How to show this function is bijective? I want to show$$ (0,1)\sim(0,1)∩(R-Q) $$
And after reading and thinking about this answer:

(1) Choose an infinite countable set of irrational numbers in $(0,1)$, call them $(r_n)_{n\geqslant0}$.
(2) Enumerate the rational numbers in $(0,1)$ as $(q_n)_{n\geqslant0}$.
(3) Define $f$ by $f(q_n)=r_{2n+1}$ for every $n\geqslant0$, $f(r_n)=r_{2n}$ for every $n\geqslant0$, >and $f(x)=x$ for every irrational number $x$ which does not appear in the sequence  >$(r_n)_{n\geqslant0}$.
Let me suggest you take it from here and show that $f$ is a bijection between $(0,1)$ and $(0,1)\setminus\mathbb Q$.

I wanted to show f is bijection (injective + surjective) ,and first I tried to show thats injective:
I must show : $$(r_{2n+1}=r_{2n+1}')~→~(q_n=q_n')$$
And: $$(r_{2n}=r_{2n}')~→~(r_n=r_n')$$
And third one is obvious.
But I stucked here because this function is new to me...and I cannot show the relation between $r_{2n+1}$ and $q_n$ in order to conclude injection statements.
And moreover , because of this problem(relation understanding) I cannot show thats a surjection too...
Could someone help me please?
 A: Denote $(\mathbb R\setminus\mathbb Q) \cap (0, 1) =: \mathbb I_1$, and let $n\in \mathbb N_0$. Here is the definition of $f$ again for convenience:
$$f:(0, 1) \to \mathbb I_1,\qquad f(x)=\cases{r_{2n+1}\quad \text{if } x =q_n, \\
r_{2n\phantom{+1}}\quad \text{if } x=r_n, \\ x \qquad \quad \!\!\text{elsewhere}.}$$
The sequences $(q_n)$ and $(r_n)$ are injective because we may choose them as such. The sequences also do not overlap because the first is a subset of $\mathbb Q$ and the second of $\mathbb I_1$.

*

*Surjectivity of $f$ is direct: we have partitioned $\mathbb I_1$ into two classes: $\operatorname{Im}(r_n)$ and $\mathbb I_1 \setminus \operatorname{Im}(r_n)$. We see that $f$ will inevitably hit everything.


*For injectivity, pick $x \neq y$ from $(0, 1)$. Try if $f(x) = f(y)$ can be possible in any of the subcases:

*

*$x, y\in \operatorname{Im}(q_n)$,

*$x, y\in \operatorname{Im}(r_n)$,

*$x \in \operatorname{Im}(q_n)$ and $y \in \operatorname{Im}(r_n)$,

*and so on...



You should find that $f(x) = f(y)$ will not happen, and so $f$ is injective, too.
