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.

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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. $a<x-\delta<x$ 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 $range(f(t): a<t<x)$. Also, $A$ is not achieved. If otherwise A is achieved at $f(y)$ with $a<y<x$, then $f((y+x)/2)$ would be larger than $A$ by monotonicity. These conclude that A is a limit point of range$(f(t): a<t<x)$, 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!

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

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

  • $\begingroup$ 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). $\endgroup$ – Mou May 16 at 18:56
  • $\begingroup$ 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). $\endgroup$ – EpsilonDelta May 16 at 20:04
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    $\begingroup$ 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? $\endgroup$ – José Carlos Santos May 16 at 20:04
  • $\begingroup$ 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! $\endgroup$ – Mou May 18 at 4:56
  • $\begingroup$ @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. $\endgroup$ – 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<y\implies f(x) \leq f(y) $ and not that $x<y\implies f(x) <f(y) $. If you want to mean the latter then use the word "strictly increasing".

Consider the set $$S=\{f(t) \mid a <t<x\} $$ 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 $a<t_0<x$ such that $f(t_0) >A-\epsilon$. Let $\delta=x-t_0$ then $\delta>0$ and we have $$A-\epsilon<f(t_0)=f(x-\delta) \leq A$$ By monotone nature of $f$ we have $$A-\epsilon<f(x-\delta) \leq f(t) \leq A$$ for all $t$ with $x-\delta<t<x$. 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 $S$.


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