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Every non-decreasing function from R to R is injective? Prove or provide counter-example if False.

False

Definition of increasing:
for all x and y, x <= y then f(x) <= f(y)

If f is not injective then there exists x1 ≠ x2 such that f(x1) = f(x2)
Example: f(x) = 0

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    "Nondecreasing" means that if $x_1$f(x_1)\leq f(x_2)$. Unfortunately this is also what "increasing" usually means without qualification. For the condition that $x_12011-03-23
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    Hint: Consider a constant function on $\mathbb{R}$2011-03-23
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    There are problems with your argument besides misinterpreting "nondecreasing". "Therefore $x_1\neq x_2$" doesn't really make sense, because that was your premise. *If* you could show that $x_1\neq x_2$ implies $f(x_1)\neq f(x_2)$, then you would be done.2011-03-23
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    So what does non-decreasing really mean?2011-03-23
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    @1337holiday: It means what I said in the first sentence of my first comment.2011-03-23
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    So this is false then because according to Alex's answer if f(x) = 0, its not injective.2011-03-23
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    @1337holiday: Correct.2011-03-23
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    What about non-increasing? It would be false then to say that all are injective because the following can be true f(x1) = f(x2) according to the definition of non-increasing?2011-03-23
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    @1337holiday: Your edited argument is incorrect. You cannot prove that all nondecreasing functions are not injective, because this is false. You want to prove that not *every* non-decreasing function is injective, for which it suffices to give an example (as Alex has) of a nondecreasing function that is not injective. Also, if $x_1=x_2$, then $f(x_1)=f(x_2)$ is true for any function, and doesn't show anything. Not being injective means there exist $x_1\neq x_2$ such that $f(x_1)=f(x_2)$.2011-03-23

2 Answers 2

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This is false. The function $f(x) = 0$ is continuous, non-decreasing, and clearly not injective.

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Alex, example works. Infact all constant functions are also an example.

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    YOu just need constant for a little while. Then you can do anything.2011-03-23