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I want to prove that $P((0,1)-\mathbb{ Q}) = 2^{\mathbb{R}}$, i have a proof which is roughly that $((0,1)-\mathbb{Q}) \cup \mathbb{Q} = (0,1)$ and there intersection is empty so assuming the AC we have that $|\mathbb{R} |= |(0,1)| = |(0,1)-\mathbb{Q}| + |\mathbb{Q}| = \max(|(0,1)-\mathbb{Q}|, \aleph_0)$ , my question is there a proof that $P((0,1)-\mathbb{ Q}) = 2^{\mathbb{R}} $ without AC ?

I have another question, what is the cardinality of the set of all real numbers whose digits are odd in the form $0.a_1 a_2 \cdots$ , i suspect its $|\mathbb{R}|$ but i don't know how to prove it?

Andrés E. Caicedo
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Ahmad
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    This doesn't use choice. – Andrés E. Caicedo Apr 16 '19 at 13:16
  • @AndrésE.Caicedo so this proof is the simpler ?! – Ahmad Apr 16 '19 at 13:17
  • @Andrés I mean you use AC implicitly to define the cardinality, right? But that's unavoidable – Sambo Apr 16 '19 at 13:18
  • You should write better titles at this point. Your current title is meaningless and tells us absolutely nothing about the question. You have 150 characters to use, and you are more than welcomed to put $\rm\LaTeX$ there as well. Writing better titles will help you find these duplicates before you even type your question. – Asaf Karagila Apr 16 '19 at 14:21

1 Answers1

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We can find a bijection between $(0,1)-\mathbb Q$ and $(0,1)$ without using choice, as was already mentioned in the comments. We can use the Cantor-Bernstein-Schröder theorem for this. Clearly we have an injection in the direction $|(0,1)-\mathbb Q| \leq |(0,1)|$. For the other direction we can do something as follows: fix your favourite irrational number $a \in (\frac{1}{2},\frac{3}{4})$ (e.g. take $a = \frac{\sqrt 2}{2}$). Now we define $f: (0,1) \to (0,1)-\mathbb Q$ by $$ f(x) = \begin{cases} \frac{x}{2} & \text{if $x$ is irrational} \\ a + \frac{x}{4} & \text{if $x$ is rational} \end{cases} $$ So essentially $f$ sends all the irrational numbers to the first half of $(0,1)-\mathbb Q$ and it sends all the irrational numbers somewhere in the second half. It should be clear that $f$ is injective.

So now that we have injections $|(0,1)-\mathbb Q| \leq |(0,1)|$ and $|(0,1)| \leq |(0,1)-\mathbb Q|$ we can conclude from Cantor-Bernstein-Schröder that $|(0,1)-\mathbb Q| = |(0,1)|$ without using choice. So now the rest of your proof goes through.

For the second question, we know that $|\mathbb R| = 2^{|\mathbb N|}$. Usually we would see elements in $2^{|\mathbb N|}$ as an infinite sequence of 1s and 0s, but we might as well see them as infinite sequences of 1s and 3s. So this way we can encode every such sequence as a real number of the form $0.a_1 a_2 \ldots$, where the $a_i$ are odd. You will have to convince yourself that no two such sequences will represent the same real number. Then we have an injection from $2^{|\mathbb N|}$ to the set of these 'odd' real numbers. Now use Cantor-Bernstein-Schröder again to conclude that its cardinality is indeed $|\mathbb R|$.

Mark Kamsma
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