Order embedding from a poset into a complete lattice

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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 <por...@narod.ru> 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.


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http://soundcloud.com/eroneity/we-berated-our-own-crapiness