Proof help - Calculus - infimum of $f(x)$

Question

I need some help proving the following question:

Let $a < b$ and let $f : [a, b] \to \mathbb R$ be an increasing function.

Prove that there exists an $s\in \mathbb R$ such that $s = \inf \{f (x) ~| ~a < x ≤ b\}$.

So, I would like to say that due to the definition of increasing function, for every $x_{1} < x_{2}$, we get $f(x_{1}) < f(x_{2})$.

From here I am a little confused, because it seems so obvious that there exists an infimum for this set.

Every hint will be so appreciated!

Thanks a lot!

Answer 1

The number $f(a)$ is a lower bound of the set $\{f(x)\,|\,a<x\leq b\}$ because $f$ is increasing.

Therefore the set has a real infimum as a consequence of the completeness of the real numbers.

Note that $f$ is not assumed to be continuous, so the infimum is not necessarily $f(a)$.