Show $F=\{ f \in \mathbb{N}^{\mathbb{N}}: \ \forall n \in \mathbb{N} \ f(2n)=n \}$ is a set with cardinality of the continuum

Question

We want to show $$F=\{ f \in \mathbb{N}^{\mathbb{N}}: \ \forall n \in \mathbb{N} \ f(2n)=n \}$$ is a set with cardinality of the continuum.

I start off by considering how the elements of $F$ look like. It seems they are of the form

$$\langle 0, 1, 1, 3, 2, 5, 3, 7, 4, 9, 5, \dots \rangle$$ We notice $F \subseteq \mathbb{N}^{\mathbb{N}} \implies \lvert F \rvert \leq \mathfrak{c}$. Now, we need to show $\lvert F \rvert \geq \mathfrak{c}$. This might be done by finding an injection from a well-known set with cardinality of the continuum to $F$. $\{0,1 \}^{\mathbb{N}}$ might be a choice but I'm not quite sure what injection to choose. Any suggestions?

Answer 1

HINT: Suppose that $f\in F$. The definition of $F$ pins down the value of $f$ at each even natural number: $f(0)=0$, $f(2)=1$, $f(4)=2$, and so on. However, it places no restrictions at all on $f(n)$ when $n$ is odd. For $f\in F$ let

$$\hat f:\Bbb N\to\Bbb N:n\mapsto f(2n+1)\;.$$

What can you say about the function

$$\varphi:F\to \Bbb N^{\Bbb N}:f\mapsto\hat f\;?$$

Answer 2

Let $O$ be the set of odd natural numbers.

There is a clear bijection between your set and $\mathbb N ^ O$.

Given a $f$ in your set send it to the function $f'$ such that $f'(n)=f(n)$ (in other words the restriction of $f$ to $O$). The inverse is the function that sends $g:O\rightarrow \mathbb N$ to the function $g'$ that sends $2^nk$ to $g(k)$. (remember that every natural number $m$ can be uniquely written in the form $m=2^nk$ with $k$ odd).