MHB Need help in understanding proof of continuity of monotone function

[kalish1]

I am reading the following proof of a proposition from Royden+Fitzpatrick, 4th edition, and need help in understanding the last half of the proof. (My comments in italics.)----------Proposition: Let $A$ be a countable subset of the open interval $(a,b).$ Then there is an increasing function on $(a,b)$ that is continuous only at points in $(a,b)$ ~ $A.$

Proof: If $A$ is finite the proof is clear. Assume $A$ is countably infinite. Let $\{q_n\}_{n=1}^{\infty}$ be an enumeration of $A.$ Define the function $f$ on $(a,b)$ by setting $$f(x) = \sum\limits_{\{n|q_n \leq x\}} \frac{1}{2^n} \ \mathrm{for \ all} \ a<x<b,$$ where the sum over the empty set is zero.

Since a geometric series with a ratio less than $1$ converges, $f$ is properly defined. Moreover,

\begin{equation}
\mathrm{if} \ a<u<v<b, \ \mathrm{then} \ f(v)-f(u) = \sum\limits_{\{n|u<q_n \leq v\}} \frac{1}{2^n}. \ \ \ \ \ \ \ \ \ \ \ (1)
\end{equation}

Thus $f$ is increasing.

I follow so far.

Let $x_0 = q_k$ belong to $A.$ Then by (1), $$f(x_0)-f(x) \geq \frac{1}{2^k} \ \mathrm{for \ all} \ x<x_0.$$ Therefore $f$ fails to be continuous at $x_0.$ Now let $x_0$ belong to $(a,b)$ ~ $C.$ Let $n \in \mathbb{N}.$ There is an open interval $J$ containing $x_0$ for which $q_n$ does not belong to $J$ for $1 \leq k \leq n.$ We infer from (1) that $|f(x)-f(x_0)|<1/2^n$ for all $x \in J.$ Therefore $f$ is continuous at $x_0.$

Why are the claims of $f$ continuous/discontinuous true? I need some clarification.

This question has been crossposted at real analysis - Need help in understanding proof of continuity of monotone function - Mathematics Stack Exchange

 

[GJA]

Hi kalish,

I am headed out to a barbeque, so I don't have a lot of time for this post, but here are some hints:

To verify the discontinuity of $$f,$$ note that in the inequality

$$f(x_{0})-f(x)\geq \frac{1}{2^{k}}~~~x<x_{0},$$

$$k$$ is fixed. That means no mattter how close you take $$x<x_{0},$$ the difference $$f(x_{0})-f(x)\geq \frac{1}{2^k}.$$

To prove continuity, use $$j$$ as the summation index in (1) (so that you don't confuse it with the fixed $$n$$ being used by Fitzpatrick in the argument for continuity). Then, for the chosen neighborhood $$J$$ of $$x_{0},$$ we have (assuming $$x_{0}<x$$)

$$\begin{align*}
\sum_{\{j\vert x_{0}<q_{j}\leq x\}}\frac{1}{2^{j}}&\leq \sum_{j=n+1}^{\infty}\frac{1}{2^{j}}\\
&=\sum_{j=0}^{\infty}\frac{1}{2^{n+1+j}}\\
&=\frac{1}{2^{n+1}}\sum_{j=0}^{\infty}\frac{1}{2^{j}}\\
&\leq\frac{1}{2^{n}}
\end{align*}$$

Hopefully those hints will get you going in the right direction. Let me know if anything is still confusing/not quite right, and I will write back when I return.