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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.

  • 3
    $2^{\mathbb{N}_0} = 10^{\mathbb{N}_0}$2012-12-01
  • 0
    I was suspecting that, but could not reason it on my own yet. Thank you.2012-12-01
  • 0
    You have in-fact reasoned that out in your question. :)2012-12-01

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