How to prove that E is nowhere dense in X if and only if every non-empty open subset of X contains a non-empty open set disjoint form E?

Question

Following what is written at the second chapter of "Elementos de Topología General" by Fidel Casarrubias Segura and Ángel Tamariz Mascarúa.

The following statements are equivalent:

1. $E\subseteq X$ is nowhere dense;
2. $X\setminus\overline{E}$ is dense in $X$;
3. for any open and non empty set $A$ there exist a non empty open set $U$ such that $U\subseteq A$ and $U\cap E=\varnothing$.

To prove it I proceeded in the following way: if $E$ is nowhere dense set so by definition $\mathscr{int}(\mathscr{cl}(E))=\varnothing$ and so the set $F=\mathscr{cl}(E)$ is co-dense and by previous theorem we can say that $X=\mathscr{cl}(X\setminus F)=\mathscr{cl}(X\setminus\mathscr{cl}(E))$, and by another theorem we know that for any non empty open set $A$ it result that $A\cap X\setminus \mathscr{cl}(E)\neq\varnothing$ and so $U=A\setminus\mathscr{cl}(E)\neq\varnothing$ is an open (non empty) set such that is contained in $A$ and $U\cap E=\varnothing$, since $U\cap E=A\setminus\mathscr{cl}(E)\cap E\subseteq A\setminus E\cap E=\varnothing$, so we proved that $1\Rightarrow2\Rightarrow3$.

Unfortunately I can't prove $3\Rightarrow1$: so could someone help me, please?

Answer 1

In 3) the phrase ' there exist an open set $U$' should be changed to ' there exist a non-empty open set $U$'. [Otherwise the result is false since we can always take $U$ to be the empty set].

Suppose 3) holds and $E$ is not nowhere dense. Then the interior of $\overline {E}$ is a non-empty open set $A$. [$\overline {E}$ is the closure of $E$]. By 3) there exist a non-empty open set $U$ such that $U \subseteq A$ and $U \cap E =\emptyset$. If $x$ is any point of $U$ then $x \in A \subseteq \overline {E}$. But $U$ is a neighborhood of $x$ which does not intersect $E$. This is a contradiction.

Answer 2

For (3) to (1). Let $A:=\operatorname{int}(\operatorname{cl}(E))$. If $A$ is non-empty, assumption $3$ gives us non-empty open $U$ such that $U \subseteq A$ and $U \cap E = \emptyset$. But take $p \in U$ then $p \in A= \operatorname{int}(\operatorname{cl}(E)) \subseteq \operatorname{cl}(E)$ so every open neighbourhood of $p$ intersects $E$. But $U$ (which is by definition an open neighbourhood of $p$) is disjoint from $U$, contradiction!

So $\operatorname{int}(\operatorname{cl}(E)) = \emptyset$, given 3, so $E$ is nowhere dense.