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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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    See also: http://math.stackexchange.com/questions/1182145/finding-an-uncountable-chain-of-subsets-the-integers2015-06-02

1 Answers 1

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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.

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    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.2012-12-08