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Santi Spadaro
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Free sequences and the cardinality of a topological space

One way of formulating Arhangel'skii's celebrated theorem about the cardinality of Lindelof first-countable spaces is the following (due to Arhangel'skii and Shapirovskii). For every Hausdorff space $X$:

$$|X| \leq 2^{t(X) \cdot \psi(X) \cdot L(X)}$$

where $L(X)$ denotes the Lindelof degree of $X$ (that is, the minimum cardinal $\kappa$ such that every open cover of $X$ has a subcover of cardinality $\leq \kappa$), $\psi(X)$ denotes the pseudocharacter of $X$ (that is, the least cardinal $\kappa$ such that every point in $X$ is a $G_\kappa$ set) and $t(X)$ denotes the tightness of $X$, that is the minimum cardinal $\kappa$ such that for every non-closed set $A \subset X$ and for every point $x \in \overline{A} \setminus A$ there is $C \subset A$ such that $x \in \overline{C}$ and $|C| \leq \kappa$.

The above inequality can be refined by making use of the notion of free sequence, introduced by Arhangel'skii in his original proof of Arhangel'skii's Theorem.

A sequence $\{x_\alpha: \alpha < \kappa\} \subset X$ is said to be free if for every $\beta < \kappa$, $\overline{\{x_\alpha: \alpha < \beta\}} \cap \overline{\{x_\alpha: \alpha \geq \beta \}}=\emptyset$. If we let $F(X)$ be the supremum of cardinalities of free sequences in $X$ it is easy to see that $F(X) \leq L(X) \cdot t(X)$.

Juhasz was the first to note that, for every Hausdorff space $X$:

(*) $$|X| \leq 2^{F(X) \cdot \psi(X) \cdot L(X)}$$

A famous problem in set-theoretic topology asks whether $F(X)$ can be dropped in the above inequality (the answer is "consistently NO" by examples of Shelah, Gorelic, Usuba, Dow, and others...)

In another direction I'd like to ask, can $L(X)$ be dropped from (*)? That is:

QUESTION: Is it true that $|X| \leq 2^{F(X) \cdot \psi(X)}$ for every (regular) space $X$?

I'd expect the answer to be NO, but Juhasz, Soukup and Szentmiklossy have proved the following partial result in the positive direction:

$$|X| \leq 2^{2^{F(X) \cdot \psi(X)}}$$

for every regular space $X$.

As a final side remark note that $\psi(X)$ cannot be dropped from (*).

Santi Spadaro
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