Order Isomorphism on Lattice preserves Lattice Structure

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