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Is there any lattice in which every element (except $0$ and $1$) has four complements?

Is the set {1,2,3,5,7,11,2310} under the relation divides a correct example?

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up vote 2 down vote accepted

The type of answer supplied by others is quite obviously correct. If you want a more "synthetic" description, consider the lattice of vector subspaces of the 2-dimensional $\mathbb{F}_4$-vector space $\mathbb{F}_4^2$ (where $\mathbb{F}_4$ is the field with four elements).

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Is the set {1,2,3,5,7,11,2310} under the relation "divides" an example? – user221458 Sep 30 '13 at 13:16
Yes, absolutely. – user43208 Sep 30 '13 at 13:17
Can please help me in this question:… – user221458 Sep 30 '13 at 13:22
I'm tempted to ask you to ask user18921 for help, since you accepted his/her answer as correct. But perhaps I'll take a look. – user43208 Sep 30 '13 at 13:24
Everyone's answer is correct. I just accepted user18921's answer because it was easier to understand. I did voted your answer as useful. But I have ticked ur answer as correct now as u answered one more question. – user221458 Sep 30 '13 at 13:30

For this problem, it would be best to write $\bot$ for the least element and $\top$ for the greatest.

Let $L = \{\bot,0,1,2,3,4,\top\},$ and define $$x \leq y \leftrightarrow x = \bot \mbox{ or } y = \top.$$

Then $(L,\leq)$ is a lattice in which every element except for $\bot$ and $\top$ has four complements.

To see this, draw a diagram, and try computing the meet and join of $0$ and $1$ (say).

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Take $L = \{0, 1, x_1, x_2, x_3, x_4, x_5 \}$ with the partial order defined by $0 < x_i < 1$ for $1 \leqslant i \leqslant 5$. Then for $i \not= j$, $x_i$ and $x_j$ are complements of each other. Therefore every element of $L$ (except $0$ and $1$) have four complements.

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