# monotonic function can only have simple discontinuity

I am self-studying Rudin, Principles of Mathematical Analysis. I am having trouble going through the theorem saying that monotonical functions can only have simple discontinuity, i.e., Suppose $$f$$ is monotonic and discontinuous at $$x$$, then $$f(x^+)$$ and $$f(x^-)$$ must exist.

In the proof, it is argued that, for any $$\epsilon>0$$, by the definition of least upper bound $$A:=\sup_{t\in(a,x)} f(t)$$, $$\exists \delta>0$$ s.t. $$a and $$A-\epsilon\le f(x-\delta)\le A$$.

From my understanding, sup is the least upper bound. By the definition itself, it doesn't mean that the least upper bound would be approached with an $$\epsilon$$-ball.

The theorem is, of course, true. I am thinking of using the facts like: $$A$$ is sup $$\Rightarrow$$ A in the closure of $$range(f(t): a. Also, $$A$$ is not achieved. If otherwise A is achieved at $$f(y)$$ with $$a, then $$f((y+x)/2)$$ would be larger than $$A$$ by monotonicity. These conclude that A is a limit point of range$$(f(t): a, which in turn is followed by the original proof.

I am wondering if my thought is necessary, or there's any quick fact to support the claim in the book.

It is sometimes a struggle for me to go through every detail of Rudin's book. It would also be much appreciated if someone can point to reference textbooks that could complement it. Thanks!

• $f$ is increasing and $\forall z<x,f(z)\le f(x)$ so $f(z)$ has a left limit. – Yves Daoust May 17 at 14:42

If $$A=\sup_{t\in(a,x)}f(t)$$, and if $$\varepsilon>0$$, then $$A-\varepsilon and therefore there is some $$x_0\in(a,x)$$ such that $$f(x_0)>A-\varepsilon$$. So, let $$\delta=x-x_0$$. Then $$a. Besides, $$x-\delta=x_0$$ and therefore $$f(x-\delta)=f(x_0)>A-\varepsilon$$. And, of course, $$f(x_0)\leqslant A$$. So, yes,$$a

• Hi, thanks for the reply. I can only convince myself that there's such x_0 only if A is a limit point of range{ f(t): a<t<x}. Otherwise, f(x) might drop suddenly from A such that f(x) is vacant in the epsilon-ball of A. I am just wondering if there's any quick fact to circumvent my 4-line reasoning (or if my reasoning is true). – Mou May 16 at 18:56
• The supremum is a limit point; this is proven in Rudin (look for the proof that says that a closed and bounded set contains it supremum). – EpsilonDelta May 16 at 20:04
• Since $A$ is the least upper bound of the set $\{f(t)\,|\,t\in(a,x)\}$ and since $A-\varepsilon<A$, then $A-\varepsilon$ is not an upper bound of that set. And this means then there is a $x_0\in(a,x)$ such that $f(x_0)>A-\varepsilon$. Where is the flaw in this argument? – José Carlos Santos May 16 at 20:04
• Hi @JoséCarlosSantos, thanks! no flaws. I now understand. I didn't get the point that $A-\epsilon$ is not an upper bound, which means there exists a slightly larger value. Thanks again! – Mou May 18 at 4:56
• @EpsilonDelta this is probably not true? Let $E=[0,1]\cup{2}$. If I understand correctly, sup E = 2, but 2 is not a limit point. But of course, $[0,1] \cup {2}$ is bounded and closed, which also contains its sup. – Mou May 18 at 5:01

You need to get clarity on a few things here. The terms "supremum" and "least upper bound" are synonymous and can be used interchangeably. Further the latter term is almost self explanatory in the sense that if $$M=\sup A$$ then $$M$$ is the least of all upper bounds of $$A$$ which further means that numbers less than $$M$$ are not upper bounds for $$A$$ and are therefore exceeded by some members from $$A$$.

Further if $$f$$ is monotonically increasing then it means that $$x and not that $$x. If you want to mean the latter then use the word "strictly increasing".

Consider the set $$S=\{f(t) \mid a which is bounded above by $$f(x)$$ and thus $$A=\sup S$$ exists and $$A\leq f(x)$$ (remember $$A$$ is the least upper bound of $$S$$ whereas $$f(x)$$ is one of the upper bounds of $$S$$). Let $$\epsilon>0$$ then $$A-\epsilon$$ is less than $$A$$ and hence is not an upper bound for $$S$$ and is therefore exceeded by some member of $$S$$. Hence we have a $$t_0$$ such that $$a such that $$f(t_0) >A-\epsilon$$. Let $$\delta=x-t_0$$ then $$\delta>0$$ and we have $$A-\epsilon By monotone nature of $$f$$ we have $$A-\epsilon for all $$t$$ with $$x-\delta. This proves that $$f(x-) =\lim_{t\to x^-} f(t) =A$$.

By the way why do you think the value $$A$$ can't be achieved? The number $$A=\sup S$$ can also be member of $$S$$ and it may also be the case $$A=f(x)$$.

Also note that the supremum is not necessarily a limit point of the set. If $$A=\sup S$$ and $$A\notin S$$ then $$A$$ is a limit point of $$S$$. If $$A\in S$$ then $$A$$ may or may not be a limit point of $$S$$.