I am self studying Functional analysis from Kreyszig and I am thinking about this exercise which is a part of section Closed Graph theoram.
Problem: let $X$ and $Y$ be normed spaces and let $T: X \to Y$ be a closed linear operator.
Show that image $A$ of a compact subset $C$ of $X$ is closed in $Y$.
Show that inverse image $B$ of a compact set $K$ which is subset of $Y$ is closed in $X$.
My attempt: for 1. $A$ is compact implies $A$ is closed. $T$ is closed but $Y$ is not Banach. I have to show that if $f(c_i ) \to f(c)$ then it is closed in $X$. But I have no idea on how to proceed
Can someone please give some hints?
Let $(c_n) \subset C$ and and $T(c_n) \to y$. We have to show that $y \in A$. Since $C$ is compact there is a subsequence $(c_{n_k})$ converging to some $c \in C$. Since $(c_{n_k},T(c_{n_k})) $ is a sequence in the graph of $T$ converging to $(c,y)$ it follows this limit belongs to the graph, so $y=Tc \in A$.
Let $(x_n)$ be sequence in $T^{-1}(B)$ converging to some $x$. We have to show that $x \in T^{-1}(B)$. Now $T(x_n)$ has a convergent subsequence $T(x_{n_k})$ in $B$. Let $T(x_{n_k}) \to y \in B$. Then $(x_{n_k},T(x_{n_k})) \to (x,y)$. Hence $(x,y)$ belongs to the graph, and $y=Tx$. But then $x \in T^{-1}(B)$ because $y \in B$.