# Cardinalities of topologies in which not each open set is a union of regular open sets

Suppose, a topological space $(X, \mathscr{T})$ consists of a set $X$ with the cardinality $\kappa$, and a topology $\mathscr{T}$ in which it is not true that each open subset of $X$ can be written as a union of a family of regular open sets(i.e. open sets that equals the interior of its closure). Then what is the cardinality of $\mathscr{T}$?

It seems to me that $|\mathscr{T}|$ can't be too "large". The "largest" topology in my mind is the cofinite topology with a cardinality which is exactly $\kappa$, since in cofinite topology, the interior of closure of a non-empty open set is $X$. Is $\kappa$ an upper bound?

Is there an example of topology, except for $\varnothing$ and $X$, some open subsets can be written as a union of a family of regular open sets, while others can not? In limited the examples I've considered, e.g. metric topology of $\mathbb R^n$, cofinite topology, $\mathscr T -\{X,\varnothing\}$, i.e. all the open sets except for $X$ and $\varnothing$, as a whole, must fall in one category or the other.

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That "not" is very poorly placed. Do you mean "...in which it is not true that each open subset of $X$..." – Thomas Andrews Jan 10 '13 at 23:54
@ThomasAndrews: Yes, that's what I mean. I will appreciate it if you could edit it to make it less ambiguous. – Metta World Peace Jan 10 '13 at 23:59
You are looking for spaces that are not semiregular. You can find a large number of examples (from Steen and Seebach) by using spacebook. There are examples of non-second countable non-semiregular spaces of cardinality $\omega$ (e.g. one-point compactification of $\mathbb{Q}$), so the answer to your question on the specific upper bound of the cardinality is no. – Martin Jan 11 '13 at 0:53
@Martin: Cool. Thank you for your answer and link. – Metta World Peace Jan 11 '13 at 0:55

Here’s a simple machine for constructing Hausdorff spaces of cardinality $\kappa$, for any infinite cardinal $\kappa$, having $2^\kappa$ open sets, not all of which can be written as unions of regular open sets. (I.e., the spaces are Hausdorff but not semiregular.)
Let $\kappa$ be any infinite cardinal, and let $X=\kappa\times\Bbb Q$, ordered lexicographically; $X$ with the order topology $\tau$ is a space of cardinality $\kappa$ with no isolated points. Let $p=\langle 0,0\rangle\in X$. Let $Y=X\times X$; the topology on $Y$ will be a refinement of the product topology. First, make each point of $X\times(X\setminus\{p\})$ isolated. Now suppose that $x\in X$; for each open nbhd $U$ of $x$ and open nbhd $V$ of $p$ in $X$ let $$B(x,U,V)=\{\langle x,p\rangle\}\cup\Big((U\times V)\setminus(X\times\{p\})\Big)\;,$$ and let $$\mathscr{B}(x)=\{B(x,U,V):x\in U\in\tau\text{ and }p\in V\in\tau\}$$ be a local base at $\langle x,p\rangle$ in $Y$.
Clearly $Y$ is Hausdorff, and it’s straightforward to check that none of the sets $B(x,U,V)$ is the union of regular open sets: any regular open set containing $\langle x,p\rangle$ must contain other points of $X\times\{p\}$. Finally, $Y$ has $\kappa$ isolated points, so it has $2^\kappa$ open (and indeed regular open) sets.