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Question: Given a bounded lattice $\mathscr{L}$ with top $T$ and bottom $B$, is it the case that any sublattice of $\mathscr{M} \subseteq \mathscr{L}$ which is bounded must also have $T$ as its top and $B$ as its bottom?

Or could other elements from $\mathscr{L}$, some $T'$ and $B'$, assume the role of the top and bottom of $\mathscr{M}$?

I think the answer to the latter question is no, but I am not sure if my idea for the proof is actually correct given the definitions of lattice and bounded lattice, which I am not sure I fully understand.

Note: This is a follow-up to a previous question.

Attempt: By definition or something else, because $\mathscr{L}$ is bounded, the empty join must be $B$ and the empty meet must be $T$. Any meet operation which has a different empty meet would be a different/non-equal meet operation. Any join operation which has a different empty join would be a different/non-equal join operation. (Because this would imply that they were defined differently.)

Thus, if the bottom of $\mathscr{M}$ were $B' \not=B$, then that could only be the case if the join operation on $\mathscr{M}$ were different from the join operation on $\mathscr{L}$. Thus the ground set of $\mathscr{M}$ may be a subset of the ground set of $\mathscr{L}$, but since the join operations are different, $\mathscr{M}$ is not a sublattice of $\mathscr{L}$.

Likewise, if the top of $\mathscr{M}$ were $T' \not=T$, then that could only be the case if the meet operation on $\mathscr{M}$ were different from the meet operation on $\mathscr{L}$. Thus even though the ground set of $\mathscr{M}$ is a subset of the ground set of $\mathscr{L}$, and both are posets with respect to the same partial order, their meet operations must necessarily be different, and therefore $\mathscr{M}$ cannot be a sublattice of $\mathscr{L}$.

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Chill2Macht
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1 Answers1

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This depends on what you mean with bounded lattice.

If you mean a lattice $\mathbf{L} = \langle L, \wedge, \vee \rangle$ which just happens to be bounded (meaning that it has a maximum and a minimum element), then a sublattice $\mathbf{M}$ of $\mathbf{L}$ doesn't even have to be bounded, but even a bounded lattice doesn't have to have the same bounds.
For example, each element of a lattice is a sublattice coinciding with its minimum and maximum.

If by a bounded lattice you mean $\mathbf{L} = \langle L, \wedge, \vee, 0, 1 \rangle$ (here the bounds are nullary operations), then every bounded sublattice of $\mathbf{L}$ must have the same $0$ and $1$ (precisely because these are fundamental operations of the algebra).

So this is essentially a matter of definition.

By the way, in your argument, you seem to be considering that if $\mathbf{L}$ is a bounded lattice and $\mathbf{M}$ a bounded sublattice of $\mathbf{L}$, then for any $B \subseteq M$, we have $\bigwedge_{M} B = \bigwedge_L B$.
This is not the case.
For example, consider the usual order on the set $$L = \{ -1, 0 \} \cup \{ 1/n : n \in \mathbb{N} \}.$$ This is a bounded lattice with $\top = 1$ and $\bot = -1$.
Now consider $M = L \setminus \{0\}$; it is a bounded sublattice of $L$, but, taking $B = L \setminus \{-1,0\}$ we have $\bigwedge_M B = -1 \neq 0 = \bigwedge_L B.$

amrsa
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  • Do you know any references that give both definitions? Or one reference which mentions one definition and another reference which mentions the other definition? It is really confusing to me that there are two non-equivalent definitions of the term "bounded lattice" (or at least of "bounded sublattice"). Maybe it makes sense to call the first definition a "bounded sublattice" and the second definition a "sub-(bounded lattice)". Although that is probably equally confusing. Maybe if one thinks of "lattices" and "bounded lattices" as being different categories? Anyway this distinction helps me. – Chill2Macht Dec 27 '17 at 15:21
  • The example with $B$ and $L$ is really good. It is something which in hindsight I wasn't aware of, hence very useful for me. Thinking about sub-lattices of $(2^E, \subseteq)$ for finite sets $E$ - whether "intersection" for a subset $A$ should be defined as "set intersection", with the empty intersection being $A$, or "set intersection followed by intersection with $E$" thus with the empty intersection being $E \supsetneq A$. The first is a "bounded sublattice" the latter a "sub-(bounded lattice)". Even in the latter case the results of meets don't have to be the same as in the full lattice.) – Chill2Macht Dec 27 '17 at 15:24
  • These pages (http://planetmath.org/boundedlattice) (https://proofwiki.org/wiki/Definition:Bounded_Lattice) (https://ncatlab.org/nlab/show/lattice) seem to use the first definition of "bounded lattice", and these two (http://mathworld.wolfram.com/BoundedLattice.html) (https://en.wikipedia.org/wiki/Lattice_(order)#Bounded_lattice) seem to use the second definition of "bounded lattice". Since the two definitions correspond to different categories, they have different sub-objects. So your explanation does seem to be entirely correct, that the difference is solely a terminological one. – Chill2Macht Dec 27 '17 at 15:47
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    In what concerns the definition that includes the bounds in the type of the algebra, this is just particular case of what happens in Universal Algebra: for any type of algebra, a subalgebra must be closed under all basic operations (see Basic Constructions; other useful reference could be, for example, the online available book A course in Universal Algebra, by Burris and Sankappanavar). – amrsa Dec 27 '17 at 16:32
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    The reason I mentioned the other possibility, is that you may have a bounded lattice (as it is the case of every finite lattice) but not including the bounds as operations. In that case, a sublattice doesn't have to include them either (unless they are finite meets/joins of other included elements). So you really have to confirm first what is your definition. Most of the times, I should think that when one refers to bounded lattices, is including the bounds as operations (as in definition you linked to). – amrsa Dec 27 '17 at 16:37
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    I think the definition from the planetmath site is a bit ambiguous, since it doesn't make clear whether or not they include the bounds as operations. – amrsa Dec 27 '17 at 16:40