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Inverse image and continuity

by im2fastfouru
Tags: continuity, image, inverse
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im2fastfouru
#1
Apr1-08, 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.
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sutupidmath
#2
Apr1-08, 07:07 PM
P: 1,633
Quote Quote by im2fastfouru View Post
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

[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.
Pere Callahan
#3
Apr1-08, 08:19 PM
P: 587
Quote Quote by sutupidmath View Post
[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]

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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