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Proof that the real numbers are countable: Help with why this is wrong

I know there are uncountably many real numbers between 0 and 1, and I am trying to figure out why this argument is wrong:

Proof that there are finitely many real numbers between 0 and 1:

Consider the string of digits after the decimal point.

We start with strings of length 0, then length 1, then length 2.

Each set of strings has only finitely many elements, so they can be numbered.

We will eventually reach every string by this process.

Therefore we can enumerate all the real numbers between 0 and 1.

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marked as duplicate by Sivaram Ambikasaran, Andres Caicedo, Zev Chonoles Dec 2 '11 at 7:03

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Correction: We will eventually reach every string of finite length by this process. Therefore we can enumerate all the dyadic numbers between 0 and 1 (i.e. those whose dyadic expansion stops eventually). –  Did Dec 2 '11 at 6:26
    
Do you mean that you are giving an incorrect proof that the real numbers from 0 to 1 are countable, not finitely many? Either way, there are problems. –  Aleks Vlasev Dec 2 '11 at 6:26
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The problem is that the real numbers consist of strings of infinite length and as such are quite different. –  Aleks Vlasev Dec 2 '11 at 6:28
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@James: As Henning points out in this (math.stackexchange.com/questions/61174/…) post, even rational numbers with denominators divisible by primes other than $2$ and $5$ are not in your list. –  user17762 Dec 2 '11 at 6:32

1 Answer 1

$1/\sqrt 2$ is not in your list. Neither is $\pi-3$. I could go on...

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But if you can enumerate the numbers not on the list then $\dots$. –  André Nicolas Dec 2 '11 at 6:44
    
@André: I'll let you do that. I'm a bit rushed today. –  TonyK Dec 2 '11 at 6:57

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