product of posets
Cartesian Ordering
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 $P$, given by
$$(a,b)\le (c,d)\mathit{\hspace{1em}\hspace{1em}}\text{iff}\mathit{\hspace{1em}\hspace{1em}}a{\le}_{1}c\text{and}b{\le}_{2}d.$$ 
Then $\le $ is a partial order^{} on $P$. $(P,\le )$ is called the product of posets $({P}_{1},{\le}_{1})$ and $({P}_{2},{\le}_{2})$. The ordering $\le $ is called the Cartesian ordering. As it is customary, we write $P$ to mean $(P,\le )$.
If ${P}_{1}$ and ${P}_{2}$ are antichains^{}, their product^{} is also an antichain. If they are both join semilattices, then their product $P$ is a join semilattice as well. The join of $(a,b)$ and $(c,d)$ are given by
$$(a,b)\vee (c,d)=(a{\vee}_{1}c,b{\vee}_{2}d).$$ 
Conversely, if the product $P$ of two posets ${P}_{1}$ and ${P}_{2}$ is a join semilattice, then ${P}_{1}$ and ${P}_{2}$ are both join semilattices. If $(a,b)\vee (c,d)=(e,f)$, then $e$ is the upper bound of $a$ and $c$. If $g\le e$ is an upper bound of $a$ and $c$, then $(g,f)$ is an upper bound of $(a,b)$ and $(c,d)$, whence $(g,f)=(e,f)$, or $g=e$. So $g=a{\vee}_{1}c$. Similarly, $f=b{\vee}_{2}d$. Dually, $P={P}_{1}\times {P}_{2}$ is a meet semilattice (and consequently, a lattice^{}) iff both ${P}_{1}$ and ${P}_{2}$ are. Equivalently, the product of (semi)lattices can be defined purely algebraically (using $\vee $ and $\wedge $ only).
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 $(a,b)$ and $(c,d)$ are comparable^{}, say $(a,b)\le (c,d)$, which implies $a{\le}_{1}c$ and $b{\le}_{2}d$. Also, $(c,b)$ and $(a,d)$ are comparable. But since $b{\le}_{2}d$, we must have $(c,b)\le (a,d)$, which means $c{\le}_{1}a$, showing $a=c$, or ${P}_{1}=\{a\}$.
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 ${\mathbb{R}}^{n}$ (http://planetmath.org/LatticeInMathbbRn), which is defined as the free abelian group^{} over $\mathbb{Z}$ in $n$ generators^{}. But from a poset perspective, it can be viewed as a product of $n$ 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
Again, let ${P}_{1}$ and ${P}_{2}$ be posets. Form the Cartesian product of ${P}_{1}$ and ${P}_{2}$ and call it $P$. There is another way to partial order $P$, called the lexicographic order^{}. Specifically,
$$(a,b)\le (c,d)\mathit{\hspace{1em}}\text{iff}\mathit{\hspace{1em}}\{\begin{array}{cc}a\le c\text{, or}\hfill & \\ a=c\text{and}b\le d.\hfill & \end{array}$$ 
More generally, if $\{{P}_{i}\mid i\in I\}$ is a collection^{} of posets indexed by a set $I$ that is linearly ordered^{}, then the Cartesian product $P:=\prod {P}_{i}$ also has the lexicographic order:
$({a}_{i})\le ({b}_{i})$ iff there is some $k\in I$ such that ${a}_{j}={b}_{j}$ for all $$ and ${a}_{k}\le {b}_{k}$.
We show that this is indeed a partial order on $P$:
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 $k,\mathrm{\ell}\in I$ we have that

(a)
${a}_{j}={b}_{j}$ for all $$ and ${a}_{k}\le {b}_{k}$, and

(b)
${b}_{j}={c}_{j}$ for all $$ and ${b}_{\mathrm{\ell}}\le {c}_{\mathrm{\ell}}$.
Since $I$ is a total order, $k$ and $\mathrm{\ell}$ are comparable, say $k\le \mathrm{\ell}$, so that ${a}_{j}={b}_{j}={c}_{j}$ for all $$ and ${a}_{k}\le {b}_{k}\le {c}_{k}$. Since ${P}_{k}$ is partially ordered, ${a}_{k}\le {c}_{k}$ as well. Therefore $({a}_{i})\le ({c}_{i})$.

(a)

•
(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 ${a}_{j}={b}_{j}$ for all $$ and $$. By the same token, $({b}_{i})\le ({a}_{i})$ implies the existence of $\mathrm{\ell}\in I$ with ${b}_{j}={a}_{j}$ for all $$ and $$. Since $I$ is linearly ordered, we can again assume that $k\le \mathrm{\ell}$. But then this means that either $$, in which case ${b}_{k}={a}_{k}$, a contradiction^{}, or $k=\mathrm{\ell}$, in which case we have that $$, another contradiction. Therefore $({a}_{i})=({b}_{i})$.
This completes^{} the proof. ∎
Title  product of posets 

Canonical name  ProductOfPosets 
Date of creation  20130322 16:33:25 
Last modified on  20130322 16:33:25 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  14 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 06A06 
Classification  msc 06A99 
Related topic  LatticeInMathbbRn 
Defines  product of lattices 
Defines  Cartesian ordering 
Defines  lexicographic ordering 