Show that the function is bounded and strictly increasing

[mathmari]

Hey!

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)

 

[I like Serena]

To show that the function is bounded, do we use the geometric sum?
$$\sum_{r_n<x}2^{-n}=\sum_{i=1}^n2^{-i}$$

Hey mathmari! (Smile)

Yes, we can use the geometric sum.
But we can't assume that $r_n$ is increasing can we?
So this first step is not correct is it? (Wondering)

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?

Additionally, $f(y)$ has all the same terms as $f(x)$.
And since it has more terms, it must indeed be greater. (Nerd)

 

[mathmari]

Yes, we can use the geometric sum.
But we can't assume that $r_n$ is increasing can we?
So this first step is not correct is it? (Wondering)

No, $r_n$ is not increasing.

But what can we do?

Does it maybe hold that the sum of $f(x)$ is less than the infinite geonmetric sum? (Wondering)

 

[I like Serena]

No, $r_n$ is not increasing.

But what can we do?

Does it maybe hold that the sum of $f(x)$ is less than the infinite geonmetric sum?

What do we need to ensure that $f(x)$ is less than the infinite geometric sum? (Wondering)

 

[mathmari]

What do we need to ensure that $f(x)$ is less than the infinite geometric sum? (Wondering)

I don't really know (Worried)

 

[I like Serena]

I don't really know (Worried)

Don't we have for any $x_0\in \mathbb R$ that:
$$
f(x_0) = \sum_{r_n<x_0} 2^{-n} < \lim_{x\to\infty} \sum_{r_n<x} 2^{-n}
$$
(Thinking)

 

[mathmari]

Don't we have for any $x_0\in \mathbb R$ that:
$$
f(x_0) = \sum_{r_n<x_0} 2^{-n} < \lim_{x\to\infty} \sum_{r_n<x} 2^{-n}
$$
(Thinking)

Yes. So, there are infinitely many $r_n$'s and so $n$ goes from $1$ to infinity, or not?

 

[I like Serena]

Yes. So, there are infinitely many $r_n$'s and so $n$ goes from $1$ to infinity, or not?

Hmm... there are actually infinitely many $r_n$'s for $f(x_0)$ as well.
But for any specific $x_0$ we still do not have all $r_n$'s. (Nerd)Taking a step back, can you tell how we can enumerate all rational numbers with a sequence $r_n$?
Can you perhaps give an example what $r_n$ would look like? (Wondering)

 

[mathmari]

Hmm... there are actually infinitely many $r_n$'s for $f(x_0)$ as well.
But for any specific $x_0$ we still do not have all $r_n$'s. (Nerd)Taking a step back, can you tell how we can enumerate all rational numbers with a sequence $r_n$?
Can you perhaps give an example what $r_n$ would look like? (Wondering)

Do we consider only the positive rationals or also the negative? (Wondering)

$f(x)$ will contain only some elements (but infinitely many) of the infinite sum, or not? (Wondering)

 

[I like Serena]

Do we consider only the positive rationals or also the negative?

It said all rationals didn't it?
But even if they were only positive, there are still infinitely many rational numbers between zero and any positive number. (Nerd)

$f(x)$ will contain only some elements (but infinitely many) of the infinite sum, or not?

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

Does this mean 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*} and so the function is bounded? (Wondering)

 

[I like Serena]

Yup. (Nod)

 

[mathmari]

Yup. (Nod)

Great! Thank you very much! (Yes)