Defining a measure of uniformity that finds the most "uniform structure" for the most cases of countably infinite sets?


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 $$\bigcup_{n=1}^{\infty}\left\{\frac{j}{k!}:j,k\in\mathbb{N},j\le k!\le n\right\}=\mathbb{Q}\cap[0,1]$$ and, 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 $\bigcup_{n=1}^{\infty} F_n$, as $t\to\infty$ we want a uniformity measure 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 uniformity measure of the discretely uniform set such as $\Delta \left(\left\{\frac{j}{k!}:j,k\in\mathbb{N},j\le k!\le t\right\}\right)$?

In most cases, 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?

  • This time it isn’t about average. It’s about finding a uniformity measure that determines the most uniform structure of A.

  • Isn't this pretty much the same question I answered a couple weeks ago?

  • I know you stated this isn’t possible. But read my whole post before you jump to conclusions.

  • I read the whole post and I still see some obvious problems, the most obvious one being that your countable set needs to be a metric space at least (you are using a concept of distance between points - you don't necessarily have that). It also needs to be bounded, covered by subset of uniformly spaced finite sets, and a lot more than that. Q \cap [0, 1] has extremely nice properties as far as countable sets go, the generic countable set is an ugly mess.

  • What do you think of my attempt? It is likely that an F_t which gives a minimum |d(F_t,A)-(e-1)/2| does not exist but if it does would this be the most uniform F_t? (The problem is this is extremely difficult to solve and I doubt you could prove this with a computer).

  • I think the specific thing you want is impossible to achieve, and the way it's worded is not rigorous enough to make sense. I showed you a few counterexamples, and at this point my suggestion would be to just give up. Try to understand exactly what you want, without ever using words like "natural", "uniform", or anything that is not well defined. Then come back to ask how to get there.

  • (See my attempt involving the power set). Why is the “specific thing” I want impossible to achieve. It doesn’t have to work for all cases but there are some cases where it should be possible. I also explained what “uniform” means. Why is it not clear enough?

  • There are cases where a unique F_t that gives the minimum |d(F_t,A)-(e-1)/2| exists. Take for example Q in [0,1]

  • Because your examples are extremely specific. There are a handful of examples where some ad-hoc solution can be found, but asking for something general that works in "most cases" is not possible. Your definition of uniform does not apply for the generic sets, it's highly specific: it requires a bounded, countable, metric space with a covering with specific properties. I already explained this. A generic countable set does not allow for such a definition of uniform.

  • I'll give you a list of countable sets (which are actually still very nice, there's much worse stuff), if you manage to give me a notion of "uniform" for each of those that makes sense to you maybe we can sort something out. 1) Finite subsets of Q; 2) Rational functions in two variables with coefficients in the algebraic closure of Q; 3) Group homomorphisms Z -> S(N); 4) Ordinals smaller than \epsilon_0; 5) Constructible sets; 6) Theorems of ZFC.

  • For finite subsets of Q, the closer the difference between consecutive elements are to equal, the more “uniform” the elements.

  • For rational functions with two variables in the algebraic closure of Q; we can use the formula involving the power set for d(Ft,A). The most uniform Ft minimizes |d(Ft,A)-(e-1)/2| since uniform Ft for arbitrary A has d(Ft,A)=(e-1)/2. If there is no way of minimizing the values than such a uniformity does not exist. The reason I’m using this is because the uniformity measure gives a unique value for each Ft.

  • The formula is in the post. I’m not sure of group homeomorphisms. Perhaps a uniform Ft does not exist for most cases.

  • I still haven’t reached a level to understand ordinals and theorems of ZFC.

  • Sorry for the late reply. That is pretty much what I was saying; a uniform Ft does not exist for most cases, and even when it does, it has to be constructed ad-hoc. Countable sets can be really, really messy. I am quite confident that there is no satisfying solution to your question; my advice is to accept that math does not always work as well as one would hope, and be content with the few special cases we already discussed.

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