# 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.

• 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... – Gaston Burrull Oct 30 '12 at 3:38
• Seems, more tricky than expected ...actually, I'm not sure wether this is true though everywhere stated – C-Star-W-Star Jan 29 '14 at 1:18
• ...though the union of interiors can be strictly smaller than the interior of unions as in $(\mathbb{Q})°\cup(\mathbb{R}\setminus\mathbb{Q})°=\varnothing\subsetneq\mathbb{R}=(\mathbb{Q}\cup\mathbb{R}\setminus\mathbb{Q})°$ – C-Star-W-Star 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

• Ty, Ill try with this definition. – Gaston 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.

Suppose A and B are any two nowhere dense subsets of a space X.

Suppose U is any non-empty open set in X.

Since A is nowhere dense, corresponding to $\;U\;$there is a non-empty open set V such that $\;V\subset U\;$and $\;V\cap A=\Phi\;.\;$............(1)

Since B is nowhere dense, corresponding to $\;V\;$there is a non-empty open set W such that $\;W \subset V\;$and $\;W \cap B = \Phi\;.\;$............(2)

From (1) and (2) we get , corresponding to any non-empty open set $\;U\;$there is a non-empty open set W such that $\;W \subset U\;$and $\;W \cap (A \cup B) = \Phi\;.\;$............(3)

From (3) we get $\;A\cup B\;$ is a nowhere dense set.

Therefore, finite union of nowhere dense sets is nowhere dense.

Let $$G = int [cl( A \cup B)]$$ where $$A$$ and $$B$$ are nowhere dense sets $$\implies G \subseteq cl(A \cup B) = cl(A) \cup cl(B) \implies G \cap (cl(B))^c \subseteq [cl(A) \cup cl(B)] \cap [cl(B)]^c = cl(A) \cap [cl(B)]^c \subseteq cl(A) \implies int [G \cap (cl(B))^c]\subseteq int [cl(A)] \implies G \cap (cl(B))^c \subseteq \emptyset$$ since $$A$$ is nowhere dense and $$G \cap (cl(B))^c$$ is open $$\implies G \subseteq cl(B) \implies G \subseteq int [cl(B)]$$ since $$G$$ is open $$\implies G = \emptyset$$ since $$B$$ is nowhere dense $$\implies A \cup B$$ is nowhere dense.

Now, I have the impression that this section is rather muddled. Having tried to find a solution myself, I would like to display a coherent, short solution which is - in essence - a detailed version of P.SUNDARAM's beautiful answer and then bring this HINT by CAMERON to a close by displaying how to use the fact he stated.

Solution not using Cameron's hint: Assuming a non-empty, open subset $U$ of $\overline{A \cup B} = \bar{A} \cup \bar {B}$, let $V = U\cap (X-\bar{A})$. Then $V$ is open and non-empty - the latter because otherwise $U\subset\bar{A}$, contradicting the assumption that $A$ is nowhere dense. Then $V\subset\bar{B}$, but that is impossible by the same reasoning. So we reached a contradiction to the assumption that $U$ was non-empty, meaning that any open subset of $\overline{A \cup B}$ is empty.

Using Camerons hint: Assuming a non-empty, open subset $U$ of $\overline{A \cup B}$, there would exist non-empty, open subsets $V_1$ and $V_2$ of $U$, s.t. $V_1\cap A = \emptyset = V_2\cap B$, by the hint. But any element in $\overline{A}$ has the property that any open neighborhood if it has non-empty intersection with $A$. But $V_1$ is an open neighborhood of any of its points, so $V_1\subset \overline{B}$ and $V_2\subset \overline{A}$. But this contradicts the assumption that $A$ and $B$ are nowhere dense. So $U$ must be empty, and hence the interior of $\overline{A \cup B}$ must be empty as well.