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Let $X$ be any topological space endowed with two topologies $\mathcal{T} \subset \mathcal{T}'$ (the later means that $\mathcal{T}'$ refines $\mathcal{T}$, that is every open subset $U \subset X$ with respect $\mathcal{T}$ (ie $U \in \mathcal{T})$ is already open with respect $\mathcal{T}'$.

Assume that $X$ is irreducible (or say more weaker connected; irreducible implies connected) with respect $\mathcal{T}$.
Connected means that if there exist two $U,V$ which are open & cloled for $\mathcal{T}$ and $X= U \dot{\cup}V $ then either $U$ or $V$ is empty. And irreducible that every two non empty open $U, V \in \mathcal{T}$ intersect properly: $U \cap V \neq \emptyset$.

Question: Are there any interesting sufficient or neccessary conditions on $X, \mathcal{T}$ and $\mathcal{T}'$ known such that following holds

every open non empty $U \in \mathcal{T}$ is 'dense' in $X$ with respect finer topology $\mathcal{T}'$

What I mean by 'interesting'? Well, everything non trivial/boring like $\mathcal{T}= \mathcal{T}'$ or $\mathcal{T}= \{X, \emptyset\}$ etc.
Note that this question generalizes this Question where $X \subset \mathbb{C}^n$ is a complex irreducible variety, $\mathcal{T}$ the Zariski topology and $\mathcal{T}'$ analytical topology on $X$ induced by $\mathbb{C}^n$

user267839
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    So you’re interested in the case in which $U\cap V\ne\varnothing$ whenever $\varnothing\ne U\in\mathscr{T}$ and $\varnothing\ne V\in\mathscr{T}'$? – Brian M. Scott Apr 30 '21 at 20:49
  • @Brian: Yes exactly, that's an equivalent characterization for $U \in \mathcal{T} $ to be dense with respect topology $\mathcal{T}'$. – user267839 Apr 30 '21 at 21:25

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