Countability of $\mathbb{N}^3$ In Sipser's book (2nd edition), Exercise 4.7 asks to show that $T = \{(i,j,k) | i,j,k \in \mathbb{N}\}$ is countable.
Here is what I did.
Let $A = (i,j,k)$.
Let $B = (i+1,j+1,k+1)$.
$A$ and $B$ are sets. $i,j,k \in \mathbb{N}$.
I create a function $f(i,j,k) = (i+1, j+1, k+1)$. $T$ is countable because for every element in the set $A$, $f$ maps it to a different element in $B$. Also $f$ hits every element in $B$.
Is it correct?
 A: The proposed proof is for the fact that $|A|=|B|$.  However it does not tell us that either set is countable.
One function that would work is $$f(i,j,k)=2^i3^j5^k$$
This is a one-to-one function from $T$ to $\mathbb{N}$.  Hence we have $|T|\le|\mathbb{N}|$, which proves that $T$ is either countable or finite.  But $T$ is obviously infinite, hence $T$ is countable.
A: To show T is coutable you need to show there is a 1-1 mapping from $\mathbb N^3$ to $\mathbb N$.
So you need to find a way to map a $(i,j,k) \rightarrow n$ and that if $f(i,j,k) =n$,  $(i,j,k)$ is the only triplet that maps to $n$.
The proof you all refer to suggests $f(i,j,k) = 2^i3^j5^k$.  This clearly maps $\mathbb N^3$ to $\mathbb N$.  But if $f(i,j,k) = 2^i3^j5^k=n$ is (i,j,k) the only triple that maps into n?
The prime factorization theorem states that there is only one way to prime factorize n.  If n only has 2,3 and 5 as prime factors then $2^i3^j5^k$ is the only way to factorize n.  So $(i,j,k)$ is the only triplet that maps to that particular $n$.
So each different triplet maps into a unique and different natural number.  So there is a one to one mapping from $\mathbb N^3$ to a subset of $\mathbb N$. So $\mathbb N^3$ is countable.
