Order Isomorphism on Lattice preserves Lattice Structure
Theorem
Let $\struct {S, \preccurlyeq_1}$ and $\struct {T, \preccurlyeq_2}$ be ordered sets.
Let $\phi: \struct {S, \preccurlyeq_1} \to \struct {T, \preccurlyeq_2}$ be an order isomorphism.
Then $\struct {S, \preccurlyeq_1}$ is a lattice if and only if $\struct {T, \preccurlyeq_2}$ is also a lattice.
Proof
Let $\struct {S, \preccurlyeq_1}$ be a lattice
Then by definition $\preccurlyeq_1$ is a lattice ordering.
We need to show that for all $x, y \in S$, the ordered set $\struct {\set {\map \phi x, \map \phi y}, \preccurlyeq_2}$ admits both a supremum and an infimum.
Let $x, y \in S$.
Then $\struct {\set {x, y}, \preccurlyeq_1}$ admits both a supremum and an infimum.
Let $c = \map \sup {\set {x, y}, \preccurlyeq_1}$.
Then by definition of supremum:
- $\forall s \in \set {x, y}: s \preccurlyeq_1 c$
- $\forall d \in S: c \preccurlyeq_1 d$
where $d$ is an upper bound of $\struct {\set {x, y}, \preccurlyeq_1} \subseteq S$.
Now consider the image of $\set {x, y}$ under $\phi$.
By definition of order isomorphism:
- $\forall \map \phi s \in \set {\map \phi x, \map \phi y}: \map \phi c \preccurlyeq_2 \map \phi s$
- $\forall \map \phi d \in T: \map \phi c \preccurlyeq_2 \map \phi d$
where $\map \phi d$ is an upper bound of $\struct {\set {\map \phi x, \map \phi y}, \preccurlyeq_2} \subseteq T$.
So by definition of supremum:
- $\map \phi c = \map \sup {\set {\map \phi x, \map \phi y}, \preccurlyeq_2}$
That is, $\struct {\set {\map \phi x, \map \phi y}, \preccurlyeq_2}$ admits a supremum.
Using a similar technique it can be shown that:
- If $c = \map \inf {\set {x, y}, \preccurlyeq}$, then:
- $\map \phi c = \map \inf {\set {\map \phi x, \map \phi y}, \preccurlyeq_2}$
Hence $\struct {\set {\map \phi x, \map \phi y}, \preccurlyeq_2}$ admits both a supremum and an infimum.
That is, $\preccurlyeq_2$ is a lattice ordering and so $\struct {T, \preccurlyeq_2}$ is a lattice.
By Inverse of Order Isomorphism is Order Isomorphism, if $\phi$ is an order isomorphism then so is $\phi^{-1}$.
So the same technique is used to show that if $\struct {T, \preccurlyeq_2}$ is a lattice then so is $\struct {S, \preccurlyeq_1}$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.4$