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Theorem

Let $I$ be a real interval. Let $f:I \to \mathbb R$ be an injective continuous real function.
Then $f$ is strictly monotone.

If the condition on continuity indeed is necessary (it most probably is but I would prefer to see it more clearly...), then we would be able to find such a function. Otherwise, I wouldn't be so convinced. I've seen the proof that uses the Intermediate Value Theorem and I do realize it requires continuity, obviously. But I don't know if we could do without it.

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  • $\begingroup$ You could also use the mean value theorem to show the converse to that statement $\endgroup$ – JB071098 Nov 17 '18 at 12:51
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    $\begingroup$ But that isn't my question? $\endgroup$ – FuzzyPixelz Nov 17 '18 at 12:53
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Consider $I = [0,2]$ and $f$ defined on $I$ via $f(x) := x$ when $x \leq 1$ and $f(x):= -x $ when $x > 1$.

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  • $\begingroup$ Well, I really feel silly right now... Thank you. $\endgroup$ – FuzzyPixelz Nov 17 '18 at 13:23

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