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 \(\displaystyle f,\) note that in the inequality

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

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

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

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

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.