Proof that one-sided limits of a monotone function exist

Proof

Here $f$ is increasing in the sense of a monotone function: $x \le y$ implies $f(x) \le f(y)$.

1. Let $S = \set{f(x) : a < x < c}$. It is nonempty, since $\frac{a + c}{2} \in (a, c)$.
2. For $a < x < c$ we have $f(x) \le f(c)$, since $f$ is increasing. So $f(c)$ is an upper bound of $S$.
3. By the completeness axiom, $S$ has a supremum $L$. Since $f(c)$ is an upper bound, $L \le f(c)$.
4. The point $c$ is a limit point of $(a, b) \cap (-\infty, c) = (a, c)$. Indeed, for $\delta > 0$ the point $c - \frac{1}{2}\min(\delta, c - a)$ lies in $(a, c)$ and within $\delta$ of $c$.
5. Let $\varepsilon > 0$. The number $L - \varepsilon$ is not an upper bound of $S$. So there is $x_0 \in (a, c)$ with $f(x_0) > L - \varepsilon$.
6. Put $\delta = c - x_0 > 0$. Let $x \in (a, b)$ with $c - \delta < x < c$. Then $x_0 < x < c$.
7. Since $f$ is increasing, $L - \varepsilon < f(x_0) \le f(x)$. Since $x \in (a, c)$, also $f(x) \le L$. So $\abs{f(x) - L} < \varepsilon$.
8. By the definition of the one-sided limit, steps 4 to 7 give $\lim_{x \to c^-} f(x) = L = \sup_{a < x < c} f(x)$.
9. Now let $T = \set{f(x) : c < x < b}$. It is nonempty, and $f(c)$ is a lower bound of $T$, since $f$ is increasing.
10. By the existence of the infimum, $T$ has an infimum $R$. Since $f(c)$ is a lower bound, $f(c) \le R$.
11. In the same way as step 4, $c$ is a limit point of $(c, b)$.
12. Let $\varepsilon > 0$. The number $R + \varepsilon$ is not a lower bound of $T$. So there is $x_1 \in (c, b)$ with $f(x_1) < R + \varepsilon$.
13. Put $\delta = x_1 - c > 0$. Let $x \in (a, b)$ with $c < x < c + \delta$. Then $c < x < x_1$, so $R \le f(x) \le f(x_1) < R + \varepsilon$. Thus $\abs{f(x) - R} < \varepsilon$.
14. So $\lim_{x \to c^+} f(x) = R = \inf_{c < x < b} f(x)$.
15. Steps 3, 8, 10 and 14 give the chain $\lim_{x \to c^-} f(x) = L \le f(c) \le R = \lim_{x \to c^+} f(x)$.