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Let $\mathfrak{A}$ is an arbitrary poset.

Does it necessarily exist an order embedding from $\mathfrak{A}$ into some complete lattice $\mathfrak{B}$, which preserves all suprema and infima defined in $\mathfrak{A}$?

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up vote 6 down vote accepted

The answer is yes.

First, given any partially ordered set $P$, there is always a complete lattice that contains $P$, for example, the collection of downward closed subsets of $P$, under containment, where $p\in P$ is identified with $\{x\in P\mid x\le p\}$.

But, in fact, any poset $\mathcal P=(P,\le)$ admits a minimal completion $\mathcal L$, in the sense that $\mathcal L$ is a complete lattice containing $\mathcal P$, $\mathcal L$ embeds (as a poset) into any complete lattice that contains $\mathcal P$ (with the embedding fixing $P$ pointwise), and $\mathcal L$ preserves all suprema and infima (including suprema and infima of infinite subsets) already present in $P$.

As with the rationals and the reals, a natural way of defining this minimal completion $\mathcal L$ of $\mathcal P$ is simply to take as $\mathcal L$ the collection of all cuts in $\mathcal P$, partially ordered by saying that $(A,B)\le(C,D)$ iff $A\subseteq C$.

Here, a cut of $\mathcal P$ is a pair $(A,B)$ of subsets of $P$ such that $B$ is the collection of upper bounds of $A$, and $A$ is the collection of lower bounds of $B$.

This construction, that generalizes Dedekind's construction of the reals as cuts of rationals, is called the Dedekind-MacNeille (or normal) completion, first introduced in

Holbrook M. MacNeille. Partially ordered sets, Trans. Amer. Math. Soc. 42 (3), (1937), 416–460. MR1501929,

where complete details can be found.

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The statement "$\mathcal L$ embeds (as a sublattice) into any complete lattice that contains $\mathcal P$" is false. It only embeds as a partially ordered set. –  Thomas Klimpel Jan 1 '13 at 23:13
    
Hi @ThomasKlimpel. You are absolutely right, of course. I've updated accordingly. Thanks! –  Andres Caicedo Jan 1 '13 at 23:27
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