$f:X\to Y$ is a continuous map, where $X$ and $Y$ are metric spaces. Show that $f(\bar E)\subset \overline{f(E)}$.
Can I say that $x\in \bar E$, there is a sequence $\{x_n\}$ in $E$ such that $x_n\to x$, and $f$ is continuous at $x$, so $f(x_n)\to f(x)$, where $\{f(x_n)\}$ is a sequence in $f(E)$. Conclude that $f(x)\in \overline{f(E)}$.
I have read some proofs but most of them do not mention this, so I'm wondering is there something wrong about this proof?
Best Answer
Alternative proof
Since f is continuous $f^{-1}(\overline{f(E)})$ is closed. Since $f(E) \subset \overline{f(E)}$ we have $E \subset f^{-1}(\overline{f(E)})$. By definition of closure $\overline{E} \subset f^{-1}(\overline{f(E)})$. Then $f(\overline{E}) \subset \overline{f(E)}$.