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I have proved the following statement and I would like to know if my proof is correct:

"If $f:B\to\mathbb{R},\ B\subseteq\mathbb{R}$ is a monotone increasing function and $f$ is not continuous at $x_0\in B$ then $f$ has a jump discontinuity at $x_0$."

Proof. We first prove that $\lim\limits_{x \to x_0^-,\ x\in B\\}f(x)$ and $\lim\limits_{x \to x_0^+,\ x\in B\\}f(x)$ always exist.

Let $x_0 \in B$: then, since $f$ is increasing $f(x_0)$ is an upper bound for $\{f(x):x<x_0, x\in B\}$ so by the least upper bound property $S:=\sup_{x<x_0,\ x\in B}f(x)$ exists. Now, let $\varepsilon>0$: then there must be $\bar{x}\in B, \bar{x}<x_0$ such that $S-\varepsilon<f(\bar{x})<S$ and being $f$ increasing it must be $S-\varepsilon < f(x)<S$ for all $\bar{x}\leq x<x_0, x\in B$ so if we set $\delta:=x_0-\bar{x}=|x_0-\bar{x}|>0$ we have that $|S-f(x)|=S-f(x)<\varepsilon$ for all $x\in B\setminus\{x_0\}, |x_0-x|=x_0-x<\delta$ thus $\lim\limits_{x \to x_0^-,\ x\in B\\}f(x)=S=\sup_{x_0<d,\ x\in B}f(x)$.

By the same reasoning $f(x_0)$ is a lower bound for $\{f(x):x>x_0, x\in B\}$ so $I:=\inf\{f(x):x<x_0, x\in B\}$ exists by the greatest lower bound property. Now, if we let $\varepsilon>0$ there must be $\bar{x}\in B, \bar{x}>x_0$ such that $f(\bar{x})<I+\varepsilon$ and being $f$ increasing it must be $I<f(x)<I+\varepsilon$ for all $x_0<x\leq\bar{x},\ x\in B$ so if we set $\delta:=|\bar{x}-x_0|=\bar{x}-x_0>0$ we have that $|f(x)-I|=f(x)-I<\varepsilon$ for all $x\in B\setminus\{x_0\}, |x-x_0|=x-x_0<\delta$ thus $\lim\limits_{x \to x_0^+,\ x\in B\\}f(x)=I=\inf_{x>x_0,\ x\in B}f(x)$.

Now, if $\lim\limits_{x \to x_0^-,\ x\in B\\}f(x)=\lim\limits_{x \to x_0^+,\ x\in B\\}f(x)=L$ it must also be the case that $f(x_0)=L$ (since if $f(x_0)>L$ by taking $\varepsilon:=|f(x_0)-L|>0$ we would have that there exists $\delta>0$ such that $|f(x)-L|<f(x_0)-L$ for all $x\in B, x>x_0, x-x_0<\delta$ ie $f(x)<f(x_0)$ for all $x\in B$ such that $x>x_0,\ x-x_0<\delta$, a contradiction, since $f$ is increasing; if $f(x_0)<L$ then by taking $\varepsilon:=L-f(x_0)>0$ we would have that there exists $\delta>0$ such that $|f(x)-L|=|L-f(x)|<L-f(x_0)$ for all $x\in B, x<x_0, x_0-x<\delta$ ie $f(x)>f(x_0)$ for all $x\in B$ such that $x<x_0,\ x-x_0<\delta$, a contradiction, since $f$ is increasing) so $f$ must be continuous.

If $\lim\limits_{x \to x_0^-,\ x\in B\\}f(x)\neq\lim\limits_{x \to x_0^+,\ x\in B\\}f(x)$ then $f$ has a jump discontinuity with jump $h:=\lim\limits_{x \to x_0^+,\ x\in B\\}f(x)-\lim\limits_{x \to x_0^-,\ x\in B\\}f(x)$.

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  • $\begingroup$ One thing that seems to be overlooked is the case of $B$ being discrete, for instance the integers. For that case, some of your inequalities will need to allow for a possible equality. $\endgroup$
    – Dunham
    Jun 17 at 12:42
  • $\begingroup$ @Dunham thank you for your interest in my question; in that case, if I allowed the possibility of taking a limit to an isolated point, it appears to me that the function would be continuous at every such point (it would be vacuously true that $\forall\varepsilon >0\exists\delta>0: |f(x)-f(x_0)|<\varepsilon\forall x\in (x_0-\delta,x_0+\delta)\setminus\{x_0\}$ if we choose $\delta >0$ to be, for example, half the distance to the nearest other point of $B$ to $x_0$) so there would be no point in talking about how $f$ would behave in a possible point of discontinuity. Does this make sense to you? $\endgroup$
    – lorenzo
    Jun 17 at 12:52
  • $\begingroup$ There is a very nice and detailed proof in Bartles' introduction to real analysis. $\endgroup$
    – Medo
    Aug 5 at 8:02

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