Is f continuous if f^-1(S) is closed for all closed sets S?

[im2fastfouru]

Any help on this problem would be appreciated

a) Show that (f^-1 S)compliment is equal to f^-1(S compliment) for any set S of reals.


Then use part a) to show The function f is continuous iff f^-1(S) is closed for every closed set S.

 

[sutupidmath]

Any help on this problem would be appreciated

a) Show that (f^-1 S)compliment is equal to f^-1(S compliment) for any set S of reals.


Then use part a) to show The function f is continuous iff f^-1(S) is closed for every closed set S.

Well, i am going to give a shot, but wait until some other people answer this as well.

Let

x\in \bar f^{-1}(S)=>x\notin f^{-1}(S)=>x\in f(S)=>y\in S=>y\notin \bar S=>x\notin f(\bar S)=>x \in f^{-1}(\bar S)


SO from here:

\bar f^{-1}(S)\subseteq f^{-1}(\bar S)

I think now you should go the other way around. I am not sure whether my reasoning is correct though. SO let someone else comment on this before you conclude on your answer.

 

[Pere Callahan]

x\in \bar f^{-1}(S)=>x\notin f^{-1}(S)=>x\in f(S)=>y\in S=>y\notin \bar S=>x\notin f(\bar S)=>x \in f^{-1}(\bar S)

Can you explain how you conclude that x must be in f(S) (third step)?

It is helpful to think of the domain and the codomain as different spaces (although here they both happen to be R, I assume).

You start with an x in the domain of f, in the statement I ask you to explain you consider the same x as an element of the codomain. I'm not saying that this can never be correct, but I think here it is not.