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Let $r_1,r_2,r_3, \ldots$ a numeration of all rational numbers and $f:\mathbb{R}\rightarrow \mathbb{R}$ with $f(x)=\sum_{r_n<x}2^{-n}$

I want to show that $f$ is bounded and increasing. I want to show also that $f$ is discontinuous at every $x \in \mathbb{Q}$ and continuous at every point $x\in \mathbb{R}\setminus \mathbb{Q}$.

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To show that the function is bounded, do we use the geometric sum?

About the monotonicity:

For $x<y$, we have that there are less rational numbers smaller than $x$ than smaller than $y$. That means that $f(x)$ has less terms at the sum than $f(y)$, and therefore we have that $f(x)<f(y)$. Is this correct?

Could you give me a hint about the continuity?

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EDIT:

About the boundness, do we have the following?

$$\sum_{r_n<x}2^{-n}=\sum_{i=1}^n2^{-i}=\sum_{i=1}^n\left (\frac{1}{2}\right )^i<\sum_{i=0}^n\left (\frac{1}{2}\right )^i=\frac{1-\left (\frac{1}{2}\right )^{n+1}}{1-\frac{1}{2}}=2\cdot \left [1-\left (\frac{1}{2}\right )^{n+1}\right ] \\ =2-\frac{1}{2^n}$$

Mary Star
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1 Answers1

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You're on the right way about boundedness and monotonicity.

For the continuity/discontinuity:

— if you are in $x_0\in \mathbb Q$, think what implies moving wathever positive distance to the right, what's the effect over the value of $f$. Is there a minimum amount by which $f$ increases no matter how small the step we give?

— if you are in $x_0\in \mathbb R \setminus \mathbb Q$, fix an $\epsilon >0$, and analyze why you can always set $\delta>0$ small enough such that for $x \in (x_0-\delta, x_0+\delta)$ you are sure that $f(x) \in (f(x_0)-\epsilon,f(x_0)+\epsilon)$ (think that you can eventually exclude all those $q_n$ such that the sum of the corresponding $2^{-n}$ is close enough to $1$.

Alejandro Nasif Salum
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  • About the boundness: Does it hold that $$\sum_{r_n<x}2^{-n}=\sum_{i=1}^n2^{-i}=\sum_{i=1}^n\left (\frac{1}{2}\right )^i<\sum_{i=0}^n\left (\frac{1}{2}\right )^i=\frac{1-\left (\frac{1}{2}\right )^{n+1}}{1-\frac{1}{2}}=2\cdot \left [1-\left (\frac{1}{2}\right )^{n+1}\right ]=2-\frac{1}{2^n}$$ ? – Mary Star Dec 17 '17 at 09:26
  • Actually the first equality is not true. To understand $f$ suppose that $$\mathbb Q = {r_1=\tfrac12,r_2=\tfrac13,r_3=\tfrac23,\ldots}.$$ Then to calculate $f(1/2)$ you have to sum $2^{-n}$ for every $n$ such that $r_n<\tfrac12$. So you have to add $2^{-2}$, but not $2^{-1}$ or $2^{-3}$... and of course there are infinitely many other negative powers of $2$ that you will have to add since there are infinitely many rational numbers smaller than $1/2$. – Alejandro Nasif Salum Dec 17 '17 at 12:04
  • But think... how bad could that sum go. Will you ever have to do the sum for every $n \in \mathbb N$? An even if that were the case (it actually "is" the case for $f(\infty)$), is this value finite or infinite? – Alejandro Nasif Salum Dec 17 '17 at 12:09
  • I haven't really understood that. Could you explain it further about the boundness? – Mary Star Dec 17 '17 at 12:29
  • Sure, I can. But first, let us do something that will immediately lead us to boundness. Supose this is the list of rationals: $$\mathbb Q = {0,\tfrac12,1,\tfrac13,\tfrac32,-1,\tfrac23,\tfrac43,-\tfrac12,2,\tfrac14,\tfrac53,-\tfrac23,\tfrac52,-2.\ldots},$$ that is, $r_1=0$, $r_2=\tfrac12$, etc. Which terms would have to be for sure when you calculate $f(1)$ (that is $f(1)=2^{-number}+2^{-other number}+\cdots$). Could you write the beginning of that sum for $f(1)$? (I mean, from the beginning to as far as you can get with this list.) – Alejandro Nasif Salum Dec 17 '17 at 12:43
  • We have that the numbers less than 1 are: $0, \frac{1}{2}, \frac{1}{3}, -1, \frac{2}{3}, -\frac{1}{2}, \frac{1}{4}, -\frac{2}{3}, -2, \ldots $ right? – Mary Star Dec 17 '17 at 12:56
  • Yes, but that means the sum would go $$f(1)=2^{-1}+2^{-2}+2^{-4}+2^{-6}+2^{-7}+\cdots, $$ do you understand why? – Alejandro Nasif Salum Dec 17 '17 at 14:39
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    Well, if you do get it, just see that $$0 \le 2^{-1}+2^{-2}+2^{-4}+2^{-6}+2^{-7}+\cdots \le$$ $$\le2^{-1}+2^{-2}+2^{-3}+2^{-4}+2^{-5}+2^{-6}+2^{-7}+2^{-8}+\cdots=1,$$ that is $$0\le f(1) \le 1.$$ But you can easily generalize this result to $0\le f(x) \le 1$ for every $x\in\mathbb R$, so $f$ is bounded.

    More interesting is the fact that $$2^{-1}+2^{-2}+2^{-4}+2^{-6}+2^{-7} \le f(1) \le$$ $$\le 2^{-1}+2^{-2}+2^{-4}+2^{-6}+2^{-7}+2^{-8}+2^{-9}+\cdots=1-2^{-3}-2^{-5}$$ since this gives much better bounds for $f(1)$ (namely, $0,8359\le f(1) \le 0,8438$).

     – Alejandro Nasif Salum Dec 17 '17 at 14:58
  • So, can we bound it above by the infinite geometric sum? – Mary Star Dec 17 '17 at 15:11
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    Yes, because for fixed $x$, $f(x)$ will omit some (actually infinitely many) terms of the infinite geometric sum, and since they're all positive, the sum will be strictly smaller. – Alejandro Nasif Salum Dec 17 '17 at 15:22
  • Ah ok! So we have that \begin{equation}0<f(x)<\sum_{n=0}^{\infty}2^{-n}\Rightarrow 0<f(x)=\frac{1}{1-\frac{1}{2}}\Rightarrow 0<f(x)<\frac{1}{\frac{1}{2}}\Rightarrow 0<f(x)<2\end{equation} right? – Mary Star Dec 17 '17 at 15:43