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Question Regarding Sets and Functions

by Diffy
Tags: functions, sets
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Diffy
#1
Sep20-07, 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?
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Diffy
#2
Sep20-07, 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?
HallsofIvy
#3
Sep20-07, 08:11 PM
Math
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Thanks
PF Gold
P: 39,502
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 "one-to-one"! If, for example, f(x)= x[sup]2[sup], where f is surely not one-to-one, 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.

Diffy
#4
Sep20-07, 09:03 PM
P: 443
Question Regarding Sets and Functions

Quote Quote by HallsofIvy View Post
Okay, I won't laugh at you too hard!
Wow, that's discouraging.

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...
Eighty
#5
Sep21-07, 10:46 AM
P: 53
Both your definition and proof are correct.


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