Is the union of nowhere dense sets which are separated by disjoint open sets nowhere dense?

Question

Proposition: In a topological space, Let F be a collection of nowhere dense sets such that each member of F is contained in a open sets and the family of all those open sets is disjoint. Then the union of members of F is nowhere dense.

We know that the finite union of nowhere dense is nowhere dense and a set But in the infinite case mentioned above I was looking for a proof of this proposition.

Thanks!

Answer 1

Let us establish a bit of notation: For each $A\in F$ there is an open set $V(A)\supseteq A$, and it is given that if $A\ne B$ then $V(A)\cap V(B)=\emptyset$.

I propose to show that $\bigcup F$ is nowhere dense by showing that its closure has empty interior. Therefore, pick an interior point $x$ of the closure, and try to get a contradiction from that.

By assumption, there is an open set $U\ni x$ consisting of points of closure of $\bigcup F$. So $U\cap A\ne\emptyset$ for at least one $A\in F$. Moreover, $U\cap V(A)\subseteq\overline{\bigcup F}$. Note carefully that if $B\in F\setminus\{A\}$ then $B\cap V(A)=\emptyset$, so $\overline{\bigcup (F\setminus\{A\})}\cap V(A)=\emptyset$.

I am tempted to leave you to take it from there. In fact, I hereby give in to that temptation.

Answer 2

HINT: Let $\mathscr{F}$ be a family of nowhere dense sets in $X$. Assume that for each $F\in\mathscr{F}$ there is an open set $U_F$ such that $F\subseteq U_F$, and the family $\{U_F:F\in\mathscr{F}\}$ is pairwise disjoint. Let $F=\operatorname{cl}\bigcup\mathscr{F}$, and suppose that $V$ is an open subset of $F$.

• Show that if $F\in\mathscr{F}$, then $V\cap U_F\subseteq\operatorname{cl}F$; conclude that $V\cap U_F=\varnothing$ for each $F\in\mathscr{F}$.
• Use the previous result to show that $V=\varnothing$.