elementary set theory - What is the set of all functions from ${0, 1}$ to $mathbb{N}$ equinumerous to?

What is the set of all functions from $\{0, 1\}$ to $\mathbb{N}$ equinumerous to? I have figured out the question when it's the other way around, but I am not making any progress here. The worst thing is, I don't know how to think of such problems, and even when it was functions from $\mathbb{N}$ to $\{0, 1\}$, I would have never gotten it myself without help. When I think about this I think of the following: it's a function, so one thing can't be sent to two+ things, but two things can be sent to the same thing. So at least, we are mapping both 0 and 1 to the same element, and at most we are mapping it to two different elements. But that doesn't really give me anything... Thanks!

Answer

$|\{\text{functions } \{0,1\} \to \mathbb{N} \}|= |\mathbb{N}^{\{0,1\}}|= |\mathbb{N}^2|$

Indeed, if you are thinking of $\mathbb{N}^2$ as the set of ordered pairs of natural numbers, then the function

$\varphi: \{\text{functions } \{0,1\} \to \mathbb{N} \} \to \mathbb{N}^2$
defined as $\varphi(f)= (f(0),f(1))$

is a bijection.

It is an injection: if $f\not=g$, then $f(0)\not=g(0)$ or $f(1)\not=g(1)$, hence $\varphi(f)\not= \varphi(g)$.

It is a surjection: given $(m,n)\in \mathbb{N}^2$, define $f:\{0,1\} \to \mathbb{N}$ as $f(0)=m$, $f(1)=n$. Then $\varphi(f)=(m,n)$.

In fact, for some intuition, think that what this bijection is saying is that a function $\{0,1\}\to \mathbb{N}$ is completely determined by a couple of natural numbers. Who could they be but $f(0)$ and $f(1)$? :)

So, now I claim that $|\mathbb{N}^2|=|\mathbb{N}|$. Indeed, it is the cartesian product of two countable sets, and the cartesian product of a finite number of countable sets is countable, hence the set of functions $\{0,1\} \to \mathbb{N}$ is countable, i.e. it is equinumerous to $\mathbb{N}$.

You comment you can't use cardinal arithmetic: a more constructive proof would be to exhibit a bijection. See this question: How does one get the formula for this bijection from $\mathbb{N}\times\mathbb{N}$ onto $\mathbb{N}$? which attacks the problem of showing that $|\mathbb{N}\times \mathbb{N}|=|\mathbb{N}|$ more directly.

ADDED: An obvious generalization of the above leads to the fact that, for $n\in \mathbb{N}$, $|\{\text{functions } \{0,\dots,n-1\} \to \mathbb{N}\}| = |\mathbb{N}^n|$. Having this in mind, now forget you ever defined $\mathbb{N}^k$ for any $k\in \mathbb{N}$. Let's have a different approach.

Let $I$ be any given set. Define $\mathbb{N}^I$ as $$\mathbb{N}^I := \{\text{functions } I \to \mathbb{N}\}$$

This is the usual notation for the set of functions from $I$ to $\mathbb{N}$.

Now, in set theory (Zermelo-Fraenkel), the number $k\in \mathbb{N}$ is defined as the set $\{0,\dots,k-1\}$. So what if we take $I=\{0,\dots,k-1\}$? We have that $$\mathbb{N}^k=\mathbb{N}^{\\{0,\dots,k-1\\}} = \\{\text{functions } \\{0,\dots,k-1\\} \to \mathbb{N}\\}$$

In particular, with $k=2$, we obtain that $\mathbb{N}^2=\{\text{functions } \{0,1\} \to \mathbb{N} \}= \{\text{functions } 2 \to \mathbb{N}\}$. Now the above proof shows that our new $\mathbb{N}^2$ is equinumerous to the set of ordered pairs of natural numbers. This is nice. Why?

Well, first observe that equinumerous sets are indistinguishable from the point of view of set theory (in fancy words: bijections are isomorphisms in Set). So if two sets are equinumerous, there is no harm done with taking any of them as the definition of the set.

Then, observe that this definition is in much bigger generality. We have defined $\mathbb{N}^I$ for any set $I$. With our previous definition, we could only extend the generality for any finite set $I$, namely defining $\mathbb{N}^I$ as the set of $|I|$-uples of natural numbers: this definition makes no sense a priori for infinite $I$.

A final observation: in this final addendum, $\mathbb{N}$ can be taken to be any set, of course.