One-sided limits of a monotonic function

Question

Let $f: \mathbb{R} \to \mathbb{R}$ be an increasing function. I am trying to prove that for any 2 distinct points of discontinuity $a,b$ of this function, if $a<b$, then $f(a^+)<f(b^-)$. I have already proven that any point of discontinuity is a jump discontinuity for such a function, i.e, for any $x$ such that $f$ is discontinuous at $x$, $f(x^-)<f(x^+)$. Any help on how to go about proving the condition I want? I can only use $\epsilon-\delta$ definitions here, no sequences, if that's even possible.

Answer 1

Suppose $a < b$, but $f(b^-) < f(a^+)$. Consider $\varepsilon = f(a^+) - f(b^-) > 0$ when applied to the definition of $f(a^+)$. That is, there exists some $\delta_1 > 0$ such that $$a < x < a + \delta_1 \implies |f(x) - f(a^+)| < f(a^+) - f(b^-) \implies f(x) > f(b^-).$$

Similarly, from the definition of $f(b^-)$, there exists some $\delta_2 > 0$ such that $$b - \delta_2 < x < b \implies |f(x) - f(b^-)| < f(a^+) - f(b^-) \implies f(x) < f(a^+).$$

Select some $c \in (a, \min\{a + \delta_1, b \})$ and $d \in (\max\{c, b - \delta_2\}, b)$. Then $a < c < d < b$. Further, from the two implications above, $$f(d) \le f(a^+) < f(b^-) \le f(c),$$ which contradicts $f(c) \le f(d)$, by monotonicity.