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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8Hint: $\Bbb{N}$ has the same cardinality as $\Bbb{Q}$. – Chris Eagle Dec 07 '12 at 17:44
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See also: http://math.stackexchange.com/questions/1182145/finding-an-uncountable-chain-of-subsets-the-integers – Martin Sleziak Jun 02 '15 at 09:52
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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.
Asaf Karagila
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1Amazing! Maybe magic is in the eyes of ignorant beholders like me. – Metta World Peace Dec 07 '12 at 23:52
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@Metta: It's a bunch of tricks, and you learn them with time. :-) – Asaf Karagila Dec 08 '12 at 00:04
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2And 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 08 '12 at 01:07