Discontinuities of a monotone increasing function $f:B\to\mathbb{R},\ B\subseteq\mathbb{R}$ can only be jump discontinuities

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 $$\{f(x):x so by the least upper bound property $$S:=\sup_{x exists. Now, let $$\varepsilon>0$$: then there must be $$\bar{x}\in B, \bar{x} such that $$S-\varepsilon and being $$f$$ increasing it must be $$S-\varepsilon < f(x) for all $$\bar{x}\leq x 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.

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

• 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. Jun 17 at 12:42
• @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? Jun 17 at 12:52
• There is a very nice and detailed proof in Bartles' introduction to real analysis.
– Medo
Aug 5 at 8:02