Are there many compact, zero-dimensional, first-countable spaces with $d(Z)=\mathfrak{c}$?

Question

Are there many compact, zero-dimensional, first-countable spaces with $d(Z)=\mathfrak{c}$, where $d(\cdot)$ stands for density (i.e. least size of a dense subspace)?

Thanks for your help!

Answer 1

Let $Y$ be any zero-dimensional, first countable, compact Hausdorff space with $1<|Y|\le 2^\omega$; then $Y^\omega$ is a zero-dimensional, first countable, compact Hausdorff space of cardinality $2^\omega$. Let $X$ be the Alexandroff double of $Y^\omega$; then $X$ is a zero-dimensional, first countable, compact Hausdorff space, and $d(X)=2^\omega$.

Moreover, we can generalize the construction of the double. Let $Y^\omega=\{y_\xi:\xi<2^\omega\}$, and let $\{Z_\xi:\xi<2^\omega\}$ be any pairwise disjointcollection of first countable compact Hausdorff spaces disjoint from $Y^\omega$. Let $X=Y^\omega\cup\bigcup_{\xi<2^\omega}Z_\xi$. For each $\xi<2^\omega$, points of $Z_\xi$ have their usual nbhds in $Z_\xi$, and $Z_\xi$ itself is a clopen subset of $X$. Basic open nbhds of $y_\xi\in Y^\omega$ are sets of the form

$$U\cup\bigcup\{Z_\eta:y_\eta\in U\text{ and }\eta\ne\xi\}\;,$$

where $U$ is an open nbhd of $y_\xi$ in $Y^\omega$. Then $X$ a zero-dimensional, first countable, compact Hausdorff space, and $d(X)=2^\omega$. ($X$ is what you get if you start with the Alexandroff double of $Y^\omega$ and replace the isolated point corresponding to $y_\xi$ with the space $Z_\xi$.)