What is a Direct image and Inverse Image in Real Analysis?

[phillyolly]

Homework Statement

I am trying to understand the definition of Direct and Inverse Images in Real Analysis I from my book, see attachment please.

 

[phillyolly]

Please explain how the set of the inverse image of G was found.

 

[Raskolnikov]

We're looking for all elements in the domain of f that can be mapped into elements of G, i.e., find all x such that f(x) is an element of G. For what values of x in the domain is 0 <= f(x) <= 4?

 

[phillyolly]

Right, but how did they find 0 and 4 so that

0 <= f(x) <= 4

?

 

[Raskolnikov]

Read the text step-by-step.

We're given E = {0 <= x <= 2}. The image of E is G = {0 <= f(x) <= 4}. Now we are finding the inverse image of G by the steps I described above. The purpose of this exercise is to just give you an example of where $$f(f^{-1}(A)) = A$$ and an example where $$f^{-1}(f(A)) \neq A.$$

 

[HallsofIvy]

The given function was $f(x)= x^2$ and $E= \{0\le x\le 2\}$.

They got $0\le f(x)\le 4$ by squaring 0 and 2.

But be careful, if it had been $E= \{-2\le x\le 2\}$ We would NOT have the image $\{4\le f(x)\le 4}$. $-2\le x\le 2$ includes all numbers from -2 to 2 and the squares of those are all between 0 and 2, also the squares of the negatives are also positive, not negative: If $E= \{-2\le x\le 2\}$ we would still have the image as $\{0\le f(x)\le 4\}$

 

[Mark44]

The given function was $f(x)= x^2$ and $E= \{0\le x\le 2\}$.

They got $0\le f(x)\le 4$ by squaring 0 and 2.

But be careful, if it had been $E= \{-2\le x\le 2\}$ We would NOT have the image $\{4\le f(x)\le 4}$. $-2\le x\le 2$ includes all numbers from 0 to 2

Correction: all numbers between -2 and 2 (inclusive).

and the squares of those are all between 0 and 2, also the squares of the negatives are also positive, not negative: If $E= \{-2\le x\le 2\}$ we would still have the image as $\{0\le f(x)\le 4\}$

 

[HallsofIvy]

Thanks for the correction, Mark44. I will now edit so I can claim I never made that mistake!