Prove that $\sup\{f(x) : x < s\} ≤ \inf\{f(y) : y > s\}$ and both values are real numbers given an increasing function

Question

Suppose there's an increasing function $f : R → R,$ indicating that $f(x) ≤ f(y)$ whenever $x ≤ y.$

My question is: given $s ∈ R,$ how would I show that $\sup\{f(x) : x < s\} ≤ \inf\{f(y) : y > s\}$ and also that both values are real numbers ?

Answer 1

You have the inequality $f(x) \leq f(y)$ for all $x \leq y$, so take $x_n$ strictly increasing to $s$ and $y_n$ strictly decreasing to $s$. Necessarily $x_n < s < y_n$ for each $n$, so $f(x_n) \leq f(y_n)$, so $\sup\{f(x): x < s\} = \lim_n f(x_n) \leq \lim_n f(y_n) = \inf\{f(y): y > s\}$.