Order Isomorphism on Lattices

Theorem

Let $\left({S, \preccurlyeq_1}\right)$ and $\left({T, \preccurlyeq_2}\right)$ be posets.

Let $\phi: \left({S, \preccurlyeq_1}\right) \to \left({T, \preccurlyeq_2}\right)$ be an order isomorphism.

Then $\left({S, \preccurlyeq_1}\right)$ is a lattice iff $\left({T, \preccurlyeq_2}\right)$ is also a lattice.

Proof

Let $\left({S, \preccurlyeq_1}\right)$ 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 $\left({\left\{{\phi \left({x}\right), \phi \left({y}\right)}\right\}, \preccurlyeq_2}\right)$ admits both a supremum and an infimum.

Let $x, y \in S$.

Then $\left({\left\{{x, y}\right\}, \preccurlyeq_1}\right)$ admits both a supremum and an infimum.

Let $c = \sup \left({\left\{{x, y}\right\}, \preccurlyeq_1}\right)$.

Then by definition of supremum:

• $\forall s \in \left\{{x, y}\right\}: s \preccurlyeq_1 c$
• $\forall d \in S: c \preccurlyeq_1 d$

where $d$ is an upper bound of $\left({\left\{{x, y}\right\}, \preccurlyeq_1}\right) \subseteq S$.

Now consider the image of $\left\{{x, y}\right\}$ under $\phi$.

By definition of order isomorphism:

• $\forall \phi \left({s}\right) \in \left\{{\phi \left({x}\right), \phi \left({y}\right)}\right\}: \phi \left({c}\right) \preccurlyeq_2 \phi \left({s}\right)$
• $\forall \phi \left({d}\right) \in S_2: \phi \left({d}\right) \succcurlyeq \phi \left({c}\right)$

where $\phi \left({d}\right)$ is an upper bound of $\left({\left\{{\phi \left({x}\right), \phi \left({y}\right)}\right\}, \preccurlyeq_2}\right) \subseteq T$.

So by definition of supremum:

$\phi \left({c}\right) = \sup \left({\left\{{\phi \left({x}\right), \phi \left({y}\right)}\right\}, \preccurlyeq_2}\right)$

That is, $\left({\left\{{\phi \left({x}\right), \phi \left({y}\right)}\right\}, \preccurlyeq_2}\right)$ admits a supremum.

Using a similar technique it can be shown that:

If $c = \inf \left({\left\{{x, y}\right\}, \preccurlyeq}\right)$, then:

$\phi \left({c}\right) = \inf \left({\left\{{\phi \left({x}\right), \phi \left({y}\right)}\right\}, \preccurlyeq_2}\right)$

Hence $\left({\left\{{\phi \left({x}\right), \phi \left({y}\right)}\right\}, \preccurlyeq_2}\right)$ admits both a supremum and an infimum.

That is, $\preccurlyeq_2$ is a lattice ordering and so $\left({T, \preccurlyeq_2}\right)$ is a lattice.

By Inverse of Order Isomorphism, if $\phi$ is an order isomorphism then so is $\phi^{-1}$.

So the same technique is used to show that if $\left({T, \preccurlyeq_2}\right)$ is a lattice then so is $\left({S, \preccurlyeq_1}\right)$.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 14$: Theorem $14.4$