
#1
Apr108, 04:41 PM

P: 6

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. 



#2
Apr108, 07:07 PM

P: 1,635

Let [tex] 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)[/tex] SO from here: [tex]\bar f^{1}(S)\subseteq f^{1}(\bar S)[/tex] 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. 



#3
Apr108, 08:19 PM

P: 588

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. 


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