This is a practice question we were given in class:
We are given that a mapping f: $R_1$ -> $R_1$ is continuous. Prove H ={$(x, f(x)) | x$ $\in$ $R_1$} is closed. Part (II) asks whether the converse is true.
Attempt at a solution:
(Part I): Let u $\in$ $R_1$ be arbitrary. Suppose {$u_k$} is a sequence that converges to u. By the continuity of f, we know that {$f(u_k)$} converges to $f(u)$. The by the Componentwise Convergence Theorem, we know ({$u_k$},{$f(u_k)$}) converges to $H(u)$=$(u,f(u))$. Since $H(u)$ is contained in $H$, this proves the graph of $H$ is closed.
First of all, is this proof correct? Also for the converse, I'm not really sure what the answer is (I think it's True/False, with no proof necessary). I know the idea is that given the closed graph $H$, $f: R_1 -> R_1$ is continuous. So I started with:
$H$ is closed. So whenever ({$u_k$},{$f(u_k)$}) converges to ($u, f(u))$, $(u,f(u))$ is in $H$. What does this imply about the continuity of $f$? Because if {$f(u_k)$} doesn't converge to $f(u)$, this doesn't invalidate the condition. Where would I go from here, or is a different approach necessary?
Any help would be appreciated.
Your proof for the first part seems fine.
Your idea for the converse is wrong. A helpful true statement is that a function on a compact set is continuous if and only if its graph is compact.
For a counterexample, consider the function $$ f(x) = \begin{cases} 1/x & x\neq 0\\ 0 & x=0 \end{cases} $$ I claim that this discontinuous function has a closed graph.