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I am temporarily banned from math stack exchange but I believe this is a statistical question. If not, wait till the ban is over in 10 days then transfer it to math stack exchange.

Question

I want to find a measure of uniformity that determines the most "uniform" "structure" for the most cases of countably infinite sets?

Definition of Structure

A "structure" of a countably infinite set $A$ is defined as follows:

If $F_1,F_2,\cdot\cdot\cdot$ are an infinite sequence of finite sets (denoted $\left\{F_n\right\}_{n=1}^{\infty}$) such that $F_1\subseteq F_2\subseteq \cdot\cdot\cdot$ and $\bigcup\limits_{n=1}^{\infty}F_n=A$ then $F_t$ (for $t\in\mathbb{N}$ between one and infinity) is a structure of $A$.

Definition of Uniform

By "uniform", I want to find an $F_t$ where the difference between pairs of consecutive elements is close to equal.

To illustrate, for $\mathbb{Q}\cap[0,1]$, the most evenly distributed $F_t$ would be $\left\{\frac{j}{k!}:j,k\in\mathbb{N},j\le k!\le t\right\}$ since

$$\bigcup_{n=1}^{\infty}\left\{\frac{j}{k!}:j,k\in\mathbb{N},j\le k!\le n\right\}=\mathbb{Q}\cap[0,1]$$

moreover; if $k=r$ is the largest positive integer where $k!\le t$ and the resulting elements (from least to greatest) is

$$F_t=\left\{\frac{0}{r!},\frac{1}{r!},\frac{2}{r!}\cdot\cdot\cdot,1\right\}$$

Then the differences of the consecutive elements or $\Delta F_r$ end up being the same.

$$\Delta F_r=\left\{\frac{1-0}{r!},\frac{2-1}{r!},\cdot\cdot\cdot,\frac{r!-(r!-1)}{r!}\right\}$$

$$ \Delta F_r=\left\{\frac{1}{r!},\frac{1}{r!},\cdot\cdot\cdot,\frac{1}{r!}\right\}$$

For most cases of $A$ the difference between consecutive elements of $F_t$ (for every $t\in\mathbb{N}$) can never be the same but if we can cover all of $A$ with $\bigcup_{n=1}^{\infty} F_n$ we want a measure of uniformity and an $F_t$ where a uniformity measure of $\Delta F_t/\sum\limits_{x\in\Delta F_t}x$ has the smallest absolute difference from the measure of uniformity of the discretely uniform $\Delta \left(\left\{\frac{j}{k!}:j,k\in\mathbb{N},j\le k!\le t\right\}\right)$ (as $t\to\infty$)?

In most cases, such a “minimum difference” can never exist; but, if it does, we want an $F_t$ that gives such a “minimum”

Therefore, I need a measure of uniformity that gives the “minimum absolute difference” for the most cases of $A$.

Attempt to Answer Question:

For every $t$ in $F_t$, organize the elements from least to greatest and take the set of the difference of pairs of consecutive elements (We denoted this $\Delta F_t$).

We divide all the elements in $\Delta F_t$ by $||F_t||=\sum\limits_{x\in\Delta F_t}x$ so the sum of all its elements is $1$. This is similar to a probability distribution which we denote as:

$$\Delta F_t/||F_t||$$

If $\mathcal{P}(\Delta F_t/||F_t||)$ is the power set of $\Delta F_t/||F_t||$ then we arrange the elements of $\mathcal{P}(\Delta F_t/||F_t||)$ into new sets based on its cardinality (denoted $n$) from $n=1$ to $n=|F_t|$.

$$M_{n,t}=\left\{X:X\in\mathcal{P}(\Delta F_t/||F_t||),|X|=n\right\}$$

Then we multiply the elements of elements in each power set and take their union:

$$S_{n,t}=\bigcup\limits_{X\in M_{n,t}}\left\{\prod\limits_{x\in X}x\right\}$$

For every $n\in\mathbb{N}$, we arrange the values in $S_{n,t}$ taking the greatest value, the greatest plus the second greatest, the greatest plus second plus third greatest and continuing till we add the greatest up to the least greatest value. We then add the mean of these values from $n=1$ to $n=|F_t|$ giving $d$.

In more rigorous terms, if ${\max}^{r}(X)$ represents the $r$-th largest element in $X$, then for every $F_t$ we want a $d(F_t,A)$ (if it exists) such for every arbitrarily small positive $\epsilon$ there exists a sufficiently large integer $N$ where for all $t\ge N$:

$$\left|d(F_t,A)-\sum_{n=1}^{|F_t|}\frac{1}{|S_{n,t}-1|}\sum\limits_{p=1}^{|S_{n,t}|-1}\sum\limits_{r=1}^{p}{\max}^{r}\left(S_{n,t}\right)\right|\le\epsilon$$

Conclusion

We want $F_t$ that minimizes $\left|d(F_t,A)-\frac{e-1}{2}\right|$ since $d(F_t,A)$ where $A=\mathbb{Q}\cap[0,1]$ and $F_t=\left\{\frac{m}{n!}:m\le n!\le t\right\}$ (which also has the most uniform distribution) is $\frac{e-1}{2}$.

Perhaps we do not need $\Delta$ but we would still need to solve for $d\left(\left\{\frac{m}{n!}:m\le n!\le t\right\},\mathbb{Q}\cap[0,1]\right)$ inorder to find an $F_t$ that minimizes:

$$\left|d(F_t,A)-d\left(\left\{\frac{m}{n!}:m\le n!\le t\right\},\mathbb{Q}\cap[0,1]\right)\right|$$

Second Question

Does my answer give what I want?

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    $\begingroup$ Your question is truly unclear, in part because you haven't provided a quantitative interpretation of "as close to equal as possible." As an illustrative example, what would your solution be for the set of triangular numbers $\{n(n-1)/2\mid n\in\mathbb{N}\}$ or, say, the set of non-negative powers of $2,$ $\{2^n\mid n\in\mathbb{N}\}$? Another way to clarify your question would be to exhibit a statistical application, which currently is not in evidence. $\endgroup$
    – whuber
    Sep 1 at 22:40
  • $\begingroup$ @whuber I clarified my post. I should have stated that my concepts don’t work when $A$ is defined in an infinite interval. $\endgroup$
    – Arbuja
    Sep 2 at 1:08
  • $\begingroup$ @whuber I did more thinking and reedited. For your example I don’t think you can find the most evenly distributed $F_t$. $\endgroup$
    – Arbuja
    Sep 2 at 23:50