2
$\begingroup$

Prove that the sets $\Bbb{R\times R}$ and $\Bbb{R}$ are equinumerous, without $\mathsf{AC}$.

(The solutions in this, this and this posts are a bit more complicated, so I'm trying a different approach).

My attempt: As $\Bbb{R}$ and $(0,1)$ are equinumerous, and as there is an obvious injection $\Bbb{R} \to \Bbb{R\times R}$, by using using Cantor–Schröder–Bernstein theorem, it is sufficient to find an injection $(0,1)\times (0,1)\to(0,1)$. For an ordered pair $\langle r,s\rangle$ of reals between $0$ and $1$, let $$ r=0.r_1 r_2 \ldots r_n\ldots \\ s=0.s_1 s_2 \ldots s_n\ldots $$ be some decimal expansion of them. Send $\langle r,s\rangle$ to the number defined by $$ 0.r_1 s_1 0 r_2 s_2 0 \ldots r_n s_n 0\ldots $$

This number must have a unique decimal expansion, as it cannot have any infinite series of nines in it. As we can send each real of form $0.a_1 a_2 0 a_3 a_4 0\ldots$ to the two reals $0.a_1 a_3 a_5 \ldots$ and $0.a_2 a_4 a_6 \ldots$, the described function must be one-to-one.

Is this solution OK?

Edit: There's a problem with the fact that $r$ and $s$ in general might not have a unique decimal expansion. We can overcome this problem by excluding from $(0,1)$ the set of numbers that have a decimal expansion with almost all digits being nines. That set is obviously countable.

$\endgroup$
4
  • $\begingroup$ @ssp thank you, I edited the post. $\endgroup$
    – Roy Sht
    Commented Jul 25, 2022 at 19:53
  • 1
    $\begingroup$ If either $r$ or $s$ has a nonunique representation, then just use the one with infinitely many zeroes (or alternatively, infinitely many nines) so that the injection is well-defined. $\endgroup$ Commented Jul 25, 2022 at 20:18
  • $\begingroup$ "That set is obviously countable." Doesn't this involve AC? $\endgroup$
    – Paul Frost
    Commented Jul 25, 2022 at 22:00
  • $\begingroup$ @PaulFrost I think $\omega^*= \omega^{<\omega}$, the set of finite sequences of natural numbers, is countable (without using $\mathsf{AC}$): send each finite series $n_1 , \ldots , n_k$ to $p_1 ^{n_1}\cdot\ldots\cdot p_k^{n_k}$, where $p_n$ is the $n$th prime. This is an injection into $\omega$, so by Cantor–Schröder–Bernstein, $|\omega^*| = \aleph_0$. There's a clear bijection between numbers in $(0,1)$ with almost all digits being $9$ to $\omega^*$ by sending such real to the sequence ending with the last non-$9$ digit. Right? $\endgroup$
    – Roy Sht
    Commented Jul 25, 2022 at 22:15

0

You must log in to answer this question.