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All spaces are assumed to be Hausdorff. Recall that a cellular family in the space $X$ is a family of pairwise disjoint non-empty open subspaces of $X$. The cellularity of $X$ ($c(X)$) is defined as the supremum of the cardinalities of the cellular families in $X$. CCC means that the cellularity is countable.

We say that a topological space $X$ is cellular-Lindelof if for every cellular family $\mathcal{C}$ there is a Lindelof subspace $Y$ of $X$ such that $U \cap Y \neq \emptyset$, for every $U \in \mathcal{C}$.

Clearly every Lindelof space is cellular-Lindelof and every CCC space is cellular-Lindelof.

The cellular-Lindelof property was introduced in our paper with Bella https://link.springer.com/article/10.1007/s00605-017-1112-4, where we note that:

FACT: Cellular-Lindelof first-countable spaces have cardinality at most $2^{\mathfrak{c}}$.

Indeed, let $X$ be a first-countable cellular-Lindelof space. Then $c(X) \leq \mathfrak{c}$ (this follows from Arhangel'skii's Theorem stating that every Lindelof first-countable space has cardinality at most continuum). Combining that with the Hajnal-Juhasz inequality $|X| \leq 2^{\chi(X) \cdot c(X)}$ (where $\chi(X)$ denotes the character of $X$) we obtain that $|X| \leq 2^{\mathfrak{c}}$.

QUESTION: Let $X$ be a first-countable cellular-Lindelof space. Is $|X| \leq \mathfrak{c}$?

A positive answer would lead to a common generalization of Arhangel'skii's Theorem and the Hajnal-Juhasz theorem stating that first-countable CCC spaces have cardinality at most continuum.

EDIT: (04/03/2019)

A partial answer can be found in a new paper with Bella: the answer is yes for normal spaces under $2^{<\mathfrak{c}}=\mathfrak{c}$.

See: https://arxiv.org/abs/1811.00660

Another partial answer can be found in yet unpublished work of Vladimir Tkachuk and Richard Wilson: the answer is yes for "cellular-compact" regular spaces in ZFC (you can guess the definition).

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