- #1
sutupidmath
- 1,630
- 4
Proposition: Let [tex] \{A_n\}_{n\in I}[/tex] be a family of countable sets. Prove that
[tex] \bigotimes_{i=1}^n A_i[/tex]
is a countable set.
Proof:
Since [tex] \{A_n\}_{n\in I}[/tex]
are countable, there are 1-1 functions
[tex] f_n:A_n->J[/tex]
(J, the set of positive integers)
Now let us define a function
[tex] h:\bigotimes_{i=1}^n A_i->J[/tex]
such that
[tex] h(a_1,a_2,...,a_n)=(p_1)^{f_1(a_1)}*(p_2)^{f_2(a_2)}*...*(p_n)^{f_n(a_n)}[/tex]
Where p_i are prime numbers such that
[tex] 2\leq p_1<p_2<...<p_n[/tex]
Now from the Fundamental Theorem of Arithmetic,it is clear that h is a 1-1 function. So, based on some previous theorems, we conclude that
[tex]\bigotimes_{i=1}^n A_i[/tex]
is a countable set.
Is this correct?
[tex] \bigotimes_{i=1}^n A_i[/tex]
is a countable set.
Proof:
Since [tex] \{A_n\}_{n\in I}[/tex]
are countable, there are 1-1 functions
[tex] f_n:A_n->J[/tex]
(J, the set of positive integers)
Now let us define a function
[tex] h:\bigotimes_{i=1}^n A_i->J[/tex]
such that
[tex] h(a_1,a_2,...,a_n)=(p_1)^{f_1(a_1)}*(p_2)^{f_2(a_2)}*...*(p_n)^{f_n(a_n)}[/tex]
Where p_i are prime numbers such that
[tex] 2\leq p_1<p_2<...<p_n[/tex]
Now from the Fundamental Theorem of Arithmetic,it is clear that h is a 1-1 function. So, based on some previous theorems, we conclude that
[tex]\bigotimes_{i=1}^n A_i[/tex]
is a countable set.
Is this correct?