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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?

etel876
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1 Answers1

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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 $\mathbb I_1$.

  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.

  2. 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.

Linear Christmas
  • 1,285
  • $x,y ∈ q_n$ or $im(q_n)$ ?which one is correct?(according to your last lines of answer) – etel876 Apr 23 '21 at 18:46
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    @solid_math Traditionally, a sequence $(x_n) := (x_n)_{n\in \mathbb N_0}$ is defined as a function from $\mathbb N_0$ to whatever set $X$ under consideration. One denotes $x(n) =: x_n$. From this point of view, the values a sequence takes on, i.e. $x_1, x_2, x_3$ and so forth, are members of $\operatorname{Im} (x_n)$. I hoped to avoid future confusion by being extra explicit here. But it is common to be less strict about one's notation in this case and assume that the reader understands what the writer is referring to. – Linear Christmas Apr 23 '21 at 18:56