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One to one function is monotone?

  1. Dec 6, 2015 #1
    1. The problem statement, all variables and given/known data
    A one to one function f: ℝ→ℝ is monotone, True or False

    2. Relevant equations


    3. The attempt at a solution
    I think the statement is false, for example: Let I =[0,1]∪[2,3] f(x)=x if x∈[0,1], f(x)=5-x,x∈[2,3]
     
  2. jcsd
  3. Dec 6, 2015 #2

    andrewkirk

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    You have not defined the value of f(x) outside of I, so the example does not meet the requirements of the question, which include that the domain be all of ##\mathbb{R}##.

    I think the statement is false, but a little more work is needed to produce a counterexample.
     
  4. Dec 6, 2015 #3
    Let f(x)=x, x is rational, f(x)=-x,x is irrational, the function is one to one,but it is jumping. Does this example apply?
     
  5. Dec 6, 2015 #4

    fresh_42

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    Yes. I thought f(x)=const. would also do, but it depends on whether f has to be strictly monotone or not. Yours is better.
     
  6. Dec 6, 2015 #5

    andrewkirk

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    Yes, that rational/irrational function is a good one.
    By the way, it's possible to extend your function in the OP to a non-monotone, injective function that has the entirety of ##\mathbb{R}## as domain. But the rational/irrational one in post 2 is easier to specify (albeit harder to visualize).
     
  7. Dec 6, 2015 #6

    fresh_42

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    Since both are dense, it's just a big X. And it leads to interesting philosophical questions: what one draws is always discrete for you put carbon atoms on the paper. (I apologize, if that remark should be regarded as improper.)
     
  8. Dec 6, 2015 #7

    WWGD

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    It is also an example of a function which is continuous when restricted to 2 dense subsets, but overall not continuous.
     
  9. Dec 6, 2015 #8

    SammyS

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    Of course you could have f(x) = x outside of the interval (-1, 1) and f(x) = -x on the interval (-1, 1) .

    or even simpler :

    f(1) = -1, f(-1) = 1, otherwise, f(x) = x.
     
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