The union of a locally finite family of nowhere dense sets is still a nowhere dense set? The union of a locally finite family of nowhere dense sets is still a nowhere dense set?
Is this right or wrong?
 A: Let $\{ A_i : i \in I \}$ be a locally finite family of nowhere dense sets. Note that the family $\mathcal B$ of all nonempty open $U \subseteq X$ which intersect only finitely many $A_i$ forms a basis for $X$. (It is a basis because given any open $U \subseteq X$ for each $x \in U$ fixing an open neighborhood $V_x$ of $x$ which meets only finitely many $A_i$, then $V_x \cap U$ is also an open neighborhood of $x$ meeting only finitely many $A_i$ — so $V_x \cap U \in \mathcal B$ — and now $U = \bigcup_{x \in U} ( V_x \cap U )$.) To show that $\bigcup_{i\in I} A_i$ is nowhere dense it suffices to show that for each $U \in \mathcal B$ there is a nonempty open $V \subseteq U$ such that $V \cap \bigcup_{i \in I} A_i = \emptyset$.
Given $U \in \mathcal B$, let $i_1 , \ldots , i_n$ enumerate those $i \in I$ such that $U \cap A_i \neq \emptyset$. Setting $V_0 = U$, use the nowhere denseness of the $A_i$ to inductively pick for each $j \leq n$ a nonempty open $V_j \subseteq V_{j-1}$ such that $V_j \cap A_{i_j} = \emptyset$. Then $V = V_n \subseteq U$ is nonempty open, and $V \cap \bigcup_{i \in I} A_i = \emptyset$.
