# Thread: if A is an open subset of R, then A is dense if and only if R - A is nowhere dense

1. ## if A is an open subset of R, then A is dense if and only if R - A is nowhere dense

i am trying to prove the => direction first.

assuming that A is dense, i know that $\displaystyle \forall (a, b) \subseteq \mathbb{R}, \exists x \in A \ s.t. \ x \in (a, b)$

i also know that A is open so $\displaystyle x \in A \Rightarrow \exists B_{r}(x) \subseteq A$

the trouble i am having is that i am not sure if $\displaystyle B_{r}(x) \subseteq (a, b)$ because if that was true then it follows that $\displaystyle \mathbb{R} - A$ is nowhere dense.

could someone help me continue the proof? thanks.

2. ## Re: if A is an open subset of R, then A is dense if and only if R - A is nowhere dens

Since $\displaystyle (a,b)$ is open, $\displaystyle A\cap (a,b)$ is open and you can choose $\displaystyle r$ such that $\displaystyle B_r(x)\subset A\cap (a,b)$.

3. ## Re: if A is an open subset of R, then A is dense if and only if R - A is nowhere dens

Originally Posted by oblixps
i am trying to prove the => direction first.
assuming that A is dense, i know that $\displaystyle \forall (a, b) \subseteq \mathbb{R}, \exists x \in A \ s.t. \ x \in (a, b)$
i also know that A is open so $\displaystyle x \in A \Rightarrow \exists B_{r}(x) \subseteq A$
the trouble i am having is that i am not sure if $\displaystyle B_{r}(x) \subseteq (a, b)$ because if that was true then it follows that $\displaystyle \mathbb{R} - A$ is nowhere dense.
$\displaystyle A\cap (a,b)$ is an open set. Is it not?