$|\mathsf{RO}(X)|$ vs. $2^{d(X)}$ for $T_3$ spaces Let $\mathsf{RO}(X)$ stand for the collection of regular open subsets of a topological space $X$ and let $d(X)$ be its density. It is well-known (see Theorem~3.3 of Hodel's chapter in the Handbook) that every regular space satisfies the inequality $|\mathsf{RO}(X)|\leq 2^{d(X)}$. What is an example of an infinite $T_3$ (regular + $T_1$) space such that $|\mathsf{RO}(X)|<2^{d(X)}$?
I think a relevant observation regarding this question is the following. From Pierce's result on the cardinality of complete Boolean algebras (see here), it can easily be deduced that if $X$ is infinite and $T_2$, then $|\mathsf{RO}(X)|\geq \omega_1$. Thus, $\textsf{CH}$ implies that if $X$ is infinite, $T_3$ and separable, then $|\mathsf{RO}(X)| = 2^{d(X)}$. For this reason, if $X$ satisfies the desired characteristics of the question above, then necessarily $d(X)\geq \omega_1$.
 A: In fact, Joseph Van Name's argument gives a ZFC example.  For $f \in \{0,1\}^{[0,1]}$ let $supp(f) = \{x \in [0,1] : f(x) = 1\}$. Let $X$ be the $\Sigma$-product given by $X = \{f \in \{0,1\}^{[0,1]}: |supp(f)| \leq \aleph_0\}$ (with the subspace topology from the product). Then $X$ satisfies the countable chain condition because it is a dense subset of a separable space.  Its weight is $\frak c$ so by Joseph Van Name's comment, $|RO(X)| \leq {\frak c}$.  The density of $X$ is also $\frak c$ because given a collection of fewer than $\frak c$ elements of $X$, there is a subset of $[0,1]$ having cardinality smaller than $\frak c$ that contains all of their supports, and therefore is clearly not dense. Therefore, $2^{d(X)} = 2^{\frak c} > {\frak c} \geq |RO(X)|$.
One further comment is that if a space $Y$ has a collection $\mathcal U$ of $d(Y)$ pairwise disjoint non-empty open subsets, then $2^{d(Y)} = |RO(Y)|$ because in this case, for every ${\mathcal V} \subset {\mathcal U}$, the interior of the closure of $\cup {\mathcal V}$ gives a regular open set, and different choices of $\mathcal V$ give different regular open sets.  (This is the reason that there is no separable example, even in ZFC.)
A: The existence of such a space $X$ is consistent with $ZFC$.
If there exists a Suslin line and $2^{\aleph_0}<2^{\aleph_1}$, then whenever $X$ is a Suslin line, we have $|\text{Ro}(X)|<2^{d(X)}$.
If $\mathcal{B}$ is a basis for a space $Z$, and $Z$ satisfies the countable chain condition, then the mapping $\phi:{}^{\omega}\mathcal{B}\rightarrow\text{Ro}(Z)$ defined by letting $\phi((U_n)_{n\in\omega})=\bigvee_{n\in\omega}(\overline{U_n})^{\circ}$ is surjective, so $|\mathrm{Ro}(Z)|\leq|\mathcal{B}|^{\aleph_0}$.
Let $X$ be a Suslin line. Then $X$ has a basis of size $d(X)=\aleph_1$. Therefore, $|\text{Ro}(X)|\leq \aleph_1^{\aleph_0}=2^{\aleph_0}$. If $2^{\aleph_0}<2^{\aleph_1}$ as well, then $|\text{Ro}(X)|<2^{d(X)}.$
If $V=L$, then GCH holds (which implies $2^{\aleph_0}<2^{\aleph_1}$) and there exists a Suslin line.
