Show that a set is dense and show that a set is open A point is called interior point of a set $E$ if there exist an $r>0$ for wich $(x-r,x+r0)$ is contained in $E$. The set of interior points of $E$ denoted by int$E$.
Show that:
1) $E$ is open if and only if $E =intE$.
2) $E$ is dense if and only if int $( \mathbb{R} \backslash E)= \emptyset$
Proof: (1) 
$\Rightarrow$ Let $E$ be an open set. Let $x \in E$ by definition there exist an $\epsilon >0$ such that $(x- \epsilon, x+ \epsilon)$. Since E is open all of its points are interior points thus $E\subset$ int $E$. and also the set of  int $E$ $ \subset E$. So E= int E.
$\Leftarrow$ if E= int E, then every point of $E$ is an interior point of $E$, thus by definition we can find an open interval that is contains in $E$. thus E is open. 
so far i am confident that proof 1 is good but i do not know how to go about prove 2.
 A: Suppose $E$ is dense in $\mathbb{R}$. Then, given any $x \in \mathbb{R}$, every open interval containing $x$ contains a point in $E$. Therefore, $x$ cannot be an interior point of $\mathbb{R} \setminus E$. Since $x$ was an arbitrary element of $\mathbb{R}$, this means that $\text{int} (\mathbb{R} \setminus E)$ is empty.
Conversely, suppose that $\text{int}(\mathbb{R} \setminus E)$ is empty. Choose any $x\in \mathbb{R}$. Consider two cases:
Case 1: $x \in E$, in which case every open interval containing $x$ contains a point in $E$, namely $x$ itself.
Case 2: $x \not \in E$. Then $x \in \mathbb{R} \setminus E$. Since $\mathbb{R} \setminus E$ has empty interior, $x$ is not an interior point of $\mathbb{R} \setminus E$. Therefore every open interval containing $x$ also contains a point in $E$.
In both cases, every open interval containing $x$ also contains a point in $E$. This is true for any $x \in \mathbb{R}$, so $E$ is dense in $\mathbb{R}$.
A: Note that the following statements are equivalent:


*

*$E$ is not dense

*There exists an open set $G$ that contains no points of $E$

*$\mathbb R \setminus E$ has interior points

