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Mar 7, 2017 at 3:18 comment added Gerhard Paseman I just noticed the proof works under the weaker condition that no two edges are comparable. So condition 2 can be replaced by e1 intersect e2 is strictly smaller than both. Possibly also condition 1 can be weakened to remove mention of size. However, some other condition maybe needed to force L to be countable. Gerhard "Adding Generality And Wider Applicability" Paseman, 2017.03.06.
Feb 21, 2017 at 16:48 comment added Gerhard Paseman I like your presentation. I have no problems with your keeping it around. Gerhard "Enough Rep To Go Around" Paseman, 2017.02.21.
Feb 21, 2017 at 16:39 vote accept Dominic van der Zypen
Feb 21, 2017 at 16:23 comment added bof @GerhardPaseman Did I just repeat your proof (which I was too lazy to read)? If so I will delete my answer.
Feb 21, 2017 at 15:40 comment added bof Since $L$ is countably infinite, we may suppose that the elements of $L$ have been enumerated as $b_1,b_2,\dots,b_n,\dots.$ Now, at each stage of the construction, having constructed a finite path $\langle e_1,e_2,\dots,e_k,e_{k+1}\rangle,$ we let $b$ be the first $b_i$ which does not yet been visited, and let $a=e_{k+1},$ and using the Lemma we extend the path $\langle e_1,\dots,e_k,a\rangle$ to a path $\langle e_1,\dots,e_k,a,b\rangle$ or $\langle e_1,\dots,e_k,a,c,b\rangle$ or $\langle e_1,\dots,e_k,a,c,d,b\rangle$ containing the vertex $b.$
Feb 21, 2017 at 15:36 comment added Gerhard Paseman Bof uses (but does not state) an enumeration of L. (Gender presumption) His a is my P(n) and his b is my U. Gerhard "To Get Personal About It" Paseman , 2017.02.21.
Feb 21, 2017 at 11:07 comment added Dominic van der Zypen That's very nice - but doesn't this prove that there is a injective map $p:\omega \to L$ with the desired property, but the $p$ you construct is not necessarily surjective? Or am I missing something?
Feb 21, 2017 at 9:17 history answered bof CC BY-SA 3.0