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Rotwang
Posts:
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From:
Swansea
Registered:
7/26/06


Re: Order embedding from a poset into a complete lattice
Posted:
Dec 30, 2012 7:29 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 downwardcomplete (i.e. any subset has an infimum). But any poset that is downwardcomplete 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 orderpreserving. 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 orderpreserving. 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.
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