Timeline for Every abelian torsion-free group is strictly totally orderable (via the compactness theorem)
Current License: CC BY-SA 3.0
8 events
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| Jun 18, 2013 at 15:43 | comment | added | Andreas Blass | It may be worth noting that, although the proof given by Emil appears to use compactness for first-order logic, it really uses compactness only for propositional logic. The point is that the relevant first-order sentences are either quantifier-free (the diagram of $G$) or universal (the axioms for ordered groups). Replacing the latter by all their instances, we get just propositional combinations of atomic sentences $a\preceq b$ for $a,b\in G$. | |
| Jun 18, 2013 at 13:24 | comment | added | Emil Jeřábek | I have added a note on generalization to other structures. I think this also illustrates the difference between the two proofs. | |
| Jun 18, 2013 at 13:18 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
generalization
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| Jun 18, 2013 at 12:54 | vote | accept | Salvo Tringali | ||
| Jun 18, 2013 at 12:52 | comment | added | Salvo Tringali | Thanks. Now the basic difference between the two approaches, far beyond the wording, is clear to me too. | |
| Jun 18, 2013 at 12:35 | comment | added | Emil Jeřábek | Well, the argument is similar, but I would not say it’s quite the same. Your argument goes by embedding the whole group into something rich enough that it turns out to be easily orderable, whereas here we reduce the problem to subgroups that are poor enough to be easily orderable. Essentially, one looks at the group as the direct limit of its finitely generated subgroups. (This does not literally work, as the orders on the subgroups are not canonically chosen. The purpose of compactness here is to make these choices in a consistent way.) | |
| Jun 18, 2013 at 11:57 | comment | added | Salvo Tringali | So essentially, the model theoretic proof, if I'm not missing anything, is sort of a "rewording" of the same argument given in the OP (I omitted some details, but it should be clear how to conclude once that the problem has been embedded into $\mathbb Q^\kappa$), right? Still, interesting and quite instructive. Thank you! | |
| Jun 18, 2013 at 11:42 | history | answered | Emil Jeřábek | CC BY-SA 3.0 |