Are there complete lattices $L, K$ such that

  • $L\not\cong K$;
  • there are injective complete lattice homomorphisms $i:L\to K$ and $j: K\to L$



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).

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  • $\begingroup$ How do you get a complete injective homomorphism $K\to L$? For a simpler example that does work, just take $K=\{0\}\cup[1,2]$ and $L=\{-1,0\}\cup[1,2]$. $\endgroup$ – Eric Wofsey Aug 7 '15 at 11:32
  • $\begingroup$ Oh, yes. Or even simpler: $K=[1,2]$ and $L=\{0\}\cup[1,2]$. Referring to your question: this is the identity $[0,1]\to\{0\}\times[0,1]$. $\endgroup$ – Adam Przeździecki Aug 7 '15 at 11:38
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    $\begingroup$ That is not a homomorphism; it does not preserve the maximal element. $\endgroup$ – Eric Wofsey Aug 7 '15 at 11:39
  • $\begingroup$ I see - I missed that point. So we need both $K$ and $L$ to be disconnected. Please write your answer and I will delete mine. $\endgroup$ – Adam Przeździecki Aug 7 '15 at 11:41
  • $\begingroup$ @Dominic van der Zypen please unaccept this answer so that I can delete it. $\endgroup$ – Adam Przeździecki Aug 7 '15 at 13:14

For a simple example with complete total orders, take $L=\{0\}\cup[1,2]$ and $K=\{-1,0\}\cup[1,2]$.

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