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I'm supposed to find an 8-element complete lattice which does not form a Boolean algebra.
Sadly, anything I can think of is either not a complete lattice or it is a Boolean algebra.

I think that what i need is a lattice in this shape:
enter image description here

Since it would not be distributive. But I still can't get my head around one thing - since a lattice is basically an ordering on a given set, I have to provide the whole set and a measure by which it is ordered. And I can't think of any set and ordering that would form this lattice.

Maybe I'm totally mistaken, honestly I don't understant it much. Any help is welcomed.

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    Any Boolean algebra is equivalent to the lattice of subsets of some set. So there is a unique Boolean algebra on 8 elements, and any complete lattice on 8 elements that isn't that one will be a complete lattice which does not form a Boolean algebra. I think your example is correct. You do not have to provide a "measure"; it is sufficient to make up a completely arbitrary partial order relation ⪯ that behaves the way you want, as long as you are careful that your definition is in fact a partial order. In your example, you have $0⪯a⪯d⪯1$ and $0⪯b⪯e⪯1$, but neither $a⪯b$ nor $b⪯a$.2012-03-18
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    @Mark: To be completely clear, any _finite_ boolean algebra is isomorphic to the powerset of a finite set. In general, only _complete atomic_ boolean algebras are isomorphic to powersets.2012-03-18
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    "I have to provide the whole set and a measure by which it is ordered." I'm not sure what you mean. The nodes of the lattice tell the underlying set you are using and the edges tell order you are using. Transitivity is implicit in these diagrams, so we know (for example) $0 \leq d$ because $0 \leq a$ and $a \leq d$, even though there is no edge directly from $0$ to $d$.2012-03-18
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    Thank you all. What I meant by the sentence - I thought that I have to always provide a method of ordering. For example that $S$ = {1,2,3} and it is ordered by divisibility. Anyway thanks for the explanation2012-03-18
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    @AlexandarŽivkovič If you wish a divisibility ordering, for your example you could just take ${0=1, a=2, b=3, c=5, d=4, e=9, f=25, 1=0}$.2012-03-18
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    There are a few references above to an "8-element complete lattice." The adjective "complete" is redundant and should be left out to avoid confusion. All finite lattices are complete.2012-03-19

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