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As the verification is left to the reader, I try to verify it and it seems it is somehow natural: when $x_n$ is a point of $E$, when $x\ge x_n$, the $c_n$ is added, while when $x_n$ is not a point of $E$, as the series is convergent, by Cauchy theorem the gap between the $x\le x_n$ and the $x\ge x_n$ can be arbitrary small.

But I found when I change this idea to the strict proof, I can reach it. Could you write down the strict proof? Thank you.

EDIT: Thanks for the comment. Now I understand how to write strictly for proving the discontinuous point, but I still can not write it down in the epsilon-delta language when proving continuous. I think the reason that I can not write it in epsilon-delta language is that (just suppose the $E$ is rational numbers in $(a,b)$ or some other countable but not finite set), for each rational number though I can enumerate and number it I still can not no what definite number in a enumeration it belongs to.

fractal
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  • See if this specific example helps. – Brian M. Scott Jul 08 '20 at 01:53
  • @BrianM.Scott Thank you! Could you write the proof of the continuous point (on your specific example) in a more strict epsilon-delta language? – fractal Jul 08 '20 at 02:42
  • Let $a\in[0,1]\setminus\Bbb Q$ and $\epsilon>0$. For $x\in[0,1]$ let $L(x)={n\in\Bbb N:q_n<x}$. If $x<a$, then $$f(a)-f(x)=\sum_{n\in L(a)\setminus L(x)}2^{-n};.$$ Given $\epsilon>0$, there is an $m\in\Bbb N$ such that $2^{-m}<\epsilon$, and there is a $\delta>0$ such that $$(a-\delta,a)\cap{q_n:n\le m}=\varnothing;.$$ Then for $x\in(a-\delta,a)$ we have $$f(a)-f(x)=\sum_{n\in L(a)\setminus L(x)}2^{-n}\le\sum_{n>m}2^{-n}=2^{-m}<\epsilon;.$$ You can make a similar argument for $x>a$. (Sorry to have taken so long: I just plain forgot about this.) – Brian M. Scott Jul 12 '20 at 19:59

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