Definition talk:Order Completion

Apparently (and I am a bit reluctant here; WP is not the greatest source) this concept does not generally exist (cf. here) because the corresponding complete poset will turn into a proper class (and hence isn't a poset). --Lord_Farin 10:59, 8 April 2012 (EDT)
 * My problem is that I never had anyone sit down and wag their finger at me sternly and explain what is so important about distinguishing between sets and classes, beyond the fact that the latter just seem to me to be an awkward workround to Russell's Paradox. --prime mover 11:11, 8 April 2012 (EDT)
 * I'm self-taught here, but I suspect that there exists no notion of a 'power class'. Simply put, a class may be identified with a certain predicate (it consists precisely of all sets having this property). A power class would be a 'class of classes' which is void. This probably gives a glimpse of why sets and classes are truly different. --Lord_Farin 17:27, 8 April 2012 (EDT)
 * Goes some way towards my own discomfort with the concept of a "class": it's "like a set" but "it's got extra stuff in it" and so "you can't do some of the stuff you can do with sets" ... it all comes across as messy and incomplete. And if it's ugly, it's probably not at its best. --prime mover 17:30, 8 April 2012 (EDT)

Alternatively, you could simply restrict to a Grothendieck universe (using a hypothetical universal set, but still satisfying all ZF axioms). Such a universe $U$ probably by definition does not contain itself, but I'm not familiar with this beyond the name. --Lord_Farin 17:33, 8 April 2012 (EDT)

Strange
This is basically defined to be unique. Most peculiar... I would expect something more along the lines of the completion being minimal relative to the embedding. That is, I would have expected that $T$ is an order completion of $S$ iff it is a complete superset of $S$ such that $S \subseteq U \subsetneqq T$ implies $U$ is not complete. --Dfeuer (talk) 00:06, 8 January 2013 (UTC)


 * It would be nice to prove the approaches equivalent. That is *true*, right? --Lord_Farin (talk) 00:11, 8 January 2013 (UTC)


 * No idea! I would hope so. --Dfeuer (talk) 00:14, 8 January 2013 (UTC)


 * Actually, it's not. Not only that, but the current definition is so wrong that no poset ever has a completion. --Dfeuer (talk) 00:16, 8 January 2013 (UTC)


 * Certainly $\overline \R$ is a completion of $\R$... Your bald claim merits a (worked) example. --Lord_Farin (talk) 00:19, 8 January 2013 (UTC)


 * My claim probably was too broad, but the extended reals are a good example. The extended reals are complete and contain the reals. The extended reals with an extra "super-infinity" tacked onto the end are also complete and contain the reals, but the super-extended reals are not isomorphic to the extender reals, so the extended reals are not a completion. --Dfeuer (talk) 00:24, 8 January 2013 (UTC)


 * Super-extended reals do not adhere to $(4)$, since there is not a unique order-preserving mapping from them to $\overline \R$. (NB: I'm off to bed.) --Lord_Farin (talk) 00:26, 8 January 2013 (UTC)
 * Order-preserving mapping is different from order isomorphism. That's where your thoughts took a wrong turn, I think. --Lord_Farin (talk) 00:28, 8 January 2013 (UTC)


 * That's near where my thoughts took a wrong turn. Whoops. --Dfeuer (talk) 01:42, 8 January 2013 (UTC)

Injections?
I switched from order-preserving mappings to order-preserving injections. The former clearly doesn't work, because any constant mapping is order preserving, but I don't actually know if the latter is the right condition either. Do we need order-preserving injections, or do we need strictly order-preserving mappings? My guess is injections, but I don't actually know yet. --Dfeuer (talk) 22:37, 9 January 2013 (UTC)


 * Coming to think of it, what we actually want is an injection, and that it is the identity on $S$. Otherwise $\overline \R$ is not going to cut it (consider any closed interval where we map $\R$ to the interior). It can be phrased in category-theoretic language but I don't want to do that right now. --Lord_Farin (talk) 22:43, 9 January 2013 (UTC)
 * Admittedly I'm not really sure; I crafted this one off the top of my head when I felt like doing so. --Lord_Farin (talk) 22:45, 9 January 2013 (UTC)


 * No, no, no, that's not right either. Is the mapping currently defined in the wrong direction???? There could be arbitrarily large complete lattices including $S$ which can have no injection into $T$. Side thought: such things are very often defined as embeddings, rather than as sets. Some things are easier that way; some harder. --Dfeuer (talk) 23:00, 9 January 2013 (UTC)


 * I refer you to the second paragraph of my preceding post. My apologies. Embedding from $S$; mapping in $(4)$ should make the triangle based at $S$ (i.e. with the embeddings) commute. Well, in any case, what we want is that in the category with objects embeddings $\iota: S \to T$ ($T$ complete) and morphisms such comm. triangles, the order completion is an initial object. --Lord_Farin (talk) 23:03, 9 January 2013 (UTC)


 * As I haven't studied category theory yet, some of that sounds vaguely familiar from studying topology and the rest sounds like alchemy --Dfeuer (talk) 23:13, 9 January 2013 (UTC)

FYI: Encyclopedia of Math has a different sort of construction. I don't quite understand it, but it sort of looks like they're forming a subset of $\mathcal P(S)$ which is complete and for which there is an embedding from $S$. I think. I was thinking something vaguely like that would be useful to prove existence of whatever we end up defining as an order completion. --Dfeuer (talk) 23:41, 9 January 2013 (UTC)


 * For some reason, I can't access the site. Is that construction similar to Existence of Dedekind Completion, without the extra condition that $L$ is non-empty and bounded above? --abcxyz (talk) 16:05, 10 January 2013 (UTC)

Am I right to guess that this is what we want? Let $(S,\le_S)$ be a poset and let $(T,\le_T)$ be a complete lattice. Then an order embedding $f: S \to T$ is an order completion of $S$ iff for any complete poset $(U,\le_U)$, if $g: S \to U$ is an order embedding then for some order embedding $h: T \to U$, $g = h \circ f$ ? --Dfeuer (talk) 00:06, 10 January 2013 (UTC)


 * Probably. I'm currently trying to do the same thing with the Dedekind completion. --abcxyz (talk) 00:50, 10 January 2013 (UTC)