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 (*).

To find a Lindelof space of arbitrarily large cardinality whose free sequences are countable let $\sigma(2^\kappa)=\{f \in 2^\kappa: |f^{-1}(1)| < \aleph_0\}$ and let $\mathbf{1} \in 2^\kappa$ denote the function which is identically equal to 1. Recall that $\sigma(2^\kappa)$ is a Frechet-Urysohn space, so, in particular, it has countable tightness. Consider the space $X=\sigma(2^\kappa) \cup \{\mathbf{1}\}$, with the topology induced from $2^\kappa$. Then

1. $X$ is $\sigma$-compact (and hence Lindelof).
2. $|X|=\kappa$.
3. $F(X)=\omega$.

To see why the first and the third claim are true let $K_n=\{f \in 2^\kappa: |f^{-1}(1)| \leq n\}$. Then $K_n$ is compact and $\sigma(2^\kappa)=\bigcup \{K_n: n < \omega\}$. If $X$ contained an uncountable free sequence $F$ then $F$ would have uncountable intersection with some $K_n$ and then $F \cap K_n$ would be an uncountable free sequence inside the compact space of countable tightness $K_n$, which is a contradiction.