# Is the union of two nowhere dense sets nowhere dense?

Is the union of two nowhere dense sets nowhere dense?

Using the following definition:

A nowhere dense set is a subset $E\subset X$ of a metric space (or topological space) $X$ such that $(\overline{E})^o=\emptyset$.

I tried using topological properties like "union of closure of sets is the closure of union", and others. I tried also using the fact that $(\overline{A})^c={(A^c)}^o$ and other complement elementary-set-theory identities.

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What do you mean by can't do a proof using $\dots$? Not allowed to? If you are allowed, the closure of a union of a finite number of sets is the union of the closures. – André Nicolas Oct 30 '12 at 3:07
Can't you prove the closure of the union is the union of the closures? – Gerry Myerson Oct 30 '12 at 3:07
Sorry, edited... – Gastón Burrull Oct 30 '12 at 3:38
Seems, more tricky than expected ...actually, I'm not sure wether this is true though everywhere stated – Alexander Frei Jan 29 '14 at 1:18
...though the union of interiors can be strictly smaller than the interior of unions as in $(\mathbb{Q})°\cup(\mathbb{R}\setminus\mathbb{Q})°=\varnothing\subsetneq\mathbb{‌​R}=(\mathbb{Q}\cup\mathbb{R}\setminus\mathbb{Q})°$ – Alexander Frei Jan 29 '14 at 1:34

Put another (equivalent) way, a set $A$ is nowhere dense iff for every non-empty open set $U$ there is a non-empty open set $V$ such that $V\subseteq U$ and $A\cap V=\emptyset$. (I leave it to you to prove the equivalence.) That version should make your task much simpler.

Added: It's worth noting that an easy inductive argument shows that any finite union of nowhere dense sets is again nowhere dense. However, we cannot in general extend this result to a countable union of nowhere dense sets. For example, the rationals are dense in the reals, but are readily a countable union of nowhere dense sets

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Ty, Ill try with this definition. – Gastón Burrull Oct 30 '12 at 3:31
Isn't this definition useless given that for any open set $U$ we have $\emptyset\subseteq U$ so it is trivial that $A\cap \emptyset=\emptyset$ – user160110 Dec 4 '15 at 23:48
@user160110: Many thanks! How embarrassing! You are quite correct. It should be fixed now. – Cameron Buie Dec 5 '15 at 0:05
@CameronBuie may I ask why it is the case that if $int(\bar{A})=\emptyset$ that when $U\cap A\neq \emptyset$ there must be an open set $V\subset U$? – user160110 Dec 5 '15 at 5:01
@user160110: Consider the set $U\setminus\bar A.$ What can you say about it? – Cameron Buie Dec 5 '15 at 14:22

Let A be a nowhere dense set. Then $(\overline{A})^o=\emptyset$. This is equivalent to saying that $(\overline{A})^c$ is dense in X. Let A and B be two nowhere dense sets. Let $S=A \cup B$. To show S is nowhere dense we will show that $(\overline{S})^c$ is dense in X, that is $(\overline{S})^c$ meets every non-empty open set. Let G be a non-empty open set. Now as $A , B$ are nowhere dense, $G\cap(\overline{A})^c\neq\emptyset$ and $G\cap(\overline{B})^c\neq\emptyset$. Also $(\overline{A})^c$,$(\overline{B})^c$ are open. Hence $G\cap(\overline{A})^c\cap(\overline{B})^c\neq\emptyset$, since $G\cap(\overline{A})^c$ is non-empty open and $(\overline{B})^c$ is dense in X. Thus $G\cap(\overline{A}\cup\overline{B})^c\neq\emptyset$, which implies $G\cap(\overline{S})^c\neq\emptyset$. Hence $(\overline{S})^c$ is dense in X. Equivalently S is nowhere dense.

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