Respuesta :
Answer:
[tex] A_n = [2^{n-1} , 3(2^{n-1}), 5(2^{n-1}), 7(2^{n-1}),....][/tex]
Step-by-step explanation:
For this case we need to produce an infinite collection of sets [tex] A_a, A_2, A_3,....[/tex] with the property that every [tex] A_i[/tex] has an infinite number of elements with [tex] A_i \cap A_j , i\neq j[/tex] and [tex] \cup_{i=1}^{\infty} A_i = N[/tex]
So for this case we need to create a set A who satisfy 3 conditions (Infinite number of elements, Disjoint and Union represent the natural numbers)
So if we define the nth term for the set A:
[tex] A_n = [2^{n-1} , 3(2^{n-1}), 5(2^{n-1}), 7(2^{n-1}),....][/tex]
We see that the set A represent all the odd multiplies of [tex] 2^{n-1}[/tex] and if we check the properties we have this:
Disjoint
If we select [tex] A_n , A_m[/tex] with [tex] n\neq m[/tex] and we can assume for example that [tex] n<m[/tex] and if we have an element a in the intersection of the sets [tex] a \in A_n \cap A_m[/tex], so then needs to exists some odd numbers k and l such that
[tex] a = 2^{n-1} k = 2^{m-1}l[/tex]
And since we assume that [tex] n<m[/tex] then we have that [tex] n\leq m-1[/tex] and we can write:
[tex] 2^{m-1} =2^m 2^i , i\geq 0[/tex]
And then [tex] 2^{n-1} k= 2^n 2^i l[/tex] and if we divide by [tex] 2^{n-1}[/tex] we got:
[tex] k = 2 2^i l[/tex] so then k is not odd since the last statement contradicts this. So then we can conclude that [tex] A_n \cap A_m = \emptyset[/tex]
Union
For this case we need to show that [tex] \cup_{i=1}^{\infty} A_i = N[/tex]
Since each element [tex] A_n[/tex] is a subset of the natural numbers then the unision of the sets represent N
For the other side of the explanation if we assume that [tex] a\in N[/tex] we can write [tex] a= 2^{n-1}k[/tex] for any [tex] n\in N[/tex] and k odd, and by this [tex] a\in A_n[/tex] and we chaek the property.
Infinite condition
For this case [tex] A_n = [2^{n-1} , 3(2^{n-1}), 5(2^{n-1}), 7(2^{n-1}),....][/tex]
is an infinite set since we don't have a limit for n so then we have infinite elements for this case.
And since all the properties are satisfied we end the problem.
