TITLE: Theorem 7.2 in General Topology by S. Willard (characterization of continuity using $f(\overline E) \subset \overline{f(E)}$
QUESTION [5 upvotes]: Theorem 7.2
If $X$ and $Y$ are topological spaces and $f:X \to Y$, then the following are all equivalent :-
I) $f$ is continuous.
II) for each $E \subset X$, $f(\bar E) \subset \overline{f(E)}$.

Proof:- (II) $\implies$ (I)

Let $x\in X$ and let $V$ be an open neighbourhood of $f(x)$.
Set $E = X-f^{-1}(V)$ and $U=X-\bar E$.
It is easy to verify that, since $f(\bar E) \subset \overline{f(E)}$, we have $x \in U $.
It is even clearer that $f(U)\subset V$
Hence $f$ is continuous at $x$

I don't understand this part of the proof. I cannot understand how  $x\in U$
Can someone help?

REPLY [2 votes]: Suppose that $x\notin U$ then $x\in\bar{E}$ and hence $f(x)\in f(\bar{E})\subseteq\overline{f(E)}\subseteq\overline{f(X)\cap(X\setminus V)}\subseteq\overline{X\setminus V}$ by definition of $E$. Therefore every neighborhood of $f(x)$ touches $X\setminus V$ and also $V$ (since $f(x)\in V$). This means that $f(x)$ is in the boundary of $V$. But since $V$ is open the boundary of $V$ is equal to $\bar{V}\setminus V$ and in particular disjoint from $V$ - a contradiction to $f(x)\in V$.