## inverse image and continuity

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.

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 Quote by 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.
Well, i am gonna give a shot, but wait untill 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.

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

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.