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Question Regarding Sets and Functions 
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#1
Sep2007, 06:38 PM

P: 443

[tex]A_0 \subset f^{1} (f (A_0)) [/tex]
This inclusion is an equality if f is injective. What I can't understand is how it is even defined if f isn't a bijection. If it is not a bijection, then there is no inverse function. Is there? 


#2
Sep2007, 07:19 PM

P: 443

Ok I think I got it. If we don't know that [tex]f:A \rightarrow B[/tex] is bijective or even surjective/injective, we want [tex]f^{1}[/tex] to be [tex] \{ a  f(a) \in B\}[/tex]
is this correct? Let [tex]f:A \rightarrow B[/tex] and [tex] A_0 \subset A [/tex] Say we want to show that [tex]A_0 \subset f^{1}( f(A_0)) [/tex] Suppose we have [tex]a \in A_0 [/tex] then by the definition of a function [tex] f(a) = b [/tex] for some [tex]b \in B[/tex] [tex]f^{1}(b) [/tex] then is [tex]\{ c  f(c) =b\}[/tex] since we have already established that [tex] f(a) = b [/tex] it is clearly the case that [tex] a \in \{ c  f(c) =b\} = f^{1}(f(a))[/tex]. Therefore, since we choose [tex]a[/tex] arbitraraly [tex] A_0 \subset f^{1}(f(A_0))[/tex] Is this right? 


#3
Sep2007, 08:11 PM

Math
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Thanks
PF Gold
P: 39,345

Okay, I won't laugh at you too hard!
The very first time I had to present a proof before the class in a graduate class it was something exactly like this! I went throught the whole thing, assured that I was exactly right! I did the whole proof assuming that f HAD an inverse! Very embarrasing! It's probably the one thing I remember more than anything else from my graduate student days! f^{1}(A), where A is a set, is defined as {x f(x) is in A}. No, it is not required that f be "onetoone"! If, for example, f(x)= x[sup]2[sup], where f is surely not onetoone, then f^{1}([1,4]= {all x such that f(x) is in that set}. That, of course is the interval [2, 2] since f(2)= f(2)= 4 and all numbers between 2 and 2 are taken to numbers between 0 and 4 and so between 1 and 4. 


#4
Sep2007, 09:03 PM

P: 443

Question Regarding Sets and Functions
Anyways, I think I said your exact definition of [tex]f^{1}[/tex] in my second post. Where I said if [tex]f:A \rightarrow B[/tex] "we want [tex]f^{1}[/tex] to be [tex]\{a  f(a) \in B \} [/tex]" How was my proof of [tex] A_0 \subset f^{1} (f(A_0))[/tex]? Was that any good? If not I hope it was at least, yet again, humorous... 


#5
Sep2107, 10:46 AM

P: 53

Both your definition and proof are correct.



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