Recall that a topological space is called sequential if a set is closed if and only if it contains all limits of convergent sequences lying inside of it. A space $X$ is called Frechet if for every non-closed set $A \subset X$ and for every point $x \in \overline{A} \setminus A$ there is a sequence lying in $A$ which converges to $x$. We assume all spaces to be Hausdorff.
Every Frechet space is of course sequential, but the converse is not true. The simplest example of a sequential non-Frechet space is the so-called Arens space. Define a topology on $X=(\omega+1) \times \omega \cup \{\infty\}$ as follows: declare every point of $\omega \times \omega$ to be isolated; for $n<\omega$, declare a basic open neighbourhood of $(\omega, n)$ to be $(k,\omega] \times \{n\}$, where $k<\omega$ and finally declare a basic open neighbourhood of $\infty$ to be $\{\infty\} \cup \bigcup_{n \geq k} (k_n, \omega] \times \{n\}$, where $k<\omega$ and $\{k_n: n < \omega \}$ is a sequence of positive integers.
Note that all points in Arens space have countable character except for $\infty$, which is the point which makes Frechetness fail. It's easy to see that the character of $\infty$ (and hence of the whole space) is the dominating number $\mathfrak{d}$. That suggests the following question:
What is the minimum character of a sequential non-Frechet space? Does it coincide with one of the known cardinal characteristics of the continuum?
Let's also look at some variations on the above question, by defining three cardinal characteristics of the continuum:
- $\mathfrak{snf}:=\min\{\chi(X): X$ is a sequential non-Frechet space $\}$
- $\mathfrak{snf}_{\aleph_0}:=\min \{\chi(X): X$ is a countable sequential non-Frechet space $\}$.
- $\mathfrak{csnf}:=\min \{\chi(X): X$ is a compact sequential non-Frechet space $\}$.
Clearly $\aleph_1 \leq \mathfrak{snf} \leq \mathfrak{snf}_{\aleph_0} \leq \mathfrak{d} \leq \mathfrak{c}$.
Moreover, the one-point compactication of a Moore-Mrowka $\Psi$-space constructed on a MAD family on $\omega$ is an example of a compact sequential non-Frechet space and this shows that:
$\mathfrak{snf} \leq \mathfrak{csnf} \leq \mathfrak{a}$.
Where $\mathfrak{a}$ is the almost-disjointness number.
From Proposition 2 of Murtinová, Eva, On (weakly) Whyburn spaces, Topology Appl. 155, No. 17-18, 2211-2215 (2008). ZBL1168.54006. it follows that:
Let $X$ be a countable sequential space such that $\chi(X) < \mathfrak{d}$. Then $X$ is Frechet.
Observing that the Arens space is countable and has character $\mathfrak{d}$ we immediately get that:
$\mathfrak{snf}_{\aleph_0}=\mathfrak{d}$.
Angelo Bella observed the following:
$\mathfrak{b} \leq \mathfrak{snf}$
where $\mathfrak{b}$ is the bounding number.
To summarise, we have that:
$\aleph_1 \leq \mathfrak{b} \leq \mathfrak{snf} \leq \mathfrak{snf}_{\aleph_0} = \mathfrak{d} \leq \mathfrak{c}$
$\hspace{3.8em} \mathfrak{snf} \leq \mathfrak{csnf} \leq \mathfrak{a}$.
I am interested in the following questions:
What are $\mathfrak{snf}$ and $\mathfrak{csnf}$? Is it true in ZFC that $\mathfrak{csnf}=\mathfrak{a}$ and $\mathfrak{snf}=\mathfrak{b}$?
Here is the proof of $\mathfrak{b} \leq \mathfrak{snf}$. We need to show that every sequential space of character smaller than $\mathfrak{b}$ is Frechet. Let $Seqcl(A)$ be the set of limit points of convergent sequences lying inside of $A$. Since the closure of a set in a sequential space is obtained by iterating $Seqcl$ (at most $\omega_1$ many times) it suffices to prove that $Seqcl(Seqcl(A)) \subset Seqcl(A)$. So let $x \in Seqcl(Seqcl(A))$. Let $\{x_n: n <\omega\} \subset Seqcl(A)$ be a sequence converging to $x$ and, for every $n<\omega$ fix a sequence $\{x^n_k: k < \omega \}$ lying inside of $A$ which converges to $x_n$. Let $\{U_\alpha: \alpha < \lambda \}$ where $\lambda < \mathfrak{b}$ be a local base at $x$. Define functions $\{f_\alpha: \alpha < \lambda \} \subset \omega^\omega$ by letting: $$f_\alpha(n):=\begin{cases} \min \{j: (\forall i \in \omega \setminus j)(x^n_i \in U_\alpha)\} \text{ if } x_n \in U_\alpha \\ 0 \text{ otherwise} \end{cases} $$
Since $\lambda < \mathfrak{b}$ there is a function $f \in \omega^\omega$ such that $f_\alpha \leq^* f$, for every $\alpha < \lambda$. Let now $S=\{x^n_{f(n)+1}: n < \omega \}$. We have that $S \setminus U_\alpha$ is finite, for every $\alpha < \lambda$ and hence $S$ is a sequence lying inside $A$ which converges to $x$.
