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Let $X$ be a regular first countable space of cellularity at most $2^\omega$. Is it true that the cardinality of $X$ is at most $2^\omega$?

A cellular family is a family of pairwise disjoint non-empty open sets. The cellularity of a space $X$ (denoted by $c(X)$) is defined as the supremum of the cardinalities of the cellular families in $X$.

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  • $\begingroup$ +1. It is known that every Hausdorff first countable space that satisfies the countable chain condition has cardinality at most $2^{\aleph_{0}}$ (a proof relies on the Erdos-Rado theorem). The question however asks for first countable spaces of cellularity up to $2^{\aleph_{0}}$ though. See the paper A. Hajnal, I. Juhász: On discrete subspaces of topological spaces, Indag. Math. 29(1967) 343–356 $\endgroup$
    – Joseph Van Name
    Commented Dec 20, 2017 at 16:52
  • $\begingroup$ By the Erdos-Rado theorem, the cardinality a regular first countable space of cellularity at most $\kappa$ must be less than or equal to $2^{\kappa}$. $\endgroup$
    – Joseph Van Name
    Commented Dec 20, 2017 at 17:05

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First of all let me note that a first-countable space of density $2^\omega$ has cardinality $\leq 2^\omega$, so a counterexample $X$ to your question must satisfy $c(X) \leq 2^\omega < d(X)$. That leads us naturally to higher Suslin Lines.

A continuous linear order $X$ (endowed with the order topology) is called a $\kappa$-Suslin line if $c(X) \leq \kappa < d(X)$.

The existence of a $\kappa$-Suslin Line is equivalent to the existence of a $\kappa$-Suslin tree and since higher Suslin trees are consistent with ZFC (Jensen proved they exist under $V=L$ for example), the existence of higher Suslin lines is also consistent with ZFC. A higher Suslin line is not necessarily first-countable but we can get something similar to a Suslin Line which is first-countable using an idea of this paper of Juhasz, Soukup and Szentmiklossy:

https://arxiv.org/pdf/math/0703728.pdf

Assume the existence of a $\mathfrak{c}$-Suslin Line $X$. Let $Y$ be the set of all points of countable cofinality in $X$. The continuity of the linear order implies that $Y$ is dense in $X$ and hence we can fix a dense subset $Z \subset Y$ of cardinality $\mathfrak{c}^+$. On $Z$ consider the topology $\tau$ generated by intervals of the form $(x,y]$, for $x < y \in Z$. Note that $(Z,\tau)$ is a first-countable space of cardinality $\mathfrak{c}^+$ and $c((Z, \tau))\leq \mathfrak{c}$, so that provides a consistent counterexample to your question.

Let me finish by noting that although it's not true that the continuum is a bound on the cardinality of first-countable spaces of cellularity $\leq 2^\omega$, the Hajnal-Juhasz inequality $|X| \leq 2^{\chi(X) \cdot c(X)}$ implies that $2^{2^\omega}$ is a bound on the cardinality of such a space (I'm assuming this is what originally motivated your question).

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