One to one function is monotone?

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Homework Statement


A one to one function f: ℝ→ℝ is monotone, True or False

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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]
 

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  • #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.
 
  • #3
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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.
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?
 
  • #4
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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.
 
  • #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).
 
  • #6
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(albeit harder to visualize).
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.)
 
  • #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.
 
  • #8
SammyS
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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?
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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