# Where the set of maximal chains is $2^{\aleph_0}$

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 8 at 21:28
@Asaf: This one? –  Brian M. Scott Feb 8 at 21:31
@Brian: Yes, I just found it myself as well! –  Asaf Karagila Feb 8 at 21:32
oops, my apologies. –  17SI.34SA Feb 8 at 21:34

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.

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For 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 8 at 21:37
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 8 at 21:38
There 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). –  Henning Makholm Feb 8 at 22:20
@Henning: Thanks, I corrected this. –  Asaf Karagila Feb 8 at 22:22
AsafKaragila, Brian M.Scott: thank you very much for all the help in last three months... :) –  17SI.34SA Feb 12 at 16:04
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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.

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