Show that the function is bounded and strictly increasing

• MHB
• mathmari
In summary, the conversation discusses a function $f$ with a numeration of all rational numbers and its properties of being bounded and strictly increasing. The group also discusses using the geometric sum and the enumeration of rational numbers to prove these properties.
mathmari
Gold Member
MHB
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)

mathmari said:
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)
mathmari said:

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)

I like Serena said:
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)

mathmari said:
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)

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

I don't really know (Worried)

mathmari said:
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)

I like Serena said:
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?

mathmari said:
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)

I like Serena said:
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)

mathmari said:
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)

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

Yep. (Nod)

I like Serena said:
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)

Yup. (Nod)

I like Serena said:
Yup. (Nod)

Great! Thank you very much! (Yes)

1. What does it mean for a function to be bounded?

A function is bounded if there exists a finite number M such that the absolute value of the function is always less than or equal to M. This means that the values of the function do not exceed a certain limit.

2. How can I determine if a function is bounded?

To determine if a function is bounded, you can evaluate the function at extreme values of the independent variable (such as positive and negative infinity). If the function approaches a finite number at both extremes, then it is bounded.

3. What is the significance of a function being strictly increasing?

A strictly increasing function means that the output of the function increases as the input increases. This can be visualized as a line that slopes upward from left to right on a graph.

4. How do I prove that a function is strictly increasing?

To prove that a function is strictly increasing, you must show that for any two inputs x1 and x2, where x1 < x2, the output of the function at x1 is less than the output at x2. This can be done algebraically or by graphing the function.

5. Can a function be both bounded and strictly increasing?

Yes, a function can be both bounded and strictly increasing. For example, the function f(x) = x on the interval [0,10] is both bounded (by 10) and strictly increasing.

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