Can someone check my proof?
If $f$ is a continuous mapping of a metric space $X$ into a metric space $Y$, prove that $$f(\overline{E}) \subset \overline{f(E)} $$
for every set $E\subset X$. ($\overline{E}$ denotes the closure of $E$.)
Proof: Suppose $x\in \overline{E} = E\cup E'$, where $E'$ is the set of limit points of $E$.
If $x\in E$, then $f(x)\in f(E)\subset \overline{f(E)}$ and we are done.
Now suppose $x\in E'$ and let $\epsilon >0$ be given. Since $f$ is continuous, there exists a $\delta >0$ such that $d(x,y)<\delta$ implies $$d(f(x),f(y))<\epsilon. $$
Since $x$ is a limit point of $E$, there is a neighborhood $N_{\delta}(x)$ of radius $\delta$ about $x$ such that $(N_{\delta}(x)\setminus \{x\}) \cap E \neq \emptyset$. Then $p\in N_{\delta}(x)$ implies that $d(f(x),f(p))<\epsilon$.
Hence, for any $\epsilon >0$, we can always find a $\delta$ such that
$$\left( N_{\epsilon}(f(x)) \setminus \{f(x)\}\right) \cap f(E) \neq \emptyset.$$
Thus $f(x)\in f(E)'\subset \overline{f(E)}$.
I'm not really sure about the last part, how can we make certain that if $p\in N_{\delta}(x) \setminus \{x\}$, then $f(p) \in N_{\epsilon}(f(x))\setminus \{f(x)\}$. Since we aren't given that $f$ is injective, we can have $x\neq p$ and $f(x) = f(p)$? Any input or alternative approaches would be greatly appreciated!
Best Answer
Since $f(E)\subset \overline{f(E)}$ then, $$E\subset f^{-1}(f(E))\subset f^{-1}(\overline{f(E)}).$$ Since $f$ is continuous and $\overline{f(E)}$ is closed, then $f^{-1}(\overline{f(E)})$ is closed. Thus, $\overline{E}\subset f^{-1}(\overline{f(E)})$ and therefore $f(\overline{E})\subset \overline{f(E)}$.