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I know that-

If a lattice is distributive then its every element has either 0 or 1 complement.

But Is it's converse true?

I am not able to find an example of a lattice which has at most 1 complement for its every element but it is not distributive. Please give an example.

user9014873
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1 Answers1

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A lattice in which each element has at most one complement may have elements with no complement at all.
It is rather easy to come up with non-distributive lattices with that property, even if we require that there is at least one pair of elements which are complements of one another.
For example, take a bounded, non-distributive lattice, in which no element except $0$ and $1$ has a complement.
For a less trivial example, one in which there are other pairs of complements, take the following example:

enter image description here

Here $\langle 0,1\rangle$ and $\langle a,b\rangle$ are the only pairs of complements, and the lattice is not distributive, since it has $M_3$ as a sublattice (see here).

If you're looking for a non-distributive lattice in which every element has a unique complement, that is not easy to point out.
I believe that there is no finite such lattice, but basically this answer tells that it exists (even non-modular).
Actually every lattice is a sublattice of a lattice with unique complements (Dilworth's paper on this).

amrsa
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  • Thanks a lot. Means, there are many such lattices exist with every element has at most one complement, but they contain $M_3 \ or \ N_5$ as sublattice. Another Example is ($N$,|) where N = {0,1,2,3,4...........} I will read Dilworth's paper and Gratzer's article for more detail :) – user9014873 Jun 11 '18 at 03:51
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    @user9014873 No, $(\mathbb N,|)$ is distributive. See this question. It has two answers more or less similar. – amrsa Jun 11 '18 at 08:41