Let $\langle\mathbf{R}^n,\mathscr{O}\rangle$ be the $n$-dimensional Euclidean space. Define $\mathbf{Q}\subseteq\mathcal{P}(\mathscr{O})$ to consist of all sets $\mathsf{Q}$ which simultanously satisfy the following four conditions:


1. $\mathsf{Q}\neq\emptyset$ and $\mathsf{Q}\neq\{\emptyset\}$
2. $\forall\_{U\in\mathsf{Q}}\exists\_{V\in\mathsf{Q}}\,\mathrm{Cl}\,{V}\subseteq U$
3. $\forall\_{U,V\in\mathsf{Q}}\,(U\neq V\rightarrow \mathrm{Cl}\,V\subseteq U\vee \mathrm{Cl}\,U\subseteq V)$
4. $\forall\_{A,B\in\mathscr{O}}\,((\forall\_{U\in\mathsf{Q}}\,A\cap U\neq\emptyset\neq B\cap U)\rightarrow\mathrm{Cl}\,A\cap \mathrm{Cl}\,B\neq\emptyset)$


So every element of the family $\mathbf{Q}$ is a set in which well-inside relation ($\mathrm{Cl}\,U\subseteq V$) is connected (in the sense of order) and which is decreasing with respect to this relation, but also satisfies the requirement according to which all open sets with non-empty intersections with all elements from the set must have a common adherent point.


Is it true that for every $\mathsf{Q}\in\mathbf{Q}$ there is a point $p\in\mathbf{R}^n$ such that $\bigcap\mathsf{Q}=\{p\}$? I will be grateful for your help.



Intersections of families of open sets ordered by well-inside relation in Euclidean space
Yes. (Assuming $\varnothing$ isn't allowed in your $Q$'s, as Joel pointed out.)


First note that for such $Q$, $\bigcap Q$ cannot contain two points. Indeed, if $a,b \in \bigcap Q$ are distinct, we can pick open neghborhoods $a \in A$, $b \in B$ such that $\operatorname{Cl} A \cap \operatorname{Cl} B = \varnothing$ and these two open sets $A$, $B$ contradict condition 4. So it suffices to rule out the case that $\bigcap Q = \varnothing$.


Suppose, for the sake of contradiction, that $\bigcap Q = \varnothing$. Pick a cofinal sequence $U\_0 \supset U\_1 \supset \cdots$ of elements of $Q$. Define $V\_n = U\_n \setminus \operatorname{Cl} U\_{n+1}$ and consider $A = \bigcup\_{n=0}^\infty V\_{4n}$ and $B = \bigcup\_{n=0}^\infty V\_{4n+2}$. Note that $\operatorname{Cl} A = \bigcup\_{n=0}^\infty \operatorname{Cl} V\_{4n}$ and $\operatorname{Cl} B = \bigcup\_{n=0}^\infty \operatorname{Cl} V\_{4n+2}$ because $\bigcap\_{n=0}^\infty \operatorname{Cl} U\_n = \varnothing$. Thus $\operatorname{Cl} A \cap \operatorname{Cl} B = \varnothing$ and yet $A \cap U \neq \varnothing \neq B \cap U$ for any $U \in Q$, which contradicts condition 4.


