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.

Solution 1:

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.

Solution 2:

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

Solution 3:

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.

Solution 4:

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.