# An Example of a Non-continuous Injection on a Real Interval, that is Not Strictly Monotonic.

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.

• You could also use the mean value theorem to show the converse to that statement – JB071098 Nov 17 '18 at 12:51
• But that isn't my question? – FuzzyPixelz Nov 17 '18 at 12:53

Consider $$I = [0,2]$$ and $$f$$ defined on $$I$$ via $$f(x) := x$$ when $$x \leq 1$$ and $$f(x):= -x$$ when $$x > 1$$.