Order Isomorphism on Lattice preserves Lattice Structure

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Theorem

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

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 is 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 $\left({S, \preccurlyeq_1}\right)$.

$\blacksquare$


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