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I read that the power of $(0,1)$ is $2^\mathbb{N_0}$ due to the fact that $(0,1)$ is equipotent, roughly speaking, to the set of all binary representations of the numbers in $(0,1)$ (and this set has itself a power of $2^\mathbb{N_0}$).

But $(0,1)$ can be also put in bijection with, say, all the base-10 number representations. Wouldn't that render a power of $10^\mathbb{N_0}$ for $(0,1)$?

Thank you in advance.

Andy.

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$2^{\mathbb{N}_0} = 10^{\mathbb{N}_0}$ –  user17762 Dec 1 '12 at 19:30
    
I was suspecting that, but could not reason it on my own yet. Thank you. –  Andy Dec 1 '12 at 19:36
    
You have in-fact reasoned that out in your question. :) –  user17762 Dec 1 '12 at 19:37
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2 Answers

The proof of $|(0,1)| = 2^{\aleph_0}$ constructs injections in each direction. If you're in base $10$ then $0.999^\sim = 1$ and yet, they don't have the same base $10$ representation.

The injection in the (probably, standard) proof (but there might be many other proofs) is that you take a map in $2^{\aleph_0}$ and interpret it as the base $10$ representation of that number. You nicely eliminate cases with infinitely many trailing same digits of $b-1$ where $b$ is your base. If you are in base $10$ and take a map in $10^{\aleph_0}$ the map described just now won't be injective.

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There is a bijection between infinite strings over $\{0,1,\ldots,9\}$ and infinite strings over $\{0,1\}$, consider: $$k\to 1^k0\text{ for }0\leq k\leq8, \qquad 9\to1^9.$$

It is a bijection because it is a prefix code (no image of single digit is a prefix of other such image), and every infinite word over $\{0,1\}$ is obviously image of some infinite word.

Therefore $2^{\aleph_0}=10^{\aleph_0}$.

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