## Order embedding from a poset into a complete lattice

Showing 1-3 of 3 messages
 Order embedding from a poset into a complete lattice Victor Porton 12/30/12 10:40 AM I asked this question at http://math.stackexchange.com but received no answer. Let A is an arbitrary poset. Does it necessarily exist an order embedding from A into some complete lattice B, which preserves all suprema and infima defined in A? -- Victor Porton - http://www.mathematics21.org Re: Order embedding from a poset into a complete lattice Butch Malahide 12/30/12 4:10 PM On Dec 30, 12:40 pm, Victor Porton wrote: > I asked this question athttp://math.stackexchange.com but received no > answer. > > Let A is an arbitrary poset. > > Does it necessarily exist an order embedding from A into some complete > lattice B, which preserves all suprema and infima defined in A? Yes, the Dedekind-MacNeille completion of A has this property. https://en.wikipedia.org/wiki/Dedekind%E2%80%93MacNeille_completion By the way, there are two grammatical errors in your post. For "Let A is" read "Let A be". For "Does it necessarily exist" read "Does there necessarily exist". Re: Order embedding from a poset into a complete lattice Rotwang 12/30/12 4:29 PM On 30/12/2012 18:40, Victor Porton wrote: > I asked this question at http://math.stackexchange.com but received no > answer. > > Let A is an arbitrary poset. > > Does it necessarily exist an order embedding from A into some complete > lattice B, which preserves all suprema and infima defined in A? I believe the answer is yes. For each a in A, let    f(a) = {b in A | b <= a}. Let B consist of all intersections of images under f of subsets of A, that is    B = {\bigcap {f(a) | a in X} | X \subset A} (where the empty intersection is taken to be the whole of A). B, ordered by the subset relation, is a poset. It's clearly closed under arbitrary intersections, and therefore downward-complete (i.e. any subset has an infimum). But any poset that is downward-complete is also upward complete, for if S is a subset of B then let    sup S = inf {T in B | T >= X for all X in S}; it is easy to show that sup S is indeed a supremum for S. So B is a complete lattice, and the map f: A -> B as defined above is order-preserving. It's also injective, for if f(a) = f(b) then    a <= a, so    a is in f(a) = f(b), so    a <= b, and similarly b <= a so that a = b. f preserves infima, for suppose {a_i | i in I} has infimum a: if b is in inf {f(a_i) | i in I} = \bigcap {f(a_i) | i in I} then b <= a_i for all i, so that b <= a and therefore b is in f(a). Conversely if b is in f(a) then b <= a <= a_i, so b is in f(a_i) for all i, and b is in \bigcap {f(a_i) | i in I}. So    inf {f(a_i) | i in I} = f(a). Showing that f preserves suprema takes a little more work; suppose that {a_i | i in I} has supremum a. The inequality    sup {f(a_i) | i in I} <= f(a) is automatic since f is order-preserving. For the reverse inequality, suppose that b is in f(a), so that b <= a. Recall that the supremum of the f(a_i) was given by    \bigcap {T in B | f(a_i) \subset T for all i in I} so in order to show that b is in sup {f(a_i) | i in I} it suffices to show that b is in T whenever T is in B and T contains every f(a_i). Any such T is in turn given by an intersection of the form    T = \bigcap {f(c_j) | j in J} for some indexing set J; it follows that f(a_i) \subset T \subset f(c_j) for every i and j. But since a_i is in f(a_i) and therefore in f(c_j), it follows that a_i <= c_j for each i. Since a is the supremum of the a_i, we have a <= c_j and b <= a, so b is in f(c_j) for all j. Therefore b is in T, as was to be proved. -- I have made a thing that superficially resembles music: http://soundcloud.com/eroneity/we-berated-our-own-crapiness