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

**oblixps** · August 27th 2011, 10:38 PM

i am trying to prove the => direction first.

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

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

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

could someone help me continue the proof? thanks.

**girdav** · August 28th 2011, 02:22 AM

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

**Plato** · August 28th 2011, 02:33 AM

Originally Posted by oblixps

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

$A\cap (a,b)$ is an open set. Is it not?

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

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