Talk:Reciprocal Function is Strictly Decreasing/Proof 2

I vote revert the changes and make this a third proof. --GFauxPas (talk) 14:08, 22 February 2013 (UTC)


 * Current proof has flaws as well, but is potentially better than the previous one. Duality incantations need to be phrased proper. There isn't as of now a page proving that $(\R,\le) \to (\R,\ge), x \mapsto -x$ is an order isomorphism. This result is, however, tacitly used. Because the approaches are slightly different both proofs need to be retained. The previous proof needs the superfluous stuff on trichotomy cut out, as Dfeuer rightly tried to point out. --Lord_Farin (talk) 16:15, 22 February 2013 (UTC)


 * We really need to get the "strictly increasing" definition sorted, and decide whether we're trying to prove strictly decreasing or dual order embedding. If the latter, we need the dual form of the order monomorphism theorem (which is where the trichotomy belongs, not in this poor theorem). Speaking of which, I really like the word "embedding". Is there a reason for "monomorphism" there? --Dfeuer (talk) 16:24, 22 February 2013 (UTC)


 * As for the negatives, that's pushed off to the theorem this one has always invoked, whose proof needs to be completed. However: I did not look into whether this way actually makes sense, or whether that theorem should be turned to a strict form to simplify matters. --Dfeuer (talk) 16:30, 22 February 2013 (UTC)


 * Why is the trichotomy stuff unneeded? --GFauxPas (talk) 17:11, 22 February 2013 (UTC)


 * Because proving an implications allows one to assume the antecedent (this is the Deduction Theorem, which is &mdash; strangely enough &mdash; not up atm). What other options there are is thus completely immaterial. --Lord_Farin (talk) 17:24, 22 February 2013 (UTC)


 * @Dfeuer, Sorry, but I fail to comprehend your last post. --Lord_Farin (talk) 17:24, 22 February 2013 (UTC)

Our current definitions of "strictly increasing mapping" and "strictly decreasing mapping" actually produce order embeddings and dual order embeddings. For totally ordered sets, these notions are equivalent, but for partially ordered sets they are not. I would suggest changing our definitions of strictly increasing and strictly decreasing to the weaker forms, so as not to force a bunch of unnecessary invocations of the equivalence theorem for tosets. --Dfeuer (talk) 17:38, 22 February 2013 (UTC)


 * Inclined to agree. PM said he'd consult his sources; let us wait for what that brings. Also, this appears to concern the first of your posts here, not the one I meant. --Lord_Farin (talk) 17:46, 22 February 2013 (UTC)


 * Cf. Talk:Mapping from Toset is Order Embedding iff Strictly Increasing. Definition is now as suggested (and as appears natural to me). --Lord_Farin (talk) 18:12, 22 February 2013 (UTC)


 * Oh, right, thanks for fixing that (regarding the unnecessary trichotomy stuff). We have Rule of Implication, that's what you mean, right? Anyway, now that there isn't a need to prove the reverse implication, there's only two proofs after all, right? --GFauxPas (talk) 18:51, 22 February 2013 (UTC)


 * Yes, Rule of Implication is what I mean. I tentatively agree that there are only two proofs. --&mdash; Lord_Farin (talk) 22:09, 22 February 2013 (UTC)