In my research the following problem appeared (and if it is true, this solves positively several my conjectures):

Let $U$ be a fixed set (usually $U$ is infinite). Let $n$ be a fixed index set ($n$ may be infinite).

I call a staroid such an $n$-ary relation $f$ between sets (which are subsets of some fixed set $U$) such that:

  1. $X\notin f$ if any component $X_i$ of $X$ is empty set.
  2. $\{(k,I\cup J)\} \cup L\in f\Leftrightarrow \{(k,I)\} \cup L\in f \vee \{(k,J)\} \cup L\in f$ for every index $k\in n$, sets $I,J\in\mathscr{P}U$ and an indexed family $L$ of subsets of $U$ such that $\operatorname{dom} L=n\setminus\{k\}$.
  3. If $X\in f$ and $\forall i\in n:Y_i\supseteq X_i$ then $Y\in f$ for every $n$-indexed family $Y$ of subsets of $U$.

Note that first two axioms can be reformulated as:

  • The set $\{ X\in U \mid L\cup\{(k,X)\}\in f \}$ is a free star for every $L\in U^{n\setminus\{k\}}$ for every $k\in n$.

As I've shown in this online note free stars are isomorphic to ideals and filters.

Conjecture: The poset of staroids (ordered by set-theoretic inclusion) is a distributive lattice. I would like to prove that it is also a co-brouwerian lattice and a co-frame.

The same question without the third axiom is also interesting.

I have only one idea to solve this: Consider functions $L$ and $L'$ equivalent iff they differ in a finite number of arguments. Then somehow use the generated equivalency relation. But I have no idea how to go further in solution of this problem.

Note that for $n=\{0,1\}$ this conjecture is true. See this my online draft and a more categorical proof ($2$-staroids are easily identified with funcoids or topogenies).


2 Answers 2


What follows uses finiteness of $n$ but I think it works for infinite $n$ too.

First let us note that for finite $n$ the third axiom follows from the first two (because for finite $n$ the third axiom follows from its particular case when $Y_j=X_j$ for all $j$ except one).

So staroids are subsets $f$ of $(\mathscr PU)^n$ whose characteristic functions $(\mathscr PU)^n\to2$ are $\textit{multi-join-homomorphisms}$.

A multi-join-homomorphism $m:\prod_{i\in n}S_i\to S$, where $S_i$ and $S$ are join-semilattices, is a map which is a join-semilattice homomorphism in each argument separately.

It is well known that multi-join-homomorphisms are (at least for finite $n$) representable by the $\textit{tensor product}$ of semilattices. That is, for any join-semilattices $(S_i)_{i\in n}$ there is a join-semilattice $\bigotimes_{i\in n}S_i$ receiving a multi-join-homomorphism $\otimes:\prod_{i\in n}S_i\to\bigotimes_{i\in n}S_i$ which is universal, i. e. for any other multi-join-homomorphism $m:\prod_{i\in n}S_i\to S$ there is a unique (ordinary) join-homomorphism $m_\otimes:\bigotimes_{i\in n}S_i\to S$ with $m=m_\otimes\circ\otimes$.

It follows that staroids may be identified with join-homomorphisms $(\mathscr PU)^{\otimes n}\to2$, which are in turn identifiable with (complements of) ideals of $(\mathscr PU)^{\otimes n}$.


I believe that one may give a more explicit description of $\bigotimes_{i\in n}\mathscr PU_i$ as follows: there must be an isomorphism $\bigotimes_{i\in n}\mathscr PU_i\cong\mathscr P_{\vee\Pi}(\prod_{i\in n}U_i)$, where $\mathscr P_{\vee\Pi}(\prod_{i\in n}U_i)$ is the sub-join-semilattice of $\mathscr P(\