(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Let [tex]x_{1}, x_{2}, ...[/tex] be a sequence of rational numbers in which each rational number in (0,1) occurs exactly once. Define the function,

[tex]H(x) = 0 [/tex] if [tex]x \leq 0[/tex], and 1 if [tex]x > 0.[/tex]

Next, define the function

[tex]F(x)= \sum^{\infty}_{k=1} 2^{-k} H(x - x_{k}). [/tex]

Prove that F is strictly increasing on (0,1), discontinuous at every rational number in (0,1) (Bonus: prove that F is continuous at every irrational number in (0,1)).

Ok, so my colleague and I understand, heuristically, why it is strictly increasing. This is because for any y1 and y2 in (0,1), such that y2>y1, the series of F(y2) will have more terms that are non zero than the series of F(y1). Thus F(y2)>F(y1). Now we need to prove that it is discontinuous at every rational number, which, up to this point, continue's to baffle us.

I understand that since it is strictly increasing, then for any point c in (0,1), the left and right hand limits must exist (i.e. F(c-) and F(c+) exist and are finite). But I can't quite make them unequal to each other at each rational point. Any insight?

thanks,

M

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# Prove that F is discontinuous at every rational number

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