I'm trying to understand how $ 2^{\aleph_0} > \aleph_0 $. I was reading through this sketch of the proof, but don't quite understand how they show that $\mathrm{card}((0,1)) = \mathrm{card}(\mathcal{P}(\mathbb{N}))$. Is there a different way of explaining this? Or maybe a different way of explaining the whole proof? I'm just trying to wrap my head around this, so any help is appreciated!
Understanding the proof for $ 2^{\aleph_0} > \aleph_0$
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elementary-set-theory
cardinals
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0Huh? I don't see any sketch here, only a reference to Cantor's proof. Also, I'm not aware of any proofs of the inequality using equinumerosity of the interval and the power set... – 2012-10-08
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0@tomasz It is easy to prove that $P(N)$ and $(0,1)$ have the same cardinality, and then Cantor diagonalization shows that $P(N)$ is uncountable.... – 2012-10-08
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1@N.S.: How so? Cantor's diagonalization, as I know it, shows that $2^{\bf N}$ is uncountable, without $(0,1)$ in-between. – 2012-10-08
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0@tomasz Usually Cantor diagonalization is used to prove that $R$ or $(0,1)$ is uncountable, without needing $2^N$. – 2012-10-08
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1@N.S.: Really? Care to explain? The only way to do that that I can think of is by binary expansion, which is really a faux bijection of the interval and $2^\bf N$, or something of the sort. – 2012-10-08
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0Here is a link to a particularly nice and intuitive proof that the power set of a set is always bigger than the original set: http://math.stackexchange.com/a/164185/28900 – 2012-10-08
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0@tomasz You assume by contradiction that $(0,1)$ is countable, you write the numbers as $x_1,...,x_n,..$ and then define $y=0.y_1y_2...y_n...$ where $y_n$ is any digit different of the n'th digit of $x_n$. – 2012-10-08
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0@N.S.: That's exactly what I have said. What you're really doing is using the almost-bijection with $2^{\bf N}$ (or $10^{\bf N}$, if you're using decimal representation). – 2012-10-08
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0Another proof of the relation $|(0,1)|=|\mathcal P(\mathbb N)|$ is [here](http://math.stackexchange.com/a/388998/462). – 2013-07-15