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.


  • $\begingroup$ Do you have any thoughts? What have you tried? $\endgroup$ – Harald Hanche-Olsen Dec 2 '14 at 8:28
  • $\begingroup$ I want to prove it by using the fact that a set is nowhere dense iff the complement of its closure is dense. But I found that members of F are just separated, their closure may not disjoint. $\endgroup$ – Jay Dec 2 '14 at 8:37

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.

  • $\begingroup$ Perfect solution. I proved this problem. Thank you! $\endgroup$ – Jay Dec 2 '14 at 10:17
  • 2
    $\begingroup$ In addition to my previous comment, it reminds me a quote from a paper of Arnie Miller "It is also hard not to show that ..." (See here, p. 14 of the PDF, section 7). $\endgroup$ – Asaf Karagila Dec 3 '14 at 19:40
  • $\begingroup$ @AsafKaragila Wonderful! Thanks. $\endgroup$ – Harald Hanche-Olsen Dec 3 '14 at 20:02

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 $V=\varnothing$.
  • $\begingroup$ I proved this problem using this hint. Thank you! $\endgroup$ – Jay Dec 2 '14 at 10:15
  • $\begingroup$ @Jacob: You're welcome! $\endgroup$ – Brian M. Scott Dec 2 '14 at 17:28

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