Here is the question:
Prove that the function $f: E \rightarrow E'$ between metric spaces is continuous iff whenever $ C\subset E'$ is closed, $f^{-1}(C)$ is a closed subset of $E$.
What I was thinking of doing is this following:
First, assume F is continuous. So we want to show that whenever $ C\subset E'$ is closed, then $f^<(C)$ is a closed subset of $E$. To do this, do we just need to show that whenever $C^c \subset E'$ is open, then $f^<(C^c)$ is open?
Best Answer
Suppose $f:E\to E'$ is continuous, then if $N\subset E'$ is open, we know that $f^{-1}(N)\subset E$ must also be open. This is the open set condition for continuity. In the case of my course this was the definition of continuity. So, we basically want to show equivalence of the open and closed set conditions.
Suppose $f$ is continuous. Suppose some $C\subset E'$ is closed. Then, we know that $C^C$ is necessarily open. This implies that $f^{-1}(C^C)$ is also open. So, $(f^{-1}(C^C))^C$ must be closed, $(f^{-1}(C^C))^C=f^{-1}(C)$. Thus, $C$ closed implies $f^{-1}(C)$ closed for all continuous $f$.
Can you prove the converse of the statement?