Hereditary separability implies both separability and countable tightness.

However, an example answering my question cannot be hereditarily separable. By Todorcevic's Theorem, every hereditarily separable regular space is hereditarily Lindelof under PFA and every hereditarily Lindelof space has cardinality bounded by the continuum.

Note also that an example answering my question cannot be compact. In fact, by Balogh's Theorem, every compact Hausdorff space of countable tightness is sequential under PFA, and it is easy to see that every separable sequential space has cardinality bounded by the continuum.

Finally note that to find a Hausdorff non-regular counterexample it suffices to take the Katětov extension of the integers. Let $U(\omega)$ be the set of all non-principal ultrafilters on $\omega$ and define a topology on $X=\omega \cup U(\omega)$ by declaring every point of $\omega$ to be isolated and a basic neighbourhood of a point $p \in U(\omega)$ to be $\{p\} \cup A$, where $A \in p$. Then $X$ is separable, has countable tightness and $|X|=2^\mathfrak{c}$.


1 Answer 1


EDIT: this argument is wrong. See Santi's comment below.

Doesn't ZFC prove there is no such space, by an argument similar to the (elementary submodel) version of the Arhangelskii theorem?

Let $X$ be a Tychonoff space that is separable and of countable tightness. Let $D$ be a countable dense set witnessing separability. Let $\mathcal{N}$ be an elementary submodel of some $H_\theta$ such that:

  1. $X$ and $D$ are in $\mathcal{N}$;
  2. $|\mathcal{N}|=2^{\aleph_0}$, $\mathcal{N}$ is closed under $\omega$ sequences, and $\mathbb{R}$ is both an element and subset of $\mathcal{N}$. (In particular, any continuum-sized element of $\mathcal{N}$ is also a subset of $\mathcal{N}$).

As with the Arhangelsii argument, countable tightness of $X$ and countable closure of $\mathcal{N}$ ensure that

(*) $\mathcal{N} \cap X$ is a closed subset of $X$.

We show that $X =\mathcal{N} \cap X$. Suppose toward a contradiction there is some $x \in X$ outside of $\mathcal{N}$. By (*) and the Tychenoff property, there is a continuous $f: X \to \mathbb{R}$ taking value 0 on $\mathcal{N} \cap X$ and value 1 at $x$. But $f$ is determined by $f \restriction D$, and ${}^D \mathbb{R}$ is a continuum-sized element of $\mathcal{N}$, hence a subset of $\mathcal{N}$ by item 2 above. So $f \in \mathcal{N}$. But then an easy elementarity argument, and the fact that $f$ is not constant, contradicts that $f \restriction \mathcal{N}$ is constantly 0.

  • 6
    $\begingroup$ Sean, I don’t see why $X \cap \mathcal{N}$ would be closed (unless you make some additional assumption, like "each point is a $G_\delta$ set", for example). Indeed, if you let $X$ be Fedorchuk’s example of a compact $S$-space from $\diamond$ then $X$ has the property that each one of its infinite closed subsets has cardinality $2^\mathfrak{c}$, so $X \cap \mathcal{N}$ would not be closed. $\endgroup$ Jul 19, 2021 at 7:26
  • 3
    $\begingroup$ Ah, thanks. I will leave it up (with a pointer to the error) as a warning! $\endgroup$
    – Sean Cox
    Jul 19, 2021 at 17:46
  • 4
    $\begingroup$ No problem, and like I said if we assume in addition that $X$ has points $G_\delta$ then (*) is true (Hint: in a regular space with points $G_\delta$, every point is a countable intersection of closed neighbourhoods). But then the second part of your argument could be simplified as follows: since $D$ is a countable element of $\mathcal{N}$ then $D$ is a subset of $\mathcal{N}$ so $X \cap \mathcal{N}$ is dense in $X$. Being both closed and dense $X \cap \mathcal{N}$ coincides with $X$. So a possible counterexample must have at least a non-$G_\delta$ point. $\endgroup$ Jul 19, 2021 at 21:03

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