Are there complete lattices $L, K$ such that
- $L\not\cong K$;
- there are injective complete lattice homomorphisms $i:L\to K$ and $j: K\to L$
?
Asked
Active
Viewed 223 times
-1
Dominic van der Zypen
- 45,374
-
2This question seems related http://mathoverflow.net/questions/1058/when-does-cantor-bernstein-hold/ – j.c. Aug 07 '15 at 09:17
2 Answers
4
For a simple example with complete total orders, take $L=\{0\}\cup[1,2]$ and $K=\{-1,0\}\cup[1,2]$.
Eric Wofsey
- 30,718
1
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).
Adam Przeździecki
- 3,031
- 1
- 18
- 25
-
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 $L={-1,0}\cup[1,2]$. – Eric Wofsey Aug 07 '15 at 11:32
-
Oh, yes. Or even simpler: $K=[1,2]$ and $L={0}\cup[1,2]$. Referring to your question: this is the identity $[0,1]\to{0}\times[0,1]$. – Adam Przeździecki Aug 07 '15 at 11:38
-
1That is not a homomorphism; it does not preserve the maximal element. – Eric Wofsey Aug 07 '15 at 11:39
-
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. – Adam Przeździecki Aug 07 '15 at 11:41
-
@Dominic van der Zypen please unaccept this answer so that I can delete it. – Adam Przeździecki Aug 07 '15 at 13:14