Prove that the set of triples $\{(a,b,c)|a,b,c \in \mathbb{N}\}$ is countable I have the following question in my textbook:
Prove that the set of triples $\{(a,b,c)|a,b,c \in \mathbb{N}\}$ is countable
Now I know that $\mathbb{N}$ is countable already, and I have completed a non-rigorous proof of this before, but I am unsure of how having a set of triples changes things, nor do I understand what a set of triples pertains to. I assume I am missing some fundamental knowledge in solving this problem?
My guess is that a set of triples is: $(a,b,c)$ so $(1,1,1),(1,1,2),(1,1,3)...(2,1,1),(2,1,2)...(3,1,1)...$ etc
I imagine I can show this to be countable via pictures if my guess is correct. But how would I go about proving it rigorously?
 A: It is possible to show this without actually constructing a bijection. Let 
$$S=\{(a,b,c)|a,b,c\in \mathbb N\}.$$
If you define $$A_n=\{(a,b,c)\in S|a+b+c=n\},$$
then it is clear that $A_n$ is finite for every value $n\in \mathbb N$. For example, $|A_1|=|A_2|=0$, $|A_3|=1$, $|A_4|=3$ and so on.
Then, you can show that $$S=\bigcup_{i\in\mathbb N} A_i.$$
Because a countable union of finite set is countable, that means that $S$ is countable.
A: Consider the function $f: \mathbb{N}^3 \to \mathbb{N}$ given by: $f(a,b,c) = 2^{a-1}\cdot 3^{b-1} \cdot 5^{c-1}$. We can prove that $f$ is one-to-one and this suffices to show that $\mathbb{N}^3$ is countable.
A: It is possible to count the triples $(a,b,c)$ such that $a+b+c$ is (not strictly increasing). The first few are $$(1,1,1),(1,1,2),(1,2,1),(2,1,1),(1,1,3),(1,2,2),(1,3,1),(2,1,2),(2,2,1), \text{etc.}$$ 
Since every triple must correspond to a sum $a+b+c$, and every triple for each sum is listed in the method for counting above, this is a bijection between $\mathbb{N}$ and $\mathbb{N}^3$.
A: Standard proof of countability of $\mathbb{Q}$ relies on proving that $\mathbb{N}\times \mathbb{N} \simeq \mathbb{N}$. With this you can show that $\mathbb{N}\times \mathbb{N} \times \mathbb{N} \simeq \mathbb{N} \times \mathbb{N} \simeq \mathbb{N}$
