PlanetMath: product of posets

Let $(P_1,\le_1)$ and $(P_2,\le_2)$ be posets. Let $P=P_1\times P_2$ , the Cartesian product of the underlying sets. Next, define a binary relation $\le$ on

, given by

and

are antichains, their product is also an antichain. If they are both join semilattices, then their product

is a join semilattice as well. The join of

and

are given by

Another simple fact about the product of posets is the following: the product is never a chain unless one of the posets is trivial (a singleton). To see this, let $P=P_1\times P_2$ and $(a,b),(c,d)\in P$ . Then

and

are comparable, say $(a,b)\le (c,d)$ , which implies $a\le_1 c$ and $b\le_2 d$ . Also,

and

are comparable. But since $b\le_2 d$ , we must have $(c,b)\le (a,d)$ , which means $c\le_1 a$ , showing

, or $P_1=\lbrace a\rbrace$ .

Remark. The product of two posets can be readily extended to any finite product, countably infinite product, or even arbitrary product of posets. The definition is similar to the one given above and will not be repeated here.

An example of a product of posets is the lattice in $\mathbb{R}^n$ , which is defined as the free abelian group over $\mathbb{Z}$ in

generators. But from a poset perspective, it can be viewed as a product of

chains, each order isomorphic to $\mathbb{Z}$ . As we have just seen earlier, this product is a lattice, and hence the name “lattice” in $\mathbb{R}^n$ .

Lexicographic Ordering

Proof. The three things we need to verify are

(Reflexivity). Clearly, $(a_i)\le (a_i)$ , since $a_i\le a_i$ for any $i\in I$ .
(Transitivity). If $(a_i)\le (b_i)$ and $(b_i)\le (c_i)$ , then for some $k,\ell\in I$ we have that
1. for all and $a_k\le b_k$ , and
2. for all $j<\ell$ and $b_{\ell}\le c_{\ell}$ .
Since is a total order, and $\ell$ are comparable, say $k\le \ell$ , so that for all $k<\ell$ and $a_k\le b_k\le c_k$ . Since is partially ordered, $a_k\le c_k$ as well. Therefore $(a_i)\le (c_i)$ .
(Antisymmetry). Finally, suppose $(a_i)\le (b_i)$ and $(b_i)\le (a_i)$ . If $(a_i)\ne (b_i)$ , then $(a_i)\le (b_i)$ implies that we can find $k\in I$ such that for all and . By the same token, $(b_i)\le (a_i)$ implies the existence of $\ell\in I$ with for all $j<\ell$ and $b_{\ell}<a_{\ell}$ . Since is linearly ordered, we can again assume that $k\le \ell$ . But then this means that either $k<\ell$ , in which case , a contradiction, or $k=\ell$ , in which case we have that $a_k<b_k=b_{\ell}<a_{\ell}=a_k$ , another contradiction. Therefore .

This completes the proof. $\qedsymbol$

Cartesian Ordering

Lexicographic Ordering