# Veryify that the union of a co-dense set and a nowhere dense set is a co-dense set

Veryify that the union of a co-dense set and a nowhere dense set is a co-dense set. Give an example to show that the union of two co-dense sets if not necessarily a co-dense set.

Note: A co-dense set $A$ in the topological $X$ denotes its complementary set is dense in $X$. And a nowhere dense set $A$ in $X$ if $\overline{A}$ is co-dense.

What I've tried: For the second question, I find an example. The real line $R$ is the whole space. $Q$ denotes all the rational numbers and $P=R-Q$ denotes all the irrational numbers. We see they are both co-dense sets. However, their union is $R$, and hence is not a co-dense. The first question is still tough for me.

Could anybody help me? New example is also welcome.

-

This is a slightly different take, using some facts from a previous answer of mine, and also the characterisation that $B \subseteq X$ is nowhere dense iff for every nonempty open $U \subseteq X$ there is a nonempty open $V \subseteq U$ such that $V \cap B = \emptyset$.

To show that $A \cup B$ is co-dense, we need only show that $\mathrm{Int} ( A \cup B ) = \emptyset$.

If $U = \mathrm{Int} ( A \cup B ) \neq \emptyset$, then as $B$ is nowhere dense there is a nonempty open $V \subseteq U$ such that $V \cap B = \emptyset$. As $A$ is co-dense, then $V \not\subseteq A$ (since $\mathrm{Int} (A) = \emptyset$), and so there is an $x \in V \setminus A$. But then $x \in U \setminus ( A \cup B )$, contradicting that $U \subseteq A \cup B$!

-
It is really new. Very appreciated your answer. I have something I don't know: Why $int A=\emptyset$ if $A$ is a codense set? Is it the reason that this equality $X-Int A=\overline{X-\overline{A}}$? – Paul Aug 7 '12 at 10:11
@Paul: You have the formula slightly wrong: $X \setminus \mathrm{Int} (A) = \overline{X \setminus A}$ (only one closure). Then $A \subseteq X$ is co-dense iff $X \setminus A$ is dense iff $X \setminus \mathrm{Int} (A) = \overline{X \setminus A } = X$ iff $\mathrm{Int} (A) = \emptyset$. – arjafi Aug 7 '12 at 10:14
O, I see. Thanks again. – Paul Aug 7 '12 at 10:28

Suppose $A$ is co-dense and $B$ is nowhere dense. So $X-A$ is dense and $X - \bar{B}$ is open and dense. So $X - (A \cup B) = (X - A) \cap (X - B) \supset (X - A) \cap (X - \bar{B})$. To show $X - (A \cup B)$ is dense, let $V$ be any open set. Since $(X - \bar{B})$ is dense $(X - \bar{B}) \cap V$ is a nonempty open set. Since $(X - A)$ is dense, there exists a $z \in (X - A)$ such that $z \in (X - \bar{B}) \cap V$. Hence there exists a $z \in (X - A) \cap (X - \bar{B}) \subset (X - A) \cap (X - B) = (X - (A \cup B))$ such that $z \in V$. Since $V$ is an arbitrary open set of $X$, $(X - (A \cup B))$ is dense. So $(A \cup B)$ is co-dense.

-
It is very importment to notice that $X-\overline{B}$ is an open set. It is what we always ignored in the process of proof. – Paul Aug 7 '12 at 7:06
@William Thanks for your answer. It is very clear. – Paul Aug 7 '12 at 7:12