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How one can find a subset of $(P(\mathbb{N}),\subseteq)$ without maximal element? ($P(\mathbb{N})$ is the power set of $\mathbb{N}$)

I think that I need point an antichain. Right?

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The empty set ? – Dan Brumleve Dec 27 '12 at 9:01
up vote 5 down vote accepted

First note that every non-empty finite subset has a maximal element. Such subset has to be infinite.

If you require the subset to be a chain then $\Big\{\{j\in\mathbb N\mid j<k\}\mid k\in\mathbb N\Big\}$ is such subset. Otherwise just taking all the finite subsets would work.

You may also note that the collection of the co-infinite, i.e. $\{A\subseteq\mathbb N\mid\mathbb N\setminus A\text{ is infinite}\}$ is also without a maximal element.

Lastly, an antichain will not work. Every element is maximal within the antichain.

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To be more explicit: Every nonempty finite subset has a maximal element. – Hagen von Eitzen Dec 27 '12 at 9:29

Just consider the collection of finite subsets of $ \mathbb{N} $. There cannot be a $ \subseteq $-maximal element because given a finite subset $ A $ of $ \mathbb{N} $, you can find a finite subset $ B $ of $ \mathbb{N} $ that properly contains $ A $.

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