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.