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](https://www.jstor.org/stable/2033324)), 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 $d(X)\geq \omega\_1$.



$|\mathsf{RO}(X)|$ vs. $2^{d(X)}$ for $T_3$ spaces
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.)


