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Let S be a bounded semilattice without maximal elements. Can we always construct an atomless boolean algebra B, containing S as a subsemilattice, such that S is cofinal in B-{1}? That is, for every x≠1 from B there is y∈S such that x≤y.

More detailed explanation since apparently my terminology was rather ambiguous:

S is a partially ordered set with the least element 0, and every pair of elements from S has an infimum (greatest lower bound). S has no maximal elements, that is for every x∈S there is y∈S greater than x.

If B is an atomless boolean algebra, B-{1} would definitely satisfy the above conditions.

The question is, if it is possible for any semilattice S satisfying the above conditions to cofinally embed it into an atomless boolean algebra B. That is, for any x∈B there is y∈S such that y is greater than x in B.

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Quibble: when you say every pair of elements has an "infimum", I think you mean "meet". The infimum would be something like the zero of the semilattice – Yemon Choi Dec 8 '09 at 23:15
Yeah, "meet" and "join" are rather confusing to me, and I already used one instead of another in the first draft of this question, so I'm sticking to the more... international terms from now on. I meant an infimum of a subset {x,y}, not the whole semilattice. – Grue Dec 8 '09 at 23:28
up vote 5 down vote accepted

The answer to the original question is that no, in fact we can never do this.

Theorem. No nontrivial Boolean algebra has a cofinal subset of B-{1} that is a join-semilattice. Indeed, B cannot have a cofinal subset of B-{1} that is even upward directed.

Proof. Suppose that B is a nontrivial Boolean algebra and S is cofinal subset of B-{1}. Let b be any element of B that is neither 0 nor 1. Thus, -b is also not 1, so both b and -b must have upper bounds in S. If S is a join-semilattice, or even only upward directed, than there will be a single element of S above both b and -b. But the only such upper bound in B is 1, which is not in S. QED

This is true whether or not B is atomic. In the atomic case, it is particularly easy to see, since the co-atoms have no upper bound in B-{1}.

There is no need in the question to insist that the embedding of S into B is a semilattice embedding, since even just an order-preserving map would give an upward directed image, which the theorem rules out.

The dual version of the question, turning the lattice upside down, is the corresponding fact that no nontrivial Boolean algebra has a filter that is also dense.

(Note: I didn't know what sense of "bounded" you meant in the first part of the question, so I ignored that...Does this matter?)

The situation with the revised version of the question is somewhat better, since as the questioner points out, there are now some examples of the phenomenon. But nevertheless, there are also counterexamples, as I explain, so the answer is still no.

The reason there are counterexamples is that the new conditions on S do not rule out the possibility that S is upward directed. Thus, the same obstacle from above still occurs. For example, any linearly ordered S is a meet-semilattice and a join-semilattice, so it can never be cofinal in a Boolean algebra. More generally, if you take any S you have in mind, and put a copy of the natural numbers on top of it (above all elements), then this will still satisfy your conditions, but since it is also upward directed, it cannot embed cofinally in a Boolean algebra.

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Ehhh... I probably confused the terminology or something. B-{1} should be itself a semilattice. Every two elements have an infimum. So, if S is B-{1} then the embedding is trivial. I'll try to update the question with more explicit explanations. – Grue Dec 8 '09 at 23:04

$\let\bez\smallsetminus$ Joel’s answer resolves the question as stated, but I feel it should be pointed out that cofinal subsets of Boolean algebras without top have a simple characterization:

  • If a poset $S$ can be embedded as a cofinal subset of $B\bez\{1\}$ for some Boolean algebra $B$, then $S$ is dually separative: for every $x,y\in S$ such that $x\nleq y$, there exists $z\ge y$ such that $x$ and $z$ have no common upper bound in $S$.
    [Take any $z\ge \neg x\lor y$.]

  • If $S$ is a dually separative poset, there exists an embedding of $S$ into a complete Boolean algebra $B$ which maps $S$ onto a cofinal subset of $B\bez\{1\}$, and preserves all existing suprema and infima in $S$. (In particular, if $S$ is a meet semilattice, it will be a semilattice embedding.)
    [Make $S$ a topological space by declaring downwards closed subsets to be closed, let $B$ be the algebra of regular closed subsets of $S$, and embed $S$ into it by mapping $x$ to $\{y\in S:x\nleq y\}$.]

For $B$ to be atomless, it is necessary and sufficient that $S$ have no maximal elements.

(I hope I dualized it correctly, these results are usually stated for downwards cofinal subsets of $B\bez\{0\}$.)

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