How to show relation between squarefree sets

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Edit: I erase the last question and post specific problems

Consider $$\text{If $t$ is prime, then set}\quad \mathcal{A}(t)=\left\{ N; \quad N=\prod_{\substack{p_i \text{ is prime}\\ 2t\ \nmid\ p_i-1}}p_i ,\text{ and $N$ is squarfree}\right\}\cup \{1\}$$ $$\text{If $t$ is composite, say $t=q_1\times q_2\times\ldots\times q_s$ for some $s$ then set} \quad\mathcal{A}(t)=\mathcal{A}(q_1)\cap\mathcal{A}(q_2)\cap\ldots\cap\mathcal{A}(q_s)$$ Where $q_i$ are distinct prime factor of $t$.

Consider set $\mathcal{B}(l)=\{ t: l\in\mathcal{A}(t)\}$ then show that,

$$\text{If $l$ is prime, then set}\quad \mathcal{B}(l)=\left\{ N; \quad N=\prod_{\substack{p_i \text{ is prime}\\ p_i\ \nmid\ l-1}}p_i \right\}$$ $$\text{If $l$ is composite and squarefree, say $l=q^`_1\times q^`_2\times\ldots\times q^`_{s^`}$ for some $s^`$ then set} \quad\mathcal{B}(l)=\mathcal{B}(q^`_1)\cap\mathcal{B}(q^`_2)\cap\ldots\cap\mathcal{B}(q^`_{s^`})$$ where $q^`_i$ are distinct prime factors of $l$.

Table for lists,

$\mathcal{A}(t)$ sets
$\mathcal{A}(1)$ $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, \ldots\}$
$\mathcal{A}(2)$ $\{1, 2, 3, 6, 7, 11, 14, 19, 22, 23, 31, 38, 43, 46, 47, 59, 62, 67, 71, 79, \ldots\}$
$\mathcal{A}(3)$ $\{1, 2, 3, 5, 6, 10, 11, 15, 17, 22, 23, 29, 30, 33, 34, 41, 46, 47, 51, 53, \ldots\}$
$\mathcal{A}(4)$ $\{1, 2, 3, 6, 7, 11, 14, 19, 22, 23, 31, 38, 43, 46, 47, 59, 62, 67, 71, 79, \ldots\}$
$\mathcal{A}(5)$ $\{1, 2, 3, 5, 6, 7, 10, 13, 14, 15, 17, 19, 21, 23, 26, 29, 30, 34, 35, 37, \ldots\}$
$\mathcal{A}(6)$ $\{1, 2, 3, 6, 11, 22, 23, 46, 47, 59, 71, 83, 94, 107, 118, 131, 142, 166, 167, 179, \ldots\}$
$\mathcal{A}(7)$ $\{1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 30, 31, 33, \ldots\}$
$\mathcal{A}(8)$ $\{1, 2, 3, 6, 7, 11, 14, 19, 22, 23, 31, 38, 43, 46, 47, 59, 62, 67, 71, 79, \ldots\}$
$\mathcal{A}(9)$ $\{1, 2, 3, 5, 6, 10, 11, 15, 17, 22, 23, 29, 30, 33, 34, 41, 46, 47, 51, 53, \ldots\}$
$\mathcal{A}(10)$ $\{1, 2, 3, 6, 7, 14, 19, 23, 38, 43, 46, 47, 59, 67, 79, 83, 86, 94, 103, 107, \ldots\}$
$\vdots$ $\vdots$

\begin{array}{|c|c|} \hline & \text{Set} \\ \hline \mathcal{B}(2) & \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \ldots, 99, 100\} \\ \mathcal{B}(3) & \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, \ldots, 97, 99\} \\ \mathcal{B}(4) & \{1\} \\ \mathcal{B}(5)& \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, \ldots, 97, 99\} \\ \mathcal{B}(6) & \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, \ldots, 97, 99\} \\ \mathcal{B}(7)& \{1, 5, 7, 11, 13, 17, 19, 23, 25, 29, \ldots, 89, 91, 95, 97\} \\ \mathcal{B}(8) & \{1\} \\ \mathcal{B}(9) & \{1\} \\ \mathcal{B}(10) & \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, \ldots, 97, 99\} \\ \mathcal{B}(11) & \{1, 3, 7, 9, 11, 13, 17, 19, 21, \ldots, 93, 97, 99\} \\ \mathcal{B}(12) & \{1\} \\ \mathcal{B}(13)& \{1, 5, 7, 11, 13, 17, 19, 23, 25, 29, \ldots, 95, 97\} \\ \mathcal{B}(14)& \{1, 5, 7, 11, 13, 17, 19, 23, 25, 29, \ldots, 95, 97\} \\ \mathcal{B}(15)& \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, \ldots, 97, 99\} \\ \mathcal{B}(16) & \{1\} \\ \mathcal{B}( 17) & \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, \ldots, 97, 99\} \\ \hline \end{array}

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There are 2 best solutions below

3
On BEST ANSWER

Let us show what is required in the question.

For each natural $n$ put $[n]=\{1,2,\dots,n\}$.

Suppose that $l$ is prime. Let $t$ be a product of distinct prime numbers $t_1,\dots, t_k$. Then $t\in \mathcal B(l)$ iff $l\in\mathcal A(t)$ iff $l\in\mathcal A(t_j)$ for any $j\in [k]$ iff $l=\prod_{\substack{p_i \text{ is prime}\\ 2t_j\ \nmid\ p_i-1}}p_i$ for any $j\in [k]$ iff $2t_j\ \nmid\ l-1$ for any $j\in [k]$. We have to compare the latter condition with $t_j\ \nmid\ l-1$ for any $j\in [k]$. Their equivalence can be easily checked for $l>2$ and for $l=2$ separately.

Suppose now that $l$ is a product of distinct prime numbers $p_1,\dots, p_n$. Let $t$ be a product of distinct prime numbers $t_1,\dots, t_k$. Then $t\in \mathcal B(l)$ iff $l\in\mathcal A(t)$ iff $l\in\mathcal A(t_j)$ for any $j\in [k]$ iff $2t_j\ \nmid\ p_i-1$ for any $i\in [n]$ and $j\in [k]$ iff $t\in\mathcal B(p_i)$ for any $i\in [n]$.

1
On

Seems false.

Pick $A_1 = \mathbb{N} / \{1\}$ and for $n>1, A_n = \varnothing$.

Check that $A_i \cap A_j = A_{ij} $ $\forall i,j \in \mathbb{N}$.

Now $B_1 = \varnothing $ since $1\notin A_i $ $\forall i \in \mathbb{N}$ and $B_2 = \{1\}$ since $2 \in A_1$. By the assumption we have:

$B_2 = B_1 \cap B_2 = \varnothing \cap \{1\} = \varnothing$. A contradiction.