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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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    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).2011-12-02
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    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.2011-12-02
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    The problem is that the real numbers consist of strings of infinite length and as such are quite different.2011-12-02
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    @James: As Henning points out in this (http://math.stackexchange.com/questions/61174/proof-that-the-real-numbers-are-countable-help-with-why-this-is-wrong) post, even rational numbers with denominators divisible by primes other than $2$ and $5$ are not in your list.2011-12-02

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$1/\sqrt 2$ 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 $\dots$.2011-12-02
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    @André: I'll let you do that. I'm a bit rushed today.2011-12-02