# For any countable totally-ordered $A$, how can mathematica show my choice function gives a unique collection of finite sets whose union is $A$?

Suppose $$A=\sqrt{\mathbb{Q}}_{>0}$$. We want a unique $$\bigcup\limits_{i\le j}A_i$$ for $$i,j\in\mathbb{N}$$ such that $$A_i$$ is finite, $$A_i\subseteq A_{i+1}$$, and $$\lim\limits_{j\to\infty}\bigcup\limits_{i\le j}A_i=A$$.

The problem is there are many $$\bigcup\limits_{i\le j}A_i$$ to choose from. Take for example

$$A_i=\left\{\sqrt{m}/\sqrt{n}:m,n\in\mathbb{N},n\neq 0, 0\le m \le n\le i\right\}$$

$$A_i=\left\{\sqrt{m}/\sqrt{2n+1}:m,n\in\mathbb{N},n\neq 0, 0\le m \le n\le i\right\}\cup\left\{\sqrt{s}/\sqrt{2k}:s,k\in\mathbb{N},k\neq 0, 0\le s \le k\le \sqrt{i}\right\}$$

$$A_i=\left\{\sqrt{m}/\sqrt{n}:m,n\in\mathbb{N},n\neq 0,0\le m\le n\le \sqrt[3]{i^2}\right\}\setminus\left(\mathbb{Q}\cap\left\{\sqrt{m}/\sqrt{n}:m,n\in\mathbb{N},n\neq 0,0\le m\le n\le \sqrt[3]{i^2}\right\}\right)\cup\left(\sqrt{m}/\sqrt{n}:m,n\in\mathbb{N},n\neq 0,0\le m\le n\le i\right)$$

$$A_i=\left\{\sqrt{m}/\sqrt{n!}:m,n\in\mathbb{N},n\neq 0,0\le m\le n!\le i\right\}\cup\left(\sqrt{m}/\sqrt{n}:m,n\in\mathbb{N},n\neq 0,0\le m\le n\le \sqrt{i}\right)$$

Or so on. We want a unique $$A_i$$ that satisfies some property/choice function of $$A$$. (In fact I want a property/choice function for any arbitrary totally-ordered $$A$$ which satisfies a unique $$A_i$$. (I heard this is not possible due to the axiom of countable choice.))

## Question

If

$$\left\{c_w\right\}_{w=1}^{\left|\bigcup_{i\le j}A_i\right|}\in \bigcup_{i\le j}A_i$$

such that $$\lim\limits_{j\to\infty}\bigcup_{i\le j}A_i=A$$ and $$\left\{c_w\right\}_{w=1}^{\left|\bigcup_{i\le j}A_i\right|}$$ is arranged from least to greatest.

I want to know if there is a unique $$A_i$$ which satisfies

$$\lim_{r\to\infty}\inf\left\{\frac{\sigma\left[\left\{\left|c_{w+1}-c_{w}\right|^{-1}\right\}_{w=1}^{\left|\bigcup_{i\le j}A_i\right|}\right]+\epsilon}{\sigma\left[\left\{\left|c_{w+1}-c_{w}\right|\right\}_{w=1}^{\left|\bigcup_{i\le j}A_i\right|}\right]+\epsilon}-1:\left|\bigcup_{i\le j}A_i\right|\le r, \ r\in\mathbb{N}\right\}$$

Where $$\sigma$$ is the standard deviation. For example, we calculate the standard deviation of $$\left\{\left|c_{w+1}-c_{w}\right|\right\}_{w=1}^{\left|\bigcup_{i\le j}A_i\right|}$$ as

$$\sqrt{\frac{1}{\left|\bigcup_{i\le j}A_i\right|-1}\sum_{w=1}^{\left|\bigcup_{i\le j}A_i\right|}\left(\left|c_{w+1}-c_{w}\right|-\frac{\sum_{w=1}^{\left|\bigcup_{i\le j}A_i\right|}\left|c_{w+1}-c_{w}\right|}{\left|\bigcup_{i\le j}A_i\right|}\right)^2}$$

# Attempt

I don't know if Mathematica can explicitly show this. Infact I don't think my property allows an infimum for example if $$A=\sqrt{\mathbb{Q}}$$ and

$$A_i=\left\{\sqrt{m}/\sqrt{n!}:m,n\in\mathbb{N},n\neq 0,0\le m\le n!\le i\right\}\cup\left(\sqrt{m}/\sqrt{n}:m,n\in\mathbb{N},n\neq 0,0\le m\le n\le \sqrt[a]{i}\right)$$

Where $$a\in\mathbb{N}$$, I don't think there is an inf since (using mathematica) if we calculated

