Proving $\mathbb{N}^k$ is countable

Question

Prove that $\mathbb{N}^k$ is countable for every $k \in \mathbb{N}$.

I am told that we can go about this inductively.

Let $P(n)$ be the statement: “$\mathbb{N}^n$ is countable” $\forall n \in \mathbb{N}$.

Base Case: $\mathbb{N}^1 = \mathbb{N}$ is countable by definition, so $\checkmark$

Inductive Step: $\mathbb{N}^{k+1}$ $“=”$ $\mathbb{N}^k \times \mathbb{N}$

We know that $(A,B)$ countable $\implies$ $A \times B$ is countable. I am stuck on the part where I have to prove the rest, but I know that, for example, $(1,2,7) \in \mathbb{N}^3 \notin \mathbb{N}^2 \times \mathbb{N}$ but instead $((1,2),7) \in \mathbb{N}^2 \times \mathbb{N}$. So how would I go about proving the statement.

Answer 1

When you want to prove something about cardinalities you actually say "I don't really care who is in the set, I just care about its size".

This means that you don't really care that $(1,2,3)\notin\mathbb N^2\times\mathbb N$, because once you proved that $\mathbb N^2\times\mathbb N$ is countable you have a natural bijection: $$\left(\left(x,y\right),z\right)\mapsto\left(x,y,z\right)$$ This function is clearly bijective, and while you cannot write "clearly" in a homework assignment - it is not hard to prove that for yourself as well. Every $(n+1)$-tuple is just an $n$-tuple with an additional element.

Alternatively, you can use the fact that $\mathbb N\times\mathbb N$ is countable as well and just "skip" the above part, by doing it implicitly as follows:

So now the induction step would be to take $f_n\colon\mathbb N^n\to\mathbb N$ which is a bijection, define $g\colon\mathbb N^{n+1}\to\mathbb N\times\mathbb N$ as follows:

$$g((a_1,\ldots,a_{n+1})) = (f_n(a_1,\ldots,a_n),a_{n+1})$$

This function is injective since two different tuples will either have a different coordinate $k\le n$ in which case the $f_n$ part would be different by injectivity of $f_n$; or different coordinate at the $n+1$ place, in which case the right coordinate of the image will be different.

It is also surjective due to a similar argument. And then $f_{n+1}$ is the composition of $g$ with a fixed bijection from $\mathbb N\times\mathbb N$ to $\mathbb N$.

Answer 2

Hint:

It is enough to prove that $A \times B$ is countable whenever $A$ and $B$ are countable. To do this, write

$$ A = \{a_0,a_1,a_2, \dots\} $$ and $$ B = \{b_0,b_1,b_2,\dots\}. $$ To find a bijection, consider the map

$$\begin{align} 0 \mapsto (a_0,b_0) \\ 1 \mapsto (a_1,b_0) \\ 2 \mapsto (a_0,b_1) \\ 3 \mapsto (a_2,b_0) \\ 4 \mapsto (a_1,b_1) \\ 5 \mapsto (a_0,b_2) \\ 6 \mapsto (a_0,b_3) \\ 7 \mapsto (a_1,b_2) \\ \vdots \qquad\quad \end{align}$$ and show this map is a bijection. This map is "natural" in the sense that if you draw out the set of ordered pairs of nonnegative integers, this map follows a logical path. It is called the Cantor pairing function and it can be described visually.

Now to see that the previous statement is enough to prove what you want, use induction.

Answer 3

Define $f:\mathbb{N}\to\mathbb{N}^k$ by just the inclusion into the first coordinate, this is clearly an injection. Define $g:\mathbb{N}^k\to\mathbb{N}:(a_1,\cdots,a_k)\mapsto p_1^{a_1}\cdots p_k^{a_k}$ where $p_i$ is the $i^{\text{th}}$ prime. This is an injection by the fundamental theorem of arithmetic. The rest follows by the Schroeder-Bernstein theorem.

Answer 4

The function $f:\mathbb{N^K}\to \mathbb{N^K\times \{m \}}$ defined by $$f(a_1,a_2,\cdots, a_k)=(a_1,a_2,\cdots, a_k,m)$$ is clearly a bijection for fixed $m\in \mathbb{N}$ and we can write $\mathbb{N^{K+1}}$ as $$\mathbb{N^{K+1}}=\bigcup_{m=1}^{\infty}\{\mathbb{N^K\times \{m \}}\}$$ and this being a countable union of countable sets is countable.

Answer 5

Hint: For your inductive step, you have assumed there is a bijection between $\mathbb{N}^k$ and $\mathbb{N}$. You now want to prove that there is a bijection between $\mathbb{N}^{k+1}$ and $\mathbb{N}$. So if you use the bijection you have on the first $k$ components of $\mathbb{N}^{k+1}$ what do you have?