One to one function is monotone?

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Homework Help Overview

The discussion revolves around the question of whether a one-to-one function defined from ℝ to ℝ is necessarily monotone. Participants are exploring the implications of this statement and examining various examples and counterexamples.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants attempt to provide counterexamples to the statement, while others question the completeness of these examples, particularly regarding the definition of the function outside specified intervals. There is also discussion about the nature of monotonicity and whether strict monotonicity is required.

Discussion Status

The discussion is active, with participants sharing examples and engaging in clarifying questions. Some guidance has been offered regarding the need for a complete definition of the function across its entire domain. Multiple interpretations of monotonicity are being explored, particularly in relation to the examples provided.

Contextual Notes

Participants note the importance of defining the function on the entirety of ℝ and consider the implications of discontinuities in the context of monotonicity. There is also mention of philosophical considerations regarding the visualization of functions.

HaLAA
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Homework Statement


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

Homework Equations

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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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.
 
andrewkirk said:
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?
 
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.
 
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).
 
andrewkirk said:
(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.)
 
It is also an example of a function which is continuous when restricted to 2 dense subsets, but overall not continuous.
 
HaLAA said:
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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