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.

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.

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. $\mathsf{Q}\neq\emptyset$
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.

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.