Exam preparation question:
How to find a partial ordered set $(P,\leq)$, such that $|P|=\aleph_0 $ and the set of all maximal chain in $(P,\leq)$ is $2^{\aleph_0}$?
Thank you!
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I am fairly certain that you asked this before and that Brian had answered that. – Asaf Karagila Feb 08 '13 at 21:28
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@Asaf: This one? – Brian M. Scott Feb 08 '13 at 21:31
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@Brian: Yes, I just found it myself as well! – Asaf Karagila Feb 08 '13 at 21:32
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oops, my apologies. – 17SI.34SA Feb 08 '13 at 21:34
2 Answers
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HINT: The complete binary tree of height $\omega$ has $\omega$ (or $\aleph_0$, if you prefer) vertices and $2^\omega$ branches.
Brian M. Scott
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Somewhat less usual than Brian's answer:
All the (nontrivial) open intervals in $\mathbb R$ with rational endpoints, ordered by reverse inclusion.
This poset is countable, and every maximal chain corresponds to a decreasing sequence of open sets whose intersection must be at most a single real number. There are at most $2^{\aleph_0}$ chains, and for every real number we can generate such maximal chain. Therefore there are exactly $2^{\aleph_0}$ maximal chains.
Asaf Karagila
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1For each real number you can generate quite a few maximal chains. You’re still okay, of course, since there are only $2^\omega$ sets of intervals. – Brian M. Scott Feb 08 '13 at 21:37
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Yes, I didn't say that each real generates a unique chain... But it's a good as any argument for showing there are exactly $\frak c$ maximal chains. – Asaf Karagila Feb 08 '13 at 21:38
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1There are maximal chains with empty intersection, such as ${(0,q)\mid q>0}$. (This does not upset the argument here, of course, for the same reason as Brian's point doesn't). – hmakholm left over Monica Feb 08 '13 at 22:20
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AsafKaragila, Brian M.Scott: thank you very much for all the help in last three months... :) – 17SI.34SA Feb 12 '13 at 16:04
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@17SI.34SA: Sure. I hope the exam wasn't too hard. Maybe I'll see you in my class in three weeks, too. – Asaf Karagila Feb 12 '13 at 16:05