Prove that $f( E^o)$ is an open set if $E$ is bounded

Question

I do not really understand how to proceed with this question; To prove a), do I need to show that every point in $f(E^0)$ is an interior point? I would greatly appreciate any help in this regard

Answer 1

a). Since Jacobian is not zero at $x$, by the inverse function theorem, $f$ is injective in a local neighborhood of $x$. Thus $f$ is an open mapping, i.e. for any $x$ in $E^o$, there is an open set $x\in G\subset E^o$ that $f(G)\subset f(E^o)$ and $f(G)$ is open. This means that $f(E^o)$ is open.

b). Since $f$ is an open mapping, $f$ maps a close set to a close set, i.e. $f(\overline{E})$ is also closed, or $\overline{f(\overline{E})}=f(\overline{E})$. By the following formulas $$ E\subset \overline{E}\quad\text{and }\quad \overline{E}=E^o\cup \partial E $$ We have $$ f(E)\subset f(\overline{E})\quad\text{and }\quad \overline{f(E)}=f^o(E)\cup \partial f(E) $$ Hence \begin{align} f^o(E)\cup \partial f(E)&=\overline{f({E})} \\ &\subset\overline{f(\overline{E})}=f(\overline{E}) \\ &=f(E^o\cup \partial E) \\ &=f(E^o)\cup f(\partial E) \end{align} c). Since $E^o\subset E$, $f(E^o)\subset f(E)$. Thus $f^o(E^o)\subset f^o(E)$. Since $f(E^o)$ is open, $f(E^o)=f^o(E^o)$. So we have $$ f(E^o)\subset f^o(E)\quad\text{and }\quad f(\overline{E})-f^o(E)\subset f(\overline{E})-f(E^o)\tag1 $$ Hence \begin{align} \partial f(E)&=\overline{f({E})}-f^o(E) \\ &\subset \overline{f({\overline{E}})}-f^o(E) \\ &=f({\overline{E}})-f^o(E) \\ &\subset f({\overline{E}})-f(E^o)\tag {by (1)} \\ &=f(E^o\cup \partial E)-f(E^o) \\ &=(f(E^o)\cup f(\partial E))-f(E^o) \\ &=f(\partial E)-f(E^o) \\ &\subset f(\partial E) \end{align}