A non-empty topological space without isolated points is called maximal if every finer topology on that space has at least an isolated point. The existence of a (Hausdorff) maximal space is a simple consequence of Zorn's Lemma.

Note that in a maximal space $(X, \tau)$, nowhere dense sets are closed (and discrete). Indeed, if there were a nowhere dense set $N$ that is not closed then $\tau \cup \{X \setminus N \}$ would be a subbase for a finer topology without isolated points on $X$. In particular, a maximal space can't contain any non-trivial convergent sequences, because a discrete set is nowhere dense in a space without isolated points.

Let $\mathfrak{max}$ be the minimal weight of a Hausdorff maximal space. Clearly $\aleph_1 \leq \mathfrak{max} \leq \mathfrak{c}$, but we can prove a better lower bound, namely: $\mathfrak{d} \leq \mathfrak{max}$.

If $X$ is a countable Hausdorff maximal space then $w(X) \geq \mathfrak{d}$.

Proof: Fix a point $x \in X$ and let $\{U_n: n < \omega \}$ be a maximal pairwise disjoint family of non-empty open sets with the property that $x \notin \overline{U_n}$, for every $n< \omega$: Clearly $x \in \overline{\bigcup \{U_n: n < \omega \}}$. Let $\{x^n_k: k < \omega \}$ be an enumeration of $U_n$.

Suppose by contradiction that $w(X)=\kappa < \mathfrak{d}$ and let $\{B_\alpha: \alpha < \kappa \}$ enumerate a local base at $x$. For every $\alpha < \kappa$, the set $B_\alpha$ intersects infinitely many $U_n$'s, so we can find an integer-valued function $f_\alpha$ with infinite domain $\subseteq \omega$ such that $x^n_{f_\alpha(n)} \in B_\alpha \cap U_n$, for every $n \in dom(f_\alpha)$.

Since $\kappa < \mathfrak{d}$, we can find a function $f: \omega \to \omega$ such that, for every $\alpha < \kappa$, there is $n \in dom(f_\alpha)$ with $f_\alpha(n) < f(n)$. Let $D=\{x^n_k: k \leq f(n), n < \omega \}$. Then $D$ is discrete and $x \in \overline{D} \setminus D$, so $D$ is a nowhere dense set in $X$ which is not closed and that contradicts maximality.

QUESTION: Is $\mathfrak{max}=\mathfrak{d}$ in ZFC?

EDIT(10/05/2019): Will Brian answered the above question in the negative, but the question of the title is still open. What is $\mathfrak{max}$? Is it equal to some product of known cardinal invariants of the continuum?


1 Answer 1


No, these cardinals are not provably equal.

This follows from a result of El'kin 1:

Let $X$ be a set and let $x \in X$. If $\tau$ is a maximal topology on $X$, then $\{U \setminus \{x\} \,:\, x \in U \in \tau \}$ is a base for a non-principal ultrafilter on $X$.

If $X$ is countable (as in your question), it follows that any local basis at $x$ has size at least as big as the ultrafilter number $\mathfrak u$, defined as the smallest possible cardinality of a base for a non-principal ultrafilter on a countable set. Hence we have found another lower bound for your cardinal: $\mathfrak u \leq \mathfrak{max}$.

The reason this answers your question is that it is consistent to have $\mathfrak d < \mathfrak u$. (This happens for example in the random real model.) By the previous paragraph, any model in which $\mathfrak d < \mathfrak u$ is also a model in which $\mathfrak d < \mathfrak{max}$.

1 A. G. El'kin, "Ultrafilters and irresolvable spaces," Vestnik Moskov. Univ. Ser. I Mat. Mekh. 24 (1969), no. 5, pp. 51-56. $\ $ I was unable to find an online version of this paper, and in fact I don't even know whether it's been translated from the Russian I presume it was written in. But you can find the result mentioned above quoted in this book (page 54), or mentioned (a little vaguely) in this paper (page 2).

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    $\begingroup$ That’s interesting. You showed that if $\kappa$ is the minimum weight of a countable irresolvable space then $\mathfrak{u} \leq \kappa$ and it is known that $\kappa \leq \mathfrak{i}$, so $\mathfrak{u} \leq \mathfrak{i}$. Now I didn’t know there was any relationship between $\mathfrak{u}$ and $\mathfrak{i}$. $\endgroup$ May 9, 2019 at 11:50
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    $\begingroup$ I didn't know that either. That inequality isn't found in Andreas' Handbook article (my go-to reference for such things), which makes me suspect something's fishy here. There are a few possibilities: (1) the theorem I quote in my post isn't true (2) the theorem you quote in your comment isn't true (3) $\mathfrak u \leq \mathfrak i$ is a known theorem, but it wasn't mentioned in Andreas' article (4) $\mathfrak u \leq \mathfrak i$ is a new theorem. $\endgroup$
    – Will Brian
    May 9, 2019 at 12:21
  • $\begingroup$ As for possibility $(1)$, I tried yesterday to prove El'kin's theorem (because I couldn't find a proof online). I was able to prove it only for the narrower class of $T_3$ submaximal spaces. This was good enough for yesterday -- I assumed that with a little more effort I might be able to weaken my assumptions to match El'kin's. But now I wonder. Do you know whether the bound $\kappa \leq \mathfrak i$ applies when the definition of $\kappa$ is restricted to this narrower class of spaces? (In principle, this could make $\kappa$ larger.) $\endgroup$
    – Will Brian
    May 9, 2019 at 12:26
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    $\begingroup$ The consistency of $\mathfrak{i} < \mathfrak{u}$ was proved by Shelah shortly after Vaughan's survey appeared: link.springer.com/article/10.1007/BF01277485. This rules out possibilities $(3)$ and $(4)$ in my list above. Your comment from 2 hours ago seems to dispense with possibility $(2)$ as well -- that argument is pretty clear and convincing. $\endgroup$
    – Will Brian
    May 9, 2019 at 18:02
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    $\begingroup$ It follows from van Douwen´s results (in "Applications of maximal topologies", 1993) that a crowded space $X$ is maximal if and only if for every $x$ the set $\{U \setminus \{x\} \,:\, x \in U \in \tau \}$ is a base for an ultrafilter on $X$. $\endgroup$ Feb 18, 2020 at 19:56

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