- #1

mathmari

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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 $\displaystyle{f(x)=\sum_{r_n<x}2^{-n}}$

I want to show that $f$ is bounded and strictly increasing.

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

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