# What's the minimal weight of a maximal space?

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?

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).

• 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 $\mathfrak{i}$. – Santi Spadaro May 9 '19 at 11:50
• 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. – Will Brian May 9 '19 at 12:21
• 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.) – Will Brian May 9 '19 at 12:26
• 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. – Will Brian May 9 '19 at 18:02
• 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 $X$. – Ramiro de la Vega Feb 18 '20 at 19:56