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