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New Edit: The Measure has several problems. My answer is stated below the post

I previously stated:

"I manipulated the Construction so that if $T_1=T_2$ then $\mu(T_1)=\mu(T_2)$. If the construction can be improved let me know."

I was wrong. I cannot manipulate the construction to have a consistent measure for equivalent sets with different non-reduced forms. However, if the equivalent sets had different reduced forms, their measures would be the same.

Moreover, to compare the size of non-equivalent sets, the sets must be simplified first. See the section "Sets Must Be Simplified Before Being Measured"

The construction of the measure is similar to Natural Density and Counting Measure.

Consider a group of countably infinite sets dense in $[a,b]$

$T_1=\left\{\left.a<\frac{M_1(n_1)}{R_1(q_1)}<b\right|n_1,q_1\in\mathbb{Z}\right\}$ $,T_2=\left\{\left.a<\frac{M_2(n_2)}{R_2(q_2)}<b\right|n_2,q_2\in\mathbb{Z}\right\},...,$ $T_m=\left\{\left.a<\frac{M_m(n_m)}{R_m(q_m)}<b\right|n_m,q_m\in\mathbb{Z}\right\}$

I want to create a measure that compares the "size" of each countable dense set to all the sets.

User @Mikhail suggested:

"It is indeed possible to get measures that behave in a nontrivial fashion on countable subsets, and indeed the axiom of choice or some weaker forms thereof are relevant here. Namely, one can get a finitely-additive (not $\sigma$-additive) measure $\xi$ on the set of subsets of $\mathbb{N}$ such that $\xi(S)=0$ for any finite set $S\subseteq\mathbb{N}$ and $\xi(S)=1$ for any cofinite subset (i.e., subset with a finite complement). Here $\xi$ takes only two values, zero or one. For sets that are neither finite nor cofinite, the measure behaves in a nontrivial way but always exactly one of a pair of complementary sets has measure one. Such measures are used in ultraproduct constructions."

However, due to my lack of understanding of ultra-product constructions and axiom of choice, I created an alternate construction which may or may not follow the principles of measure theory. The construction below involves cardinality and ratios.

The Construction

We begin by taking the cardinality of $T_c=\left\{\left.a<\frac{M_c(n_c)}{R_c(q_c)}<b\right|n_c,q_c\in\mathbb{Z}\right\}$ (where $c$ is an integer and $1\le c \le m$)

\begin{equation} \left|\left\{\left.a<\frac{M_c(n_c)}{R_c(q_c)}<b\right|n_c,q_c\in\mathbb{Z}\right\}\right| \end{equation}

The countably infinite set can divided into an infinite union of finite sets.

\begin{equation} \left|\lim_{t\to\infty}\bigcup_{q_c=-t}^{t}\left\{\left.a<\frac{M_c(n_c)}{R_c(q_c)}<b\right|n_c\in\mathbb{Z}\right\}\right| \end{equation}

The union can be modified inorder for the "compactness" of $R_c(q_c)$ to be included in the measure; however, $T_c$ must be simplified first (There is a section below defending simplification). Here are the steps to fully reducing the set.

  1. One must be able to factor $h_c$ out of all coefficients of $M_c(n_c)$ and factor $p_c$ out of all coefficients of $R_c(q_c)$.

  2. $h_c/p_c$ must be fully reduced to get ${h_c}'/{p_c}'$. If the denominator is irrational then it must be rationalized.

  3. After multiplying the ${h_c}'$ to the numerator ${p_c}'$ to the denominator we end up with $M_c^{'}(n_c)$ and $R_c^{'}(q_c)$. (All coefficients must not be in the form of fractions)

We end up using ${R_c}^{'}(q_c)$ to modify the previous equation.

$$s_c(t)=\left|\bigcup_{v_c\in \left\{\left.-t\le {{R_c}^{'}(q_c)} \le t \right|q_c\in \mathbb{Z} \right\}}\left\{\left.a<\frac{M_c(n_c)}{v_c}<b\right|n_c\in\mathbb{Z}\right\}\right|$$

The function $s_c(t)$ is the modified cardinality of $T_c$. Similarly $s_1(t),s_2(t),..s_m(t)$ are the modified cardinalities of $T_1,T_2,..,T_m$.

If the finite sets in the union of $s_c(t)$ were disjoint, $s_c(t)$ would equal

$$=\sum_{v_c\in \left\{\left.-t\le {{R_c}^{'}(q_c)}\le t \right|q_c\in \mathbb{Z} \right\}}\left|\left\{\left.a<\frac{M_c(n_c)}{v_c}<b\right|n_c\in\mathbb{Z}\right\}\right|$$

However, in most cases, the finite sets are not disjoint. Hence $s_c(t)$ would be less than the sum above. This makes the modified cardinalities countably sub-additive.

The "construction" of the measure of $T_c$ ends up being

$$\mu(T_c)=\lim_{t\to\infty}\frac{s_{c}(t)}{s_1(t)+s_2(t)+...+s_m(t)}$$

$$\mu(T_c)=\lim_{t\to\infty}\frac{s_c(t)}{\sum_{p=1}^{m} s_{p}(t)}$$

Sets must be Simplified before being measured

In order to show one set is a subset of another set, both sets must be simplified.

Consider sets $T_1=\left\{\left.\frac{2n+1}{4p+2}\right|n,p\in\mathbb{Z}\right\}$ and $T_2=\left\{\left. \frac{8r+6}{16v+12}\right|r,v\in\mathbb{Z}\right\}$.

We can prove $T_2\subset{T_1}$, if the numerator of $T_2$ is a subset of the numerator of $T_1$ and the denominator of $T_2$ is a subset of the denominator of $T_1$.

However $\left\{\left.8r+6\right|r\in\mathbb{Z}\right\}\not\subset\left\{\left.2n+1\right|n\in\mathbb{Z}\right\}$ and $\left\{\left.16v+12\right|v\in\mathbb{Z}\right\}\not\subset\left\{\left.4p+2\right|p\in\mathbb{Z}\right\}$.

We have to simplify both sets! Set $T_1$ is fully reduced but $T_2$ can be simplified into $\left\{\left.\frac{4r+3}{8v+6}\right|r,v \in \mathbb{Z}\right\}$. We find that $\left\{\left.4r+3\right|r\in\mathbb{Z}\right\}\subset\left\{\left.2n+1\right|n\in\mathbb{Z}\right\}$ and $\left\{\left.8n+6\right|v\in\mathbb{Z}\right\}\subset\left\{\left.4p+2\right|p\in\mathbb{Z}\right\}$.

Hence $T_2\subset T_1$ but both sets must be simplified before being compared to one another.

If we set $T_1=T_2$, with the sets in different unreduced forms, the sets must be simplified into reduced forms. Otherwise the sets cannot be properly compared to one another. Infact, if the equivalent reduced sets were measured, both set would have equal measure.

I will give an example but not go in detail with the calculations. If we set $T_1=\left\{\left.\frac{2p+1}{2q+1}\right|p,q\in\mathbb{Z}\right\}$ and $T_2=\left\{\left.\frac{2p+3}{2q+5}\right|p,q\in\mathbb{Z}\right\}$ and compare the size of the sets between [1,2]; both sets are already reduced and should have a measure of $1/2$. This is easy to prove, since the cardinalities of the sets divided into finite unions are the same and so are the elements of the sets' denominators when restricted by $[-t,t]$.

Example for Applying the Measure (If $T_2\subset T_1$ then $\mu(T_2)<\mu(T_1)$)

Say we set $T_1=\left\{\left.\frac{\sqrt{n}}{q^2}\right|n,q\in\mathbb{Z}\right\}$ and $T_2=\left\{\left.\frac{\sqrt{2m+1}}{r^4}\right|m,r\in\mathbb{Z}\right\}$ and compare the size of the set between $[1,2]$ (Note that $T_2\subset{T_1}$.)

Since $T_1$ is fully simplified, we can apply the measure.

$$s_1(t)=\left|\bigcup_{v_1\in\left\{\left.-t<q^2<t\right|q\in\mathbb{Z}\right\}}\left\{\left.1<\frac{\sqrt{n}}{v_1}<2\right|n\in\mathbb{Z}\right\}\right|$$

$$=\sum_{q=1}^{\left\lfloor\sqrt t\right\rfloor}\left\lfloor(2q^2)^2 \right\rfloor-\left\lceil(q^2)^2 \right\rceil+1$$

(Note we do not sum elements that are undefined or repeated in other finite sets.)

Since $T_2$ is simplified, we can apply the measure.

$$s_2(t)=\left|\bigcup_{v_2\in\left\{\left.-t<{r^4}<t\right|q\in\mathbb{Z}\right\}}\left\{\left.1<\frac{\sqrt{2m+1}}{v_2}<2\right|m\in\mathbb{Z}\right\}\right|$$

$$=\sum_{r=1}^{\lfloor\sqrt[4]{t}\rfloor}\left\lfloor\frac{(2r^4)^2-1}{2}\right\rfloor-\left\lceil\frac{(2r^4)^2-1}{2}\right\rceil+1$$

The measure of each set ends up being

$$\mu(T_1)=\lim_{t\to\infty}\frac{s_1(t)}{s_1(t)+s_2(t)}=1$$

$$\mu(T_2)=\lim_{t\to\infty}\frac{s_2(t)}{s_1(t)+s_2(t)}=0$$

Hence $T_1$ has a measure of 1 and $T_2$ has a measure of 0.

Properties of the Construction

  1. If $T_1 \subset T_2$ then $\mu(T_1)<\mu(T_2)$

  2. If $T_1$ and $T_2$ are fully reduced and $T_1=T_2$ then $\mu(T_1)=\mu(T_2)$. (Note that $T_1$ and $T_2$ can be in different reduced forms and still be equal)

(If $>>$ means "growth rate is infinitely greater than as $x\to\pm\infty$", $<<$ means "growth rate is infinitely smaller than as $x\to\pm\infty$", $\sim$ means "growth rate is porportional as $x\to\pm\infty$", and any listed function eventually increases without any bound, the following is also true.)

  1. If ${|M_1(x)|}>>{|M_{2}(x)|}$ and $|R_1(x)|\sim|R_2(x)|$; $|M_{1}(x)|\sim |M_{2}(x)|$ and $|R_{1}(x)|>>|R_{2}(x)|$; or $|M_{1}(x)|>>|M_{2}(x)|$ and $|R_{1}(x)>>R_{2}(x)|$, then $\mu(T_1)=0$ and $\mu(T_2)=1$.

  2. If ${|M_1(x)|}<<{|M_{2}(x)|}$ and $|R_1(x)|\sim|R_2(x)|$; $|M_{1}(x)|\sim |M_{2}(x)|$ and $|R_{1}(x)|<<|R_{2}(x)|$; or $|M_{1}(x)|<<|M_{2}(x)|$ and $|R_{1}(x)<<R_{2}(x)|$, then $\mu(T_1)=1$ and $\mu(T_2)=0$.

  3. If $|M_{1}(x)|>>|M_{2}(x)|$, $|R_{1}(x)|<<|R_{2}(x)|$, and $|R_{1}(x)|>>|M_{2}(x)|$, then $\mu({T_1})=0$ and $\mu({T_2})=1$.

  4. If $|M_{1}(x)|>>|M_{2}(x)|$, $|R_{1}(x)|<<|R_{2}(x)|$, and $|R_{1}(x)|<<|M_{2}(x)|$, then $\mu({T_1})=1$ and $\mu({T_2})=0$

However, as suggested in a previous question, there is no "best way" of measuring the size of countable sets.

Can the following construction be used to measure countable sets? If not, how can we alter the construction, to make a measure possible?

What other measures can be used?

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  • $\begingroup$ Whoever is downvoting please explain. $\endgroup$ – Arbuja Jan 7 '17 at 20:18
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    $\begingroup$ This construction doesn't seem to be well-defined at all. First off, the measure of a set should depend on the set only, and not on the presentation of the set in terms of some functions $M$ and $R$. It is not clear at all that your construction is at all well-defined: if I gave you the same set in terms of different functions $M'$ and $R'$, would you be able to show that they give the same measure of the set? The question is hard to answer unless the construction is clarified to the point where its well-definedness may be understood. $\endgroup$ – Joshua Mundinger Jan 7 '17 at 21:17
  • $\begingroup$ @Alqatrkapa If so then why is the following true. If $\left\{\left.M_1(p_1)\right|p_1\in\mathbb{Z}\right\}\subset\left\{\left.M_2(p_2)\right|p_2\in\mathbb{Z}\right\}$ and $\left\{\left.R_1(q_1)\right|q_1\in\mathbb{Z}\right\}\subset\left\{\left.R_2(q_2)\right|q_2\in\mathbb{Z}\right\}$ then $$\left\{\left.\frac{M_1(p_1)}{R_1(q_1)}\right|p_1,q_1 \in \mathbb{Z}\right\}\subset\left\{\left.\frac{M_1(p_2)}{R_2(q_2)}\right|p_2,q_2 \in \mathbb{Z}\right\}$$. $\endgroup$ – Arbuja Jan 7 '17 at 22:14
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    $\begingroup$ Maybe include some of these example calculations to help us understand then? And do it in such a way where you first chose $M_1=2x+1$, and then do the calculation again but now chose $M_1=\ln x$, to show us that you get the same answers no matter which functions you chose for the $M_i$ $\endgroup$ – user2520938 Jan 7 '17 at 23:07
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    $\begingroup$ I still don't understand why the construction is well-defined. Maybe could you label in your question where the proof that it is well-defined is? $\endgroup$ – Joshua Mundinger Jan 10 '17 at 2:54
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I found a counter-example to the credibility of my measure that I should have found a long time ago. If we set $T_1=\left\{\left.\frac{m}{2n+1}\right|m,n \in \mathbb{Z}\right\}$ and $T_2=\left\{\left.\frac{v}{4r+2}\right|v,r\in\mathbb{Z} \right\}$, then according to my measure, $\mu(T_1)=\frac{2}{3}$ and $\mu(T_2)=1/3$.

However, if we change $T_1$ into $\left\{\left.\frac{2m}{4n+2}\right|m,n \in \mathbb{Z}\right\}$ we find that $T_1 \subset T_2$. Hence for my measure to have some credibility $\mu(T_1)<\mu(T_2)$. Morover $T_1$ had to be non-reduced to be comapared to $T_2$, which goes against my argument of simplification.

Hence, my measure cannot be used.

Hoefully someday there will be a measure that can be applied to any countable dense set.

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