> Can two lattices be isomorphic and their associated posets not?

If two lattices are isomorphic preserving *infima* and *suprema*, ie *limits*, then they are order isomorphic.

The reals and hyperreals provide a rather confusing counter example to the converse. I am admittedly struggling with this myself, as it is highly non-constructive.

From model theory we have two maps \$$\phi : \mathbb{R} \to\, ^\ast \mathbb{R} \$$ and \$$\psi :\, ^\ast\mathbb{R} \to \mathbb{R} \$$ such that:

- if \$$x \leq_{\mathbb{R}} y\$$ then \$$\phi(x) \leq_{^\ast \mathbb{R}} \phi(y)\$$
- if \$$p \leq_{^\ast \mathbb{R}} q\$$ then \$$\psi(q) \leq_{\mathbb{R}} \psi(q)\$$
- \$$\psi(\phi(x)) = x\$$ and \$$\phi(\psi(p)) = p\$$

Now consider \$$\\{ x \, : \, x \in \mathbb{R} \text{ and } 0 < x \\}\$$.

The hyperreals famously violate the [Archimedean property](https://en.wikipedia.org/wiki/Archimedean_property). Because of this \$$\bigwedge_{^\ast \mathbb{R}} \\{ x \, : \, x \in \mathbb{R} \text{ and } 0 < x \\}\$$ does not exist.

On the other than if we consider \$$\bigwedge_{\mathbb{R}} \\{ \psi(x) \, : \, x \in \mathbb{R} \text{ and } 0 < x \\}\$$, that *does* exist by the completeness of the real numbers (as it is bounded below by \$$\psi(0)\$$).

Hence

$$\bigwedge_{\mathbb{R}} \\{ \psi(x) \, : \, x \in \mathbb{R} \text{ and } 0 < x \\} \neq \psi\left(\bigwedge_{^\ast\mathbb{R}} \\{ x \, : \, x \in \mathbb{R} \text{ and } 0 < x \\}\right)$$

So \$$\psi\$$ *cannot* be a complete lattice homomorphism, even though it is part of an order isomorphism.

However, just to complicate matters, I believe that \$$\phi\$$ and \$$\psi\$$ are a mere *lattice* isomorphism, preserving finite meets and joints.