What is the meaning of $\mathbb{N}^\mathbb{N}$?

Question

Here, $\mathbb{N}$ is the set of natural numbers. I was reading why there is no bijection from $\mathbb{N}^\mathbb{N}$ to $\mathbb{N}$, and I understood we took any subset from $\mathbb{N}^\mathbb{N}$ and proved it not bijective to $\mathbb{N}$ by Cantor's Diagonalization argument.

What am I unable to understand is, what exactly is meant by $\mathbb{N}^\mathbb{N}$?

I read somewhere that it is recognized as the space of all sequences of natural numbers. I didn't understand why? What is the meaning of that notation?

It is the infinite cartesian product $$\mathbb N \times \mathbb N \times \mathbb N \times \cdots$$ This corresponds $1-1$ with a sequence of natural numbers. — Peter, Jul 26 '18 at 21:41
For any set $X$, $X^{\mathbf N}$ denotes the set of all maps from $\mathbf N$ to $X$,i.e. the set of all infinite sequences of elements of $X$. — Bernard, Jul 26 '18 at 21:44

score 1 · Accepted Answer · answered Jul 26 '18 at 21:41

$A^B$ is the set of all functions from $B$ to $A$. When $A=B=\mathbb{N}$, $\mathbb{N}^\mathbb{N}$ is the set of all functions $f:\mathbb{N}\to\mathbb{N}$. Any such function can be viewed as a sequence $f(1),f(2),f(3),\ldots$, so changing notation a bit with $a_n=f(n)$ we see that the set of all functions $\mathbb{N} \to \mathbb{N}$ is the set of all sequences $a_1,a_2,\ldots$ of natural numbers.

score 1 · Answer 2 · answered Jul 26 '18 at 21:44

Let $k\in\mathbb{N}$. One could write an element $x\in\mathbb{N}^k$ as the $k$-tuple $$x=(a_1,\ldots,a_k),$$ which is as saying that for each $i\in\{1,\ldots,k\}$, there exists an entry $a_i\in\mathbb{N}$. Similarly, when we write $\mathbb{N}^{\mathbb{N}}$ we mean that if $x\in\mathbb{N}^{\mathbb{N}}$, then for each $i\in\mathbb{N}$ there is an entry $a_i\in\mathbb{N}$. That translates to $$x=(a_i)_{i\in\mathbb{N}},$$ which, as you have said, is a sequence of natural numbers.

What is the meaning of $\mathbb{N}^\mathbb{N}$?

2 Answers2