Countability of Finite Product Sets: Foundational Issues?

[WWGD]

TL;DR

It seems obvious that $A \subset B$ implies $|A|<|B|$. Does it need proof?

Hi,
I gave my friend a proof that the set of pairs $\ mathbb N \times \mathbb N \times...\mathbb N$ (Finite product) is countable. I gave an injection :

$(a_1, a_2,...,a_n) \right arrow p_1_^{a_1} p_2^{a_2}...p_n^{a_n}$

Where the $p_i$ are distinct primes. My friend is telling me this is not enough and I can't see why,but I may be missing some foundational issues. I should each such product injects into the Naturals and any subset of the Naturals is countable. Am I missing anything here? Seems he's missing something or blowing it out of proportion.

 

[Infrared]

Well, it's not true. $\mathbb{N}\subset\mathbb{Z},$ but $|\mathbb{N}|=|\mathbb{Z}|.$

It is true that $A\subset B$ implies $|A|\leq |B|$, but this is just a matter of definition: $|A|\leq |B|$ means that there is an injection $A\hookrightarrow B$, and if $A$ is a subset of $B$, then just use inclusion.

Anyway, since your map, let's call it $f$, is an injection, $\mathbb{N}\times\ldots\times\mathbb{N}$ is in bijection with $f(\mathbb{N}\times\ldots\times\mathbb{N})$, which is an infinite subset of $\mathbb{N}$, and hence countable.

 

[WWGD]

Well, it's not true. $\mathbb{N}\subset\mathbb{Z},$ but $|\mathbb{N}|=|\mathbb{Z}|.$

It is true that $A\subset B$ implies $|A|\leq |B|$, but this is just a matter of definition: $|A|\leq |B|$ means that there is an injection $A\hookrightarrow B$, and if $A$ is a subset of $B$, then just use inclusion.

Anyway, since your map, let's call it $f$, is an injection, $\mathbb{N}\times\ldots\times\mathbb{N}$ is in bijection with $f(\mathbb{N}\times\ldots\times\mathbb{N})$, which is an infinite subset of $\mathbb{N}$, and hence countable.

Thanks. My bad, I was assuming finite subsets, so strict inclusion.