This question about finite posets is a special case of my research problem about filters (because for finite posets the set of filters is (dually) isomorphic to the set of elements).

Consider a (finite) lattice $\mathfrak{A}$ with least element. I denote binary meet, binary join, least, and greatest elements correspondingly as $\sqcap$, $\sqcup$, $\bot$, $\top$.

I call star of an element $a$ the set $$\star a = \{ x\in\mathfrak{A} \mid x\sqcap a\ne\bot \}.$$

It is easy to show that for boolean lattices $\star x = \{ \overline X \mid X\in\mathfrak{A},X\not\geq x\}$. Thus (as this is a lattice isomorphism) for boolean lattices we have $$\forall a,b\in\mathfrak{A}: (\star a=\star b \Rightarrow a=b).$$

Can the last formula hold for non-boolean (finite) (distributive?) lattices? What are necessary and/or sufficient conditions for our lattice for this formula to hold?


1 Answer 1


This is not true for a general finite, distributive lattice. Indeed, consider the lattice which has 6 elements: $\perp,\top,x,y,x\sqcap y, x\sqcup y$, which is distributive. Notice that $\star x=\{\top,x,y,x\sqcup y,x\sqcap y\}=\star y$.

Edit: Here's a proof that if $L$ is a finite, distributive lattice with $\star x=\star y\implies x=y$, then $L$ must be boolean. Instead of working with $\star$, let's work with the complement, call it $\circ x=L\setminus\star x=\{y\in L:y\sqcap x=\perp\}$. I first want to claim that $\circ x$ has a largest element. Indeed, this follows from distributivity: if $y,z\in\circ x$, then $x\sqcap(y\sqcup z)=(x\sqcap y)\sqcup(x\sqcap z)=\perp\sqcup\perp=\perp$, so $y\sqcup z\in\circ x$. Hence, since $L$ is finite, we can define $x^\circ=\max(\circ x)$. Now, notice that $\circ x$ is a down-set in the poset induced by the meet and join, so we have $\circ x=\circ y\iff x^\circ=y^\circ$. We need one extra result which follows from definition chasing, so I'll omit the proof: $(x\sqcap y)^\circ=x^\circ\sqcup y^\circ$.

Now, We know that $\top=\perp^\circ=(x\sqcap x^\circ)^\circ=x^\circ\sqcup x^{\circ\circ}$. But, of course, by definition $x^\circ\sqcap x^{\circ\circ}=\perp$. Thus, if $y=x^\circ$ for some $x\in L$, we have $y\sqcap y^\circ=\perp$ and $y\sqcup y^\circ=\top$; in other words, all of these elements have complements. Finally, remember that we assumed $x^\circ=y^\circ\implies x=y$, so actually $f:L\to L:x\mapsto x^\circ$ is an injection. But since $L$ is finite, $f$ must actually be a bijection, so every element $y\in L$ is of the form $y=x^\circ$ for some $x\in L$. In other words, all elements of $L$ have complements, so since $L$ is distributive, it must be boolean.

As a final thought, it is worth pointing out that non-distributive lattices can satisfy your formula. For instance, consider the lattice with elements $\perp,\top,x,y,z$ where $\{x,y,z\}$ is an antichain. Certainly this lattice isn't distributive. On the other hand, $\star\top=\{x,y,z,\top\}$, $\star x=\{x,\top\}$, $\star y=\{y,\top\}$ and $\star z=\{z,\top\}$.

  • $\begingroup$ Thanks for the answer. But I have also asked about other possible necessary and/or sufficient conditions. You have not answered this part of the question. Well, it is too vague question, but I will not (yet) accept your answer in the hope of other interesting answers. $\endgroup$
    – porton
    Jun 26, 2018 at 18:40
  • 1
    $\begingroup$ Understandable. I don't know an answer to your question but will think about it. Also, there is an easier counterexample for distributive lattices, namely $\perp,\top,x$ where $\star x=\star\top=\{x,\top\}$. $\endgroup$ Jun 26, 2018 at 18:48
  • $\begingroup$ Your example provides a case when the formula does not hold, while I asked for the case when the formula does hold for a non-boolean lattice. So you in a sense answered a question reverse to what I asked $\endgroup$
    – porton
    Jun 26, 2018 at 19:17
  • $\begingroup$ I'm not sure if stackexchange notifies you about edits to answers, but I've now included a proof that your formula holds in a finite, distributive lattice if and only if that lattice is boolean $\endgroup$ Jun 26, 2018 at 19:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.