- #1
What is a Direct image and Inverse Image in Real Analysis?
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Homework Help Overview
The discussion revolves around the concepts of Direct and Inverse Images in Real Analysis, specifically focusing on the definitions and examples provided in a textbook. Participants are exploring how to determine the inverse image of a set based on a given function.
Discussion Character
- Conceptual clarification, Assumption checking, Problem interpretation
Approaches and Questions Raised
- Participants are attempting to understand how the inverse image of a set is derived from a function, particularly questioning the values that define the range of the function. There are discussions about the implications of different domains on the resulting images.
Discussion Status
The conversation is active, with participants providing insights into the definitions and examples related to the topic. Some guidance has been offered regarding the interpretation of the function and its domain, but there is no explicit consensus on the best approach to understanding the inverse image.
Contextual Notes
There are references to specific sets and functions, such as E = {0 <= x <= 2} and f(x) = x^2, which are central to the discussion. Participants are also considering how changes in the domain affect the image and inverse image, indicating a need for careful consideration of definitions and assumptions.
- #2
- #3
Raskolnikov
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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?
- #4
phillyolly
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- 0
Right, but how did they find 0 and 4 so that
0 <= f(x) <= 4
?
0 <= f(x) <= 4
?
- #5
Raskolnikov
- 193
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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 [tex]f(f^{-1}(A)) = A[/tex] and an example where [tex]f^{-1}(f(A)) \neq A.[/tex]
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 [tex]f(f^{-1}(A)) = A[/tex] and an example where [tex]f^{-1}(f(A)) \neq A.[/tex]
- #6
HallsofIvy
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The given function was [itex]f(x)= x^2[/itex] and [itex]E= \{0\le x\le 2\}[/itex].
They got [itex]0\le f(x)\le 4[/itex] by squaring 0 and 2.
But be careful, if it had been [itex]E= \{-2\le x\le 2\}[/itex] We would NOT have the image [itex]\{4\le f(x)\le 4}[/itex]. [itex]-2\le x\le 2[/itex] 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 [itex]E= \{-2\le x\le 2\}[/itex] we would still have the image as [itex]\{0\le f(x)\le 4\}[/itex]
They got [itex]0\le f(x)\le 4[/itex] by squaring 0 and 2.
But be careful, if it had been [itex]E= \{-2\le x\le 2\}[/itex] We would NOT have the image [itex]\{4\le f(x)\le 4}[/itex]. [itex]-2\le x\le 2[/itex] 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 [itex]E= \{-2\le x\le 2\}[/itex] we would still have the image as [itex]\{0\le f(x)\le 4\}[/itex]
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- #7
Mark44
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Correction: all numbers between -2 and 2 (inclusive).HallsofIvy said:
HallsofIvy said:
- #8
HallsofIvy
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Thanks for the correction, Mark44. I will now edit so I can claim I never made that mistake!
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