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How can I find a chain of length $2^{\aleph_0}$ in $ (P(\mathbb{N}), \subseteq )$.

The only chain I have in mind is

$$\{\{0 \},\{0,1 \},\{0,1,2 \},\{ 0,1,2,3\},...,\{\mathbb{N} \} \}$$

But the chain is of length $\aleph_0$, right?

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Hint: $\Bbb{N}$ has the same cardinality as $\Bbb{Q}$. –  Chris Eagle Dec 7 '12 at 17:44

1 Answer 1

up vote 8 down vote accepted

Hint: Since there is a bijection between $\mathbb Q$ and $\mathbb N$ there is an order isomorphism between their power sets with inclusion.

Now think about Dedekind-cuts.


Also, your chain is indeed countable.

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Amazing! Maybe magic is in the eyes of ignorant beholders like me. –  Metta World Peace Dec 7 '12 at 23:52
    
@Metta: It's a bunch of tricks, and you learn them with time. :-) –  Asaf Karagila Dec 8 '12 at 0:04
    
And this trick's really pretty natural: it's hard to directly solve this for $\Bbb{N}$, because $\Bbb{N}$ is in some vague sense too "small". So replace $\Bbb{N}$ with a "bigger" countable set and the problem becomes easier. –  Chris Eagle Dec 8 '12 at 1:07

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