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If a lattice is countable, prove that it has a subset that is both totally ordered and cofinal in the lattice. Cofinal means that for each $l$ in the lattice, there is some $a$ in the subset such that $l\le a$.

My idea was to try to use Zorn's lemma on the set of all totally ordered subsets and prove it has a maximal element which must be cofinal, but this hasn't helped much.

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  • $\begingroup$ Do a direct construction. $\endgroup$ – Kris Sep 30 '13 at 20:05
  • $\begingroup$ How do I do a direct construction?9 $\endgroup$ – Jsjnevsjska Sep 30 '13 at 20:11
  • $\begingroup$ If your lattices are bounded, then the singleton consisting of the top element is cofinal. But in fact all you need here is a join semilattice. $\endgroup$ – Zhen Lin Sep 30 '13 at 20:15
  • $\begingroup$ The lattice isn't bounded. What is a join semi lattice? Would it be totally ordered? $\endgroup$ – Jsjnevsjska Sep 30 '13 at 20:19
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    $\begingroup$ The title "Every lattice" is false. Consider all finite sets of real numbers ordered by inclusion, as a counter example. The question in the body has a positive answer, though, as Andreas Blass shows below. $\endgroup$ – Asaf Karagila Oct 1 '13 at 5:39
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First list the elements of the lattice in a sequence (possible as the lattice is countable). Then define a new sequence whose $n$-th term is the join of the first $n$ terms of your original sequence. This new sequence is totally ordered and cofinal in the lattice.

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