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Question: If $f$ is increasing on $(a,b)$, continuous at $a$, $b$, then $f$ is increasing on $[a,b]$.

My Work:

It remains to prove that $f(a) < f(x) < f(b)$ for $x \in (a,b)$.

Assume the contrary and without loss of generality, assume there exists $x_0 \in (a,b)$ such that $f(a) \geq f(x_0)$.

I have proved when $f(a) > f(x) \implies f(a) - f(x) > 0$

Let $\varepsilon = f(a) - f(x)$ and $\displaystyle x = \min\left(\frac{\delta}{2}+ a, \frac{x + x_0}{2}\right) \implies f(x_0) > f(x)$ since $f$ is increasing on $(a,b)$.

Then $\forall \delta > 0$

If $0 < x - a < \delta$ then $\varepsilon = f(a) - f(x_0) < f(a) - f(x) \implies |f(x) - f(a)| > \varepsilon$.

Therefore, $f$ is not continuous at $a$. Contradiction.

Is there anything wrong with this proof?

Also, how would you get the contradiction if $f(a) = f(x_0)$ for some $x_0 \in (a,b)$?

Setting $\varepsilon = f(a) - f(x_0) + 1$ does not do much.

Thanks.

2 Answers 2