Question about injection, surjection, bijection, and mapping

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Discussion Overview

The discussion revolves around the concepts of injection, surjection, bijection, and mapping in the context of the function f(x) = x³ - x + 1. Participants explore the conditions under which this function is bijective and discuss specific intervals where it may or may not meet the criteria for being a bijection.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant states that f(x) is a bijection if it is both a surjection and an injection, providing the function f(x) = x³ - x + 1 as an example.
  • Another participant points out that f(x) is not injective because f(1) = f(-1) = 1, but claims it is surjective over all real numbers.
  • A participant questions whether the function is bijective on the interval (-∞, -1] in addition to [1, +∞) and suggests that the union of these intervals could represent a bijection.
  • Several participants clarify that (-∞, -1] ∪ [1, +∞) is not a valid interval, suggesting that other intervals, such as {1}, could be considered.
  • One participant proposes that the intervals [-∞, -√3/3], [-√3/3, √3/3], and [√3/3, ∞) are maximal intervals where f(x) is injective.
  • There is a discussion about the requirement for the question to specify an interval greater than a single point, with some participants emphasizing the importance of this condition.

Areas of Agreement / Disagreement

Participants generally agree that (-∞, -1] ∪ [1, +∞) is not a valid interval for the function to be bijective. However, there is no consensus on the specific intervals that should be considered acceptable for the function to be bijective, as different intervals are proposed and debated.

Contextual Notes

Some participants note the significance of the "turning points" at x = ±√3/3 in determining the injectivity of the function, but the implications of these points remain unresolved in the discussion.

Who May Find This Useful

This discussion may be useful for students or individuals studying mathematical functions, particularly those interested in the properties of bijections, injections, and surjections in real analysis.

Nok1
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f(x) is a bijection if and only if f(x) is both a surjection and a bijection. Now a surjection is when every element of B has at least one mapping, and an injection is when all of the elements have a unique mapping from A, and therefore a bijection is a one-to-one mapping.

Let's say that f(x) = x3-x+1.

It's easy to see that it is not injective by showing that f(1)=f(-1)=1
Since the function is defined for all x, it is surjective (-inf, +inf)

Then the example finds an interval such that the function is a bijection on a mapping of S to S such that S is a subset of R.

The example problem says that the function is a bijective on the interval [1,+inf). As I see it, it would also be surjective on the interval (-inf,-1].

Is that correct and the book just doesn't mention it since it just asks for "a single" interval? or is there a reason that this interval does not work?

It would seem that the union of [-inf, -1) (1, +inf] would be the interval on which the function is a bijection.

Thanks a lot for your time.
 
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[itex](-\infty,-1]\cup[1,+\infty)[/itex] is not an interval, so it doesn't work. But other answers are possible, for example S = {1}.
 
CRGreathouse said:
[itex](-\infty,-1]\cup[1,+\infty)[/itex] is not an interval, so it doesn't work. But other answers are possible, for example S = {1}.
Okay cool. The question does state an interval which is more than a single point.

So [itex](-\infty,-1][/itex] would be an acceptable interval then is the conclusion?
 
Yes, it would. In fact, since the "turning points" are at [itex]x= \pm\sqrt{3}/3[/itex], "maximal" intervals on which f(x)= x3- x+ 1 is an injection are
[tex]\left[-\infty, -\frac{\sqrt{3}}{3}\right][/tex]
[tex]\left[-\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right][/tex]
and
[tex]\left[\frac{\sqrt{3}}{3}, \infty\right][/tex]
 
Nok1 said:
Okay cool. The question does state an interval which is more than a single point.

{1} = [1, 1] is an interval.
 
Yes, of course. The point was that the question asked to find "an interval which was more than a single point". It wouldn't be necessary to add "which was more than a single point" if there were no intervals containing only a single point.
 
HallsofIvy said:
Yes, of course. The point was that the question asked to find "an interval which was more than a single point". It wouldn't be necessary to add "which was more than a single point" if there were no intervals containing only a single point.

Yes, but that was posted only after my post.
 
Sorry about being unclear in the beginning. Thanks for the replies.
 

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