How do you show that $\mathbb{R} \backslash \mathbb{Q} \cong \mathbb{N}^{\mathbb{N}}$? What is a good starting point in showing this?
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1Do you know that $\mathbb{R}$ has that cardinality? Do you know that $\mathbb{Q}$ has strictly smaller cardinality? – Tobias Kildetoft Jul 19 '13 at 12:50
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5Isomorphism in what sense? – Ido Jul 19 '13 at 12:50
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Iso in Sets, I suppose. – Edoardo Lanari Jul 19 '13 at 13:06
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$|\mathbb{R}|>|\mathbb{Q}|$ $\implies$ $|\mathbb{R} \setminus\mathbb{Q}|=|\mathbb{R}|=|2^{\mathbb{N}}|=|\mathbb{N}^{\mathbb{N}}|.$
Edoardo Lanari
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You didn't explain any step. Also, he asked for a starting point, not a solution. – Ido Jul 19 '13 at 13:15
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1The amount of explanations needed strongly depends on the knowledge of the reader; if he doesn't know anything about cardinality then every answer is useless, if he knows a bunch of theorems in cardinality computation this is really clear. Moreover the problem is so easy that starting point=ending point. – Edoardo Lanari Jul 19 '13 at 13:21
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1You could at least note what theorems you use and whether they are proved with axiom of choice or not. – Ido Jul 19 '13 at 13:32
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2+1: consise answer, which still requires the questioner to think a bit. – wildildildlife Jul 19 '13 at 13:37
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I think that each step can be done in ZF: the first is additivity plus the fact that k+l=max{k,l}, the second is surely in ZF and the third uses already mentioned facts. – Edoardo Lanari Jul 19 '13 at 14:28