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Where does this argument fail? (my opinion; the conclusion is a non sequitur)

If you just map the reals to the naturals in a mirror image of each other you will eventually have all the reals. Like this.

1 -> 0.1

2 -> 0.2

3 -> 0.3

...

9 -> 0.9

10 -> 0.01

11 -> 0.11

12 -> 0.21

...

99 -> 0.99

100 -> 0.001

101 -> 0.101

102 -> 0.201

...

...999 -> 0.999...

If you keep going forever you will eventually have all the real numbers mapped.

Friedrich
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    "Then we can take another real number (in any arbitrary order) and extract a natural number from it that is not in our set." That works if you think that previously you've only chosen finitely many natural numbers - then you know you haven't chosen them all yet. But that would mean you are assuming you've got a sequence $x_1,x_2,\dots$ of real numbers which covers all the real numbers. If there was such a sequence, your argument would be correct. – Thomas Andrews Jun 06 '16 at 00:35
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    (But there is no such sequence.) – Thomas Andrews Jun 06 '16 at 00:35

1 Answers1

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Consider then the interval $[0,1]$. By this argument, every natural number can be extracted from some real number in this interval. So you've shown an injection $f: \mathbb{N} \rightarrow [0,1]$, i.e. $|\mathbb{N}| \leq |[0,1]|=\mathfrak{c}$, i.e. you haven't really shown $|\mathbb{N}|> \mathfrak{c}$.

M10687
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    The correct answer is "which number is mapped to $\frac13$?", otherwise you're arguing in a somewhat circular fashion. Another option would be to say "Let's trace Cantor's proof on this specific list of real numbers" and generate a real number which will provably not be in the enumerated list. Just saying "well, it contradicts a known theorem" is sort of a cop out that shouldn't really convince anyone who is trying to come into terms with the failure of this proof. – Asaf Karagila Aug 09 '16 at 11:18
  • The question as it stands now has been edited as is pretty obvious if you read my response. The original question asked why the argument didn't show $|\mathbb {N}| > \mathfrak {c}$, and the first paragraph of my answer says why. It's pretty obvious I'm not answering exactly the same question here. – M10687 Aug 09 '16 at 12:29