Vtyjkyuk

On the one hand we know that an infinite product of countable sets is uncountable, so $mathbb Z_+timesmathbb Z_+timesldots=mathbb Z_+^omega$ is uncountable.

On the other hand, the finite product of countable sets is countable and the countable union of countable sets is countable, so $displaystylecup_n=1^inftymathbb Z_+^n$, where $mathbb Z_+^n=mathbb Z_+timesldotstimesmathbb Z_+$ ($n-$times), is countable.

So this is my question: why $mathbb Z_+^omeganeqcup_n=1^inftymathbb Z_+^n$?

It seems to me that, since the union's index runs to infinity, these sets should be the same.

There not even of the same type: $cup_n=1^infty mathbbZ^n_+$ is a set of vectors that all have finite length: each $x$ in the union must be in some $mathbbZ^n_+$ and is thus of the form $(a_1,ldots, a_n)$ for that $n$. A vector has a unique length so it is in exactly one such set. While an $x in mathbbZ_+^infty$ is by definition a sequence (countably infinite vector) of the form $(a_1, a_2 ,a_3, ldots)$. So a member of the right hand side is never in the left hand side or vice versa. So certainly no equality of sets (where $A = B$ for sets $A,B$ by definition means that $x in A$ iff $x in B$).