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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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1  
"Nondecreasing" means that if $x_1<x_2$, then $f(x_1)\leq f(x_2)$. Unfortunately this is also what "increasing" usually means without qualification. For the condition that $x_1<x_2$ implies $f(x_1)<f(x_2)$, it's safest to go with "strictly increasing". –  Jonas Meyer Mar 23 '11 at 2:31
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Hint: Consider a constant function on $\mathbb{R}$ –  Brandon Carter Mar 23 '11 at 2:31
    
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. –  Jonas Meyer Mar 23 '11 at 2:33
    
So what does non-decreasing really mean? –  1337holiday Mar 23 '11 at 2:35
    
@1337holiday: It means what I said in the first sentence of my first comment. –  Jonas Meyer Mar 23 '11 at 2:36

2 Answers 2

up vote 3 down vote accepted

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. –  Ross Millikan Mar 23 '11 at 5:06

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