Prove that $f(\overline C ) \subset \overline{ f(C)}$ . Prove that $f: A \to B$ is continuous iff its graph is compact where $A$ is compact and $A,B$ are metric spaces.
My attempt: I have already proved it. But somehow i am not satisfied with my proof. Implies part is Ok. But converse part i want to prove f is continuous by using $f(\overline C) \subset \overline{ f(C)}$.
For this i took $C\subset E$ arbitrary. Let $T=\{(x, f(x)):\, x  \in C\}$, $F=\{(x, f(x)): x \in \overline{C}\}$. Now $\overline T$ is closed in graph(f),  hece it is compact. I cant proceed further. Please give me hint.
 A: I'll try proving what the title states when $f$ is continuous.
If $x\in f(\overline{A})$ then there exists $a\in \overline{A}$ such that $f(a)=x$, now since it is in the closure, we have
$$\forall U \mbox{ open set containing } a \mbox{, } U\cap A\not=\emptyset$$
Now, taken an open set $V$ containing $x$, since $f$ is continuous, $f(V)^{-1}$ is a an open set containing $a$, so as we saw $A\cap f(V)^{-1} \not=\emptyset$.
Now, taking it's immage, $f(\emptyset)\not=f(A\cap f(V)^{-1}) \subseteq f(A)\cap V$, so every $V$ intersects $f(A)$: by definition $x$ is in the closure of $f(A)$
I have show that $x\in f(\overline{A}) \Rightarrow x\in \overline{f(A)}$; this means $f(\overline{A})\subseteq \overline{f(A)}$.
A: As noted in the comments, I don't think what you're asking to prove will actually solve the desired question.  So I will use sequences to prove your question.  So suppose your graph is compact.  Note that since we are dealing with Metric spaces then we have the Bolzano-Weierstrass property (in the sense that every sequence in our graph has a convergent subsequence).
Let $C \subset f(A)$ be closed and let $x_n \in f^{-1}(C)$ with $x_n \rightarrow x \in A$.  Since our graph is compact then the sequence $(x_n, f(x_n))$ has a convergent subsequence in the graph of $f$, i.e. $(x_{n_k}, f(x_{n_k})) \rightarrow (x,y)$, which must be in the graph of $f$ since our graph is closed.  But then $(x,y) = (x,f(x))$, and $f(x_{n_k}) \rightarrow f(x)$ so then $f(x) \in C$.  Finally, this means that $x \in f^{-1}(C)$, thus $f^{-1}(C)$ is closed and $f$ is continuous.
