f is continuous, show that f(closure) is a subset of closure of f If $f:X\rightarrow Y$ is continuous and $E\subset X$, prove that $f(\overline{E})\subset \overline{f(E)}$. Provide an example to show that the inclusion does not have to be equality.
So far what I have is that the preimage of a closed set in $Y$ is closed in $X$. So $f^{-1}(\overline{f(E)})$ is closed and contains $E$.
 A: A counterexample to the equality: $f(x)=\arctan(x)$, $X={\bf R}$.
A: You're doing fine.  You've shown $E\subseteq f^{-1}(\overline{f(E)})$ and $f^{-1}(\overline{f(E)})$ is closed.  Now recall that the closure of $E$ is the smallest closed set containing $E$; so it follows from what you've shown that $\overline E\subseteq f^{-1}(\overline{f(E)})$.  Now recall (or prove) that $A\subseteq f^{-1}(B)\iff f(A)\subseteq B$, and apply it with $A=\overline E$ and $B=\overline{f(E)}$.
A: You're off to a great start.  $f^{-1}(\overline{f(E)})$ is a closed set containing $E$.  Then, since $\overline{E}$ is by one definition $E \cup E'$, where $E'$ is the set of limit points of $E$, then it must be the case that $E' \subseteq f^{-1}(\overline{f(E)})$ (why?).  But then that means $\overline{E} \subseteq f^{-1}(\overline{f(E)})$, which implies $f(\overline{E}) \subseteq \overline{f(E)}$.
A: Solution:
Let $x \in f(\overline{E})$ there is some $y \in \overline{E}$ such that $f(y) = x$, now because its in the closure we have some $y_n \in E$ with $y_n \rightarrow y$, now by continuity this means that
$$\lim f(y_n) = f(\lim y_n) = f(y) = x $$
So $x \in \overline{f(E)}$, because $(f(y_n)) \in f(E)$ for all $n$ and $f(y_n)_n \rightarrow x$
A: A counterexample, consider $X=Y=\mathbb{R}$, $E=\overline{E}=[0,\infty) \subset \mathbb{R} $ and a continuos function $ f(x) = \frac{1}{1+x^2} $ 
Note that $f(E) = f(\overline{E})=(0,1] \neq \overline{f(E)} = [0,1] $
A: First, we need this Theorem 1:

$\overline A$ is the smallest closed set containing $A$.

By Theorem 1, we know that: $f(E) \subset \overline{f(E)}$. As $f^{-1}(f(E)) = E$, we have: $E \subset f^{-1}(\overline{f(E)})$.
Then we need this Theorem 2:

$f$ is continuous if and only if for each closed set $C$ of $Y$, $f^{-1}(C)$ is closed in $X$.

By Theorem 2, we know that $f^{-1}(\overline{f(E)})$ is closed in $X$. By Theorem 1, we know that: $E \subset \overline E \subset f^{-1}(\overline{f(E)})$.
Thus: $f(\overline E) \subset \overline{f(E)}$.
