What does $\overline{f(E)}$ mean exactly?

Question

I am prove that if $f$ is continuous, then $f(\overline{E}) \subset \overline{f(E)}$.

I am not clear on what $\overline{f(E)}$ is?

Answer 1

If $f:X \to Y$, and $E \subseteq X$, it means the closure of the set $f(E) \subseteq Y$

Answer 2

In your case, $\overline{f(E)}$ means the closure of $f(E)$, the image of $E$ by $f$.

The closure $\overline F$ of a set $F$, is defined as follow: it is the smallest (in the sens of $\subseteq$) set $G=\overline F$ such that $F\subseteq G$.

It exists since:

$$\overline F=\bigcap_{G\supseteq F\text{ and $G$ is closed}}G.$$

For instance, if $F$ is closed, you cans see that $\overline F=F$.

Answer 3

Given any subset $A$ of a topological space $(X,\tau)$, $\bar{A}$ (the closure of $A$) is the smallest closed set containing $A$. Equivalently, it is the set of the elements $x\in X$ such that, for every neighbourhood $V$ of $x$, $V \cap A \neq \emptyset$.

Answer 4

Considering your tag (real analysis).

The elements of $F(E)$, are in the form $F(x_n)$ given $x_n \in E$. Suppose $F(x_n)=:y_n$ define a convergent sequence $(y_n)$, the set $\overline{F(E)}$ is the set of all limits in the form $F(x_n)$.