F[i_] := DeleteDuplicates[
Sort[Flatten[
Table[(Sqrt[m]/Sqrt[n]), {n, 1, Floor[i]}, {m, 1, Floor[n]}]]]]
G1[i_] := DeleteDuplicates[
Sort[Flatten[
Table[Sqrt[m]/Sqrt[n!], {n, 1, Floor[i^(1/5)]}, {m, 1,
Floor[n!]}]]]]
G2[i_] := DeleteDuplicates[
Sort[Flatten[
Table[Sqrt[m]/Sqrt[n!], {n, 1, Floor[i^(1/3)]}, {m, 1,
Floor[n!]}]]]]
G3[i_] := DeleteDuplicates[
Sort[Flatten[
Table[Sqrt[m]/Sqrt[n!], {n, 1, Floor[i^(1/2)]}, {m, 1,
Floor[n!]}]]]]
G4[i_] := DeleteDuplicates[
Sort[Flatten[
Table[Sqrt[m]/Sqrt[n!], {n, 1, Floor[i^(2/3)]}, {m, 1,
Floor[n!]}]]]]


And get the maximum i1, i2, i3, i4 where

Length[Differences[Sort[Union[(F[i1]),(G1[i1])], Less]]]<r
Length[Differences[Sort[Union[(F[i2]),(G2[i2])], Less]]]<r
Length[Differences[Sort[Union[(F[i3]),(G3[i3])], Less]]]<r
Length[Differences[Sort[Union[(F[i4]),(G4[i4])], Less]]]<r
Length[Differences[Sort[F[i5],Less]]]<r


Then we calculate $$\sigma\left[\left\{\left|c_{w+1}-c_{w}\right|^{-1}\right\}_{w=1}^{\left|\bigcup_{i\le j}A_i\right|}\right]$$ as c, c1, c2, c3 and $$\sigma\left[\left\{\left|c_{w+1}-c_{w}\right|\right\}_{w=1}^{\left|\bigcup_{i\le j}A_i\right|}\right]$$ as d, d1, d2, d3.

c = N[StandardDeviation[
Sort[Differences[Sort[Union[(F[i1]), (G[i1])], Less]]^(-1), Less]],
15]
d = N[StandardDeviation[
Sort[Differences[Sort[Union[(F[i1]), (G[i1])], Less]], Less]], 15]

c1 = N[StandardDeviation[
Sort[Differences[Sort[Union[(F[i2]), (H[i2])], Less]]^(-1), Less]],
15]
d1 = N[StandardDeviation[
Sort[Differences[Sort[Union[(F[i2]), (H[i2])], Less]], Less]], 15]
c2 = N[StandardDeviation[
Sort[Differences[Sort[Union[(F[i3]), (J[i3])], Less]]^(-1), Less]],
15]
d2 = N[StandardDeviation[
Sort[Differences[Sort[Union[(F[i3]), (J[i3])], Less]], Less]], 15]
c3 = N[StandardDeviation[
Sort[Differences[Sort[Union[(F[i3]), (K[i3])], Less]]^(-1), Less]],
15]
d3 = N[StandardDeviation[
Sort[Differences[Sort[Union[(F[i4]), (K[i4])], Less]], Less]], 15]
c4 = N[StandardDeviation[
Sort[Differences[Sort[F[i5], Less]]^(-1), Less]], 15]
d4 = N[StandardDeviation[Sort[Differences[Sort[F[i5], Less]], Less]],
15]


and r approaches infinity we caclculate:

c/d
c1/d1
c2/d2
c3/d3
c4/d4


The problem is I can't calculate these fractions at high values such as r=100000 or 10000000 to see assume whether an infimum exists.

For some $$r\in\mathbb{N}$$, as $$a\to\infty$$, it appears $$\frac{\sigma\left[\left\{\left|c_{w+1}-c_{w}\right|^{-1}\right\}_{w=1}^{\left|\bigcup_{i\le j}A_i\right|}\right]+\epsilon}{\sigma\left[\left\{\left|c_{w+1}-c_{w}\right|\right\}_{w=1}^{\left|\bigcup_{i\le j}A_i\right|}\right]+\epsilon}-1$$ approaches closer to zero.

Since there is no specific $$a$$ which gives an infimum, I don't think there is an infimum. Is this the case? If so what property/choice function can we find of $$\left|c_{w+1}-c_{w}\right|$$ which gives an infimum for $$A=\mathbb{Q}_{>0}$$ or any totally-ordered countable $$A$$.

• could you boil down your problem? I don't understand your question – Mr Puh Sep 30 '20 at 15:04
• @MrPuh I want a general property of $A$ which gives a unique union of $A_i$. – Arbuja Sep 30 '20 at 16:51