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From Cantor's theorem follows that the set of all subsets of $N$ is uncountable.

Cantor's theorem:

If $A$ is any set,then there is no bijection of $A$ onto the set $2^A$

But why this set is continuum? Cardinality of $R$ is continuum but I don't know how prove from this that set given above is also continuum.

Hint: think of an element of $P(N)$ as corresponding to a binary expansion-for a given subset A, let $x=\sum_{i\in A}2^i $ This gives you (almost) a bijection between P(N) and [0,1]

Can you explain why this given bijection?

Henno Brandsma
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unit 1991
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  • @JMoravitz How $\frac12=0.1\bar0_2=0.0\bar1_2$? – unit 1991 Feb 03 '22 at 15:22
  • for the same reason that 1 = 0.999999... – JMoravitz Feb 03 '22 at 15:22
  • Yes but $\frac{1}{2} = 0.5\bar0$? – unit 1991 Feb 03 '22 at 15:24
  • ....... In decimal you have things like $512.36 = 5\cdot 10^2 + 1\cdot 10^1+2\cdot 10^0+3\cdot 10^{-1}+6\cdot 10^{-2}$. In binary you have things like $110.10_2 = 1\cdot 2^2 + 1\cdot 2^1+0\cdot 2^0+1\cdot 2^{-1}+0\cdot 2^{-2}$. Here... $\frac{1}{2}=0.1\overline{0}_2$ is saying that one half is equal to $0$ plus one half plus zero quarters plus zero eighths plus zero for each other power of two... Of course one half is equal to "one half plus nothing else" – JMoravitz Feb 03 '22 at 15:26
  • Thanks I am now reading similiar question and if problems arise I will edit this question – unit 1991 Feb 03 '22 at 15:30

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A quick proof would be (if you know that the Cantor set has cardinality $\mathfrak{c})$ every branch of the full binary tree of height $\omega$ is in bijection with $\mathcal{P}(\mathbb{N})$, and every branch corresponds to a single element of the Cantor set by decimal representation in base $3$. Moreover, that correspondence is a bijection.

Good Morning Captain
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