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Is there a simple, easy to understand, proof of the following?

$$ |\mathcal{P} (\mathbb{N}) | = |\mathbb{R}|$$

That is, the cardinality of the power set of the natural numbers is the same as the cardinality of the real numbers.

The suggestions question has so much discussion, I am not sure I can take the answer as settled. The simplest way of proving that $|\mathcal{P}(\mathbb{N})| = |\mathbb{R}| = c$

Penelope
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  • There's a (relatively) easy way to define an explicit bijection betaeeen the irrational reals and the infinite subsets of $\mathbb N,$ by considering the continued fraction expansion of the irrational. Since the set of f8nite subsets of $\mathbb N$ is countable, and the set of rationals is countable, you can thus show an explicit bijection between $\mathbb R$ and $P(\mathbb N).$ The key is proving that the continued fraction for an irrational number is unique. – Thomas Andrews Jul 16 '23 at 23:27
  • First show that $[0,1]$ has the same cardinality as $\mathbb R$. Then represent elements of $[0,1]$ as binary sequences which are equivalent to subsets of natural numbers. – John Douma Jul 16 '23 at 23:28
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    @JohnDouma but the binary representation is not always unique. – Thomas Andrews Jul 16 '23 at 23:29
  • @ThomasAndrews sure, but it's just two representatives at most for each number. Just mod out by the equivalent binary representations. – Brevan Ellefsen Jul 17 '23 at 01:50
  • @ThomasAndrews: Continued fractions establish a bijection between sequences of positive numbers and irrationals in the interval $(0,1)$. The sequence can even have finite range. – Chad K Jul 17 '23 at 06:13
  • @Penelope: It's easier to show that the Cantor set and $\mathcal{P}(\Bbb{N})$ are equinumerous - because there's a bijection defined by one formula. For $\Bbb{R}$ you need to stitch together several maps or use Schröder-Bernstein. – Chad K Jul 17 '23 at 06:19
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    To keep things simple, you can see $\mathcal{P}(\mathbb{N})$ as the decimal representation of the real numbers. See e.g. https://en.wikipedia.org/wiki/Cardinality_of_the_continuum#Alternative_explanation_for_%F0%9D%94%A0_=_2%D7%900 – Abezhiko Jul 17 '23 at 11:18

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