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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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    @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