Question about injection, surjection, bijection, and mapping

  • Thread starter Nok1
  • Start date
  • #1
18
0

Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
CRGreathouse
Science Advisor
Homework Helper
2,820
0
[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}.
 
  • #3
18
0
[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?
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,769
911
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]
 
  • #5
CRGreathouse
Science Advisor
Homework Helper
2,820
0
Okay cool. The question does state an interval which is more than a single point.
{1} = [1, 1] is an interval.
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,769
911
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.
 
  • #7
CRGreathouse
Science Advisor
Homework Helper
2,820
0
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.
 
  • #8
18
0
Sorry about being unclear in the beginning. Thanks for the replies.
 

Related Threads for: Question about injection, surjection, bijection, and mapping

Replies
5
Views
2K
Replies
8
Views
4K
Replies
5
Views
2K
Replies
10
Views
2K
Replies
13
Views
2K
  • Last Post
Replies
3
Views
2K
Replies
3
Views
9K
Top