$ [0,1] \cup \mathbb{N} \sim [0,1] $

Question

Prove that $ [0,1] \cup \mathbb{N} \sim [0,1] $ ( meaning, I need to prove equipotency ) by showing an explicit bijection ( Without Cantor-Schroder-Bernstein theorem ).
I tried giving a function of the form $ f(x)= \begin{cases} \frac{1}{n} &\text{if}\, \exists n \in \mathbb{N^+}.x=n \\ x &\text{otherwise} \end{cases} $ , for all $ x \in [0,1] \cup \mathbb{N} $ ; However, this function is a mistake since it's not a bijection ( it's not one-to-one since for-example $ f(2)= f(1/2) $ ). I've been breaking my head over this problem, do you have any ideas?

Answer 1

For an explicit bijection, you may interleave $1,2,3,...$ with $1/2, 1/3, ...$, by defining:

$f(1)=\frac{1}{2}, f\left(\frac{1}{2}\right)=\frac{1}{3}, f(2)=\frac{1}{4}, f\left(\frac{1}{3}\right)=\frac{1}{5}, f(3)=\frac{1}{6}, f\left(\frac{1}{4}\right)=\frac{1}{7}$ and so on, in general:

$$f(x)= \begin{cases} \frac{1}{2n}, & \text{if } x\in \{1, 2, 3, ...\}\\ \frac{1}{2n-1}, & \text{if } x=\frac{1}{n}. n\in \{2, 3, ...\}\\ x, &\text{otherwise} \end{cases} $$

Answer 2

Hint: let $X = \{ \frac{1}{n} \mid n \in \Bbb{N}^+\}$, then $[0, 1]$ is the disjoint union $([0, 1] \setminus X) \cup X$ and $X \sim \Bbb{N}$. Now use an explicit bijection of $\Bbb{N}$ and the disjoint union $\Bbb{N} \sqcup \Bbb{N}$ to get a bijection of $[0, 1] \cup \Bbb{N}$ with $\Bbb{N}$.