Ideas for defining a "size" which informally measures subsets of rationals to eachother?

Question

EDIT: I posted an answer to this question. Can somone check?

Consider set $T_1,T_2,...T_p$ which are subsets of $\mathbb{Q}$

I want to create a new definition of "size" that distinguishes between subsets of $\mathbb{Q}$ that have "significantly" more elements inside of $\mathbb{Q}$ than other subsets. (Note that I want this size to apply to countably dense and countably finite subsets of $\mathbb{Q}$) .

For example, if we compared rational numbers to integers, we know there are significantly more rationals in rationals than integers in rationals. Morover, there are infinite rationals between every integer, further proving the previous statment.

Unfortunately, formal measures, which assigns a weight for each singleton $\left\{x\right\}$ in $T_p$,

$$\mu(T_p)=\sum_{x \in T_p}\mu(\left\{x\right\})$$

would not be meaningful since assigning zero or positive weight would give subsets of $\mathbb{Q}$ zero or infinite measure. This does not distinguish which of those subsets could have significantly more or less elements inside of $\mathbb{Q}$.

Cardinality is also a problem, since it counts the number of elements, rather than determine which subsets $\mathbb{Q}$ have more elements in $\mathbb{Q}$. Morover, the cardinality of any countably infinite set is infinity.

However, informal measures can be used. For example, the asymptotic density of a subset of $\mathbb{N}$ is the number elements the subset "fills" of natural numbers between $[0,b]$ as $b\to\infty$.

How can we create a new definition of "size" that constructs an informal measure of the subsets of $\mathbb{Q}$ and meets the following requirments?

-If $T_1=T_2$ and $\mu(T_1)=\mu(T_2)$

-If $T_1\subseteq T_2$ then $\mu(T_1) \le \mu(T_2)$

Answer 1

There is a "natural" extension of the natural density to the rationals, but I am not sure how useful it is. The set of rationals in $[-b,b]$ with denominator less than $b$ is finite, so we can compute the fraction of these rationals that are in $T$ and ask if there is a limit as $b \to \infty$. We want to consider the rationals when expressed in lowest terms. This clearly meets your request that the density of a superset is at least as great as that of a subset and that sets which are negatives of each other get the same density.

There are $\phi(k)$ fractions with denominator $k$ in each unit interval. Wikipedia states that $\sum\limits_{i=1}^b \phi(k) \approx \frac 3{\pi^2}b^2$ so we have about $\frac 6{\pi^2}b^3$ rationals in $[-b,b]$ with denominator $b$ or less.
${}{}{}{}$ It seems one density that would be interesting would be the density of rationals with odd numerators. This is all the ones with even denominators and half of the ones with odd denominators. We need the sum of the totient function over the even numbers to get this, but I haven't found one.

Answer 2

I strongly recommend that you read Mancosu's article on counting principles, as he reviews the interesting literature that has emerged out of this question. In particular, following the work of Benci and Di Nasso (which is the most promising line of attack), he analyzes the question as being equivalent to the existence of a "numerosity" function $v$ from $S$ (the set of sets to be counted) to $N$ (the set which will be used for counting, typically, in the case of countable sets, $\mathbb{N}$) such that:

(1) If $A, B \in S$ are such that $A \subset B$, then $v(A) < v(B)$; (2) If $v(A) = v(A')$ and $v(B) = v(B')$, then $v(A \sqcup B) = v(A' \sqcup B')$ and $v(A \times B) = v(A' \times B')$ (it preserves disjoint unions and products).

Ideally, we would want it to preserve bijections, but we know that is not compatible with (1) for infinite sets. Nevertheless, there may happen that we are able to impose some other restriction on the bijections to make it work, i.e. something like

(3) If $A \simeq B$ (an isomorphism under an appropriate relation), then $v(A) = v(B)$.

In the second article mentioned above, the authors show that the existence of (an interesting example of) such a function is equivalent to the existence of a certain type of ultrafilter on the natural numbers (Ramsey or selective ultrafilters), which incidentally shows that this question is independent of $\mathsf{ZFC}$. If you're familiar with ultrafilters, their appearance in this context is not very surprising, since they typically give rise to a measure. There are also interesting connections to non-standard analysis which are explored in the article.

Answer 3

How about $\mu: \mathcal{P}(\mathbb{Q})\to \mathbb{N}\cup \{\infty\}$ defined by $\mu(T)=\limsup_{a<b\in \mathbb{Q}} |T\cap (a,b)|$. It measures the asymptotic behavior of local density and incorporates your requirement.

Answer 4

The theory of sets asserts that sets of integers and rational numbers are countable sets, that is, they have the same cardinality. At the same time, our experience suggests the opposite:

For example, if we compared rational numbers to integers, we know there are significantly more rationals in rationals than integers in rationals. Moreover, there are infinite rationals between every integer, further proving the previous statement.>

To reconcile the theory with common sense helps the particular case of lemma to the theorem of Matiyasevich on the existence of a one-to-one correspondence $$ \left(\mathbb Z^+\right)^2\leftrightarrow \mathbb Z^+,$$ using the functions $$\begin{cases} z(x,y) = \dfrac12\left((x+y)^2 - x - 3y +2\right)\\ x(z) = z-\dfrac12(s(z)-1)(s(z)-2)\\ y(z) = s(z)-x(z), \end{cases}\tag1$$ where $$s(z) = \left[\sqrt{2z-{7\over4}}-{1\over2}\right] + 2\tag2$$

The functions $(1-2)$ realize the Cantor ordering of the set - by increasing the sum $s=x + y$, and if the sums are equal, by increasing $x$ (see also Wolfram Alfa). Therefore, the Cantor's consequence is $$q_i = \left\{ {1\over1}, {1\over2}, {2\over1}, {1\over3}, {2\over2}, {3\over1}, {1\over4}, {2\over3}, {3\over2}, {4\over1}, {1\over5}, {2\over4}, {3\over3}, {4\over2}, {5\over1}, {1\over6}, {2\over5}, {3\over4}, {4\over3}, {5\over2}, {5\over2},\dots \right\}$$

In my opinion, this order can be used as the "natural" order of positive rational numbers.

For example, natural series displayed in consequence of Cantors indexes $$N_n = {1\over2}\left((n+1)^2 - n - 3 +2\right) = {n(n+1)/2},$$ fractions with odd numerator and even denominator - in consequence $$H_n = \left\{2, 7, 9, 16, 18, 20,\dots \right\} .$$

Thus, each infinite subset of rational numbers can be displayed as the increasing sequence of Cantor indexes, allowing to form representative samples of a given volume. Reciprocal to the selective expectancy of $x\over y$, or its limit (when it converges), or some another selective statistics can be used as the required measure of subsets.

At the same time, more informative (and therefore more promising) direction is the construction of empirical probability distributions.

Answer 5

Here is another definition my professor offered.

However, I believe it does not follow my second requirement for an appropriate measure/density.

Requirement 1: If $T_1$ and $T_2$ are subset of $\mathbb{Q}$ and $T_1=T_2$, then $D(T_1)=D(T_2)$

Requirement 2: If $T_1$ and $T_2$ are subset of $\mathbb{Q}$ and $T_1\subset{T_2}$, then $D(T_1)<D(T_2)$

We can turn $\mathbb{Q}$ into a finite set by restricting its numerator and denominator.

$$V(r)=\left\{\left.\frac{m}{n}\right|1<m<r,1<n<r,\gcd(m,n)=1\right\}$$

$$\lim_{r\to\infty}V(r)=\mathbb{Q}$$

We then take the number of elements in $V(r)$ that $T$ intersects with.

$$\frac{\left|T\bigcap{V(r)}\right|}{|V(r)|}$$

As $r$ becomes larger

$$D(T)=\lim_{r\to\infty}\frac{\left|T\bigcap{V(r)}\right|}{|V(r)|}$$