The question is as follows:
Let $f:E\rightarrow \mathbb{R}$ where $E\subset \mathbb{R}$. The function $f$ is continuous if and only if $\forall$ closed sets $A\subset \mathbb{R}, \exists$ a closed set $B\subset \mathbb{R}$ such that $f^{-1}(A)=B \cap E.$
I proved this for the same question but with open sets. I started with the fact that $$\exists \ \epsilon _{x} > 0 \ \ \ \ \text{such that} \ \ \ \ N_{\epsilon _{x}}(f(x)) \subset A.$$ Since $f$ is continuous, $$\exists \ \delta _{x}> 0 \ \ \ \ \text{such that} \ \ \ \ N_{\delta _{x}}(x)\cap E\subset f^{-1}(N_{\epsilon _{x}}(f(x)).$$ Hence $$B\equiv \bigcup_{x \in f^{-1}(A)}^{ } N_{\delta _{x}}(x) \ \ \ \text{is open.}$$ Therefore we can conclude $$f^{-1}(A) \subset \bigcup_{x \in f^{-1}(A)}^{ } N_{\delta _{x}}(x)\cap E = B\cap E\subset f^{-1}(A).$$ So, $f^{-1}(A)=B \cap E.$ And proving the other way around is similar.
For the question I am trying to answer, I am trying to use the same idea just with the notion of closed sets. However, I am stuck. I think I will have to somehow integrate the complement of the set $A$ into the proof in one way or another, and then make a conclusion. I am still not sure.
Finally, can this even be proved in a similar manner? Any help would be much appreciated.
Thanks in advance!
Using your results for open sets, the
closed set proposition is simple.
Assume f is continuous.
If K is closed, then R - K is open.
Thus f$^{-1}$(R - K) = R - f$^{-1}$(K)) = R is open.
Hence f$^{-1}$(K)) is closed.
Use the same method to show if the inverse images of
closed sets are closed, then the function is continuous.