Regular space and regular open sets A topological space $(X,\tau)$ is said to be regular if for every $x\in X$ and every closed set $F$ not containing $x$ there exists $U,V\in\tau$ such that $x\in U$, $F\subset V$ and $U\cap V=\emptyset$.
A regular open set is an open set $U$ such that $\operatorname{int}(\operatorname{cl}(U))=U$.
I tried to find a regular topological space whose only regular open set is $X$ and $\emptyset$, but I tried all the topological spaces I know, but I could not find it except indiscrete topology with more thab2 elements, maybe there is no such thing as this topological space which is regular and admits only $X$ and $\emptyset$ as regular open set except indiscrete topology.
I am so confused and my heart is broken actually. Any help will greatly be appreciated.
 A: $(X, \tau) $ be a topological space.

*

*$\emptyset, X$ are two trivial regular open sets.


*Every regular open sets are open.
Consider indidcerete topology on $X$ where $|X|>2$.
Then $(X, \tau) $ is a regular space having only two trivial regular open sets ( as only open sets are $\emptyset, X$)
A: Claim. A topological space $X$ has no regular opens other than $X$ and $\emptyset$ if and only if there is no pair of non-empty disjoint opens in $X$.
Proof. Left to right, if there are no regular opens other than $X$ and $\emptyset$, then for each non-empty open $U$ we have $\mathrm{cl}(U) = X$, since otherwise $\emptyset \subsetneq U \subseteq \mathrm{int} (\mathrm{cl} (U)) \subsetneq X$. In other words, $\mathrm{int}(X \setminus U) = \emptyset$, so there is no non-empty open $V$ disjoint from $U$. Right to left, if there is no pair of disjoint opens, then for each non-empty open $U$ we have $\mathrm{int}(X \setminus U) = \emptyset$, so $\mathrm{cl}(U) = X$ and $\mathrm{int}(\mathrm{cl}(U)) = X$. Thus $U$ is not regular unless $U = X$.
Non-example. No regular space which is not indiscrete satisfies this condition: if $F$ is a non-empty and non-total closed set, then regularity produces disjoint non-empty opens.
Non-example. No Hausdorff space with at least two points satisfies this condition: if $x$ and $y$ are distinct points, then the Hausdorff axiom produces disjoint non-empty opens.
Example. Each space where the specialization preorder is upward directed satisfies this condition. For example, the Alexandrov topology of an upward directed partially ordered set satisfies this condition, as does the topology on the reals consisting of intervals of the form $(a, +\infty)$ plus the empty and total set.
