# Is the Cartesian product of an infinite number of $\mathbb{Z}^+$ countable?

We know the Cartesian product of a finite number of $\mathbb{Z}^+$ is countable, for example, $\mathbb{Z}^+ \times \mathbb{Z}^+ \times \mathbb{Z}^+$, because, let $m, n, p$ be from each of the $\mathbb{Z}^+$, we can find a one-to-one function $2^m 3^n 5^p$ that maps them to a subset of $\mathbb{Z}^+$. But what if we increase the number of $\mathbb{Z}^+$ to (countable) infinity, i.e.

$$\mathbb{Z}^+ \times \mathbb{Z}^+ \times \mathbb{Z}^+ \times \cdots$$ ?

• No. This set is uncountable, and has precisely the same size as the set of real numbers. – Andrés E. Caicedo Oct 8 '13 at 18:15
• Even if you replace $\mathbb{Z}^+$ by a set of cardinality $2$, the resulting set will be uncountable. – Tobias Kildetoft Oct 8 '13 at 18:15
• The first hint is to search the site better. – Asaf Karagila Oct 8 '13 at 18:15
• @TobiasKildetoft Thanks. I now see this can be shown by the diagonal method. – qed Oct 8 '13 at 21:21

For your infinite product, map $(n_1,n_2,\ldots,)$ to the subset $\{n_1,n_2,\ldots\}$. Then you will get a surjection onto ${\mathbb P}({\mathbb Z}^+)$, which has cardinality greater than the cardinality of ${\mathbb Z}^+$ by Cantor's theorem. Thus your infinite product has cardinality greater than or equal to that of ${\mathbb P}({\mathbb Z}^+)$, and it is in fact equal, which can be shown with a little more work. It is the same cardinality as the set of real numbers, which can also be shown with a little more work. What is true is that the union of countably many countable sets is countable, but not the product.