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We can show that $\mathbb{Q}$ is isomorphic to $\mathbb{N}$ by constructing a bijection between the two with Cantor's diagonalization argument. Is there a similar way we can show that $\mathbb{Q}^c$ is isomorphic to $\mathbb{R}$?

I was thinking we could do it in the following way, if we can show that the cardinality of the reals and the irrationals are the same then we are done.

It is pretty obvious that the size of $\mathbb{Q}^c$ can be at most the size of $\mathbb{R}$, but we also know that $\mathbb{Q}^c\not\simeq \mathbb{N}$ so the size of $\mathbb{Q}^c$ is not $\aleph_0$. But then this is just the continuum hypothesis.

Is there another way to construct a bijection that gives us an answer that is decidable?

Ltoll
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  • What is the c supposed to mean? Is $c\in\mathbb{N}$? Since $\mathbb{Q}^n$ is countable for every $n$, so there can not be a bijection from $\mathbb{Q}^n\to\mathbb{R}$ – Cornman Oct 01 '17 at 01:34
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    I think $\mathbb Q^c = \mathbb R \setminus \mathbb Q$? – Anthony Carapetis Oct 01 '17 at 01:35
  • Yeah It means the complement of $\mathbb{Q}$, or $\mathbb{R}\backslash\mathbb{Q}$. – Ltoll Oct 01 '17 at 01:36
  • Anyway, assuming only ZF I think this is a little tricky - as you note, the obvious argument tells us that $|\mathbb Q^c| > |\mathbb N|$, but this does not show it is equal to $|\mathbb R|$ without assuming CH. With the axiom of choice I think there is an easy proof that $|A \setminus B| = |A|$ when $|A| > |B|$ are both infinite. WIthout, see e.g. this answer. – Anthony Carapetis Oct 01 '17 at 01:41
  • Let $\mathbb Q={r_n:n\in\mathbb N}$ be an enumeration of the rational numbers, and define an injection $f:\mathbb R\to\mathbb R\setminus\mathbb Q$ by setting $$f(x)=\sum_{r_n\lt x}10^{-n^2}.$$ – bof Oct 01 '17 at 02:39
  • Isomorphism is the wrong term. Your are asking about being equinumerous. – William Elliot Oct 01 '17 at 03:24

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