MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am interested in algebras whose subalgebra lattice is supremum-founded. Let us call those algebras small.

A complete lattice $(L, \leq)$ is called supremum-founded, if for any two elements $x < y$ from $L$ there is an element $s \in L$ which is minimal with respect to $s \leq y$, $s \not\leq x$.

Of course, every finite lattice is supremum-founded whence every finite algebra is small. Moreover, it is not very difficult to show that every $1$-locally finite algebra (that is, an algebra whose $1$-generated subalgebras are finite) is small.

Does anybody know some more/bigger classes of examples? Is there a suitable reference?

share|cite|improve this question
It may be useful to look at the contrast. Let Y be a subuniverse which properly contains another subuniverse X, and pick an element z from their difference. The subalgebra generated by z is a candidate to exhibit supremum foundness; it fails only if there is a chain of decreasing subalgebras which stay partly outside of X. Thus, looking at algebras like natural numbers with successor are a first step toward looking at the contrast. Gerhard "Ask Me About System Design" Paseman, 2012.01.31 – Gerhard Paseman Jan 31 '12 at 20:36
(It helps me and perhaps others to have expanded upon your observation regarding 1-locally finite algebras.) Gerhard "Ask Me About System Design" Paseman, 2012.01.31 – Gerhard Paseman Jan 31 '12 at 20:41
Gerhard, I am somewhat surprised how your second question arises after typing your first comment. If we assume 1-locally finiteness and look at the scenario you describe in your first comment, then the subalgebra $\langle z \rangle$ generated by the single element $z$ will most certainly be finite. But then, it is immediate that an infinite chain of decreasing subalgebras of $\langle z \rangle$ that stay partly outside of X cannot exist. Thus, one of the subalgebras of $\langle z \rangle$ will be the algebra we are looking for. – Niemi Feb 1 '12 at 12:53
Yes. I was originally thinking that small implied 1-locally finite, and was going to conclude with that, but it does not quite work that way. So I started with a general Y, X and z to figure out where I could go with it. It seems I ended up expanding upon your observation without creating something new, thus my parenthetical comment. I was going to follow up with a few stupid questions, but caught myself in time. When I have something worthy of an answer, I won't stay silent, believe me. Gerhard "Also Noisy At Other Times" Paseman, 2012.02.01 – Gerhard Paseman Feb 1 '12 at 18:35
I asked JB Nation for a property that a subalgebra lattice should have in order that the algebra be "supremum-founded" as you define it. He suggested the (sufficient) condition that all elements of the lattice be dually compact. Note: if x > y and x is compact, then you can find a maximal element above y not above x. Dually, if x < y and x is "dually compact" (which means x above meet S implies x above meet F for some finite subset F of S) then there is a minimal element below y that is not below x. So, which algebras have subalgebra lattices in which every element is dually compact? – William DeMeo Feb 9 '12 at 5:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.