Are there complete lattices $L, K$ such that
- $L\not\cong K$;
- there are injective complete lattice homomorphisms $i:L\to K$ and $j: K\to L$
?
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Sign up to join this communityAre there complete lattices $L, K$ such that
?
Let $A=[0,\omega_0]$ and $K=[0,1]$ be closed intervals of the ordinals and the real numbers with natural orderings. Let $L=A\times K$ be equipped with the lexicographic order where the $A$ coordinate is more important. Then both $L$ and $K$ are complete and the injective homomorphisms are easy to construct. To see that $L\not\cong K$ as ordered sets observe that $K$ is connected in its order topology while $L$ is not.
Edit: Please unaccept this answer so that I can delete it. The correct answer is provided by Eric Wofsey. The $K$ admits no complete embedding into $L$ as such an embedding would have to preserve the least and the largest elements (inf and sup of the empty set).
For a simple example with complete total orders, take $L=\{0\}\cup[1,2]$ and $K=\{-1,0\}\cup[1,2]$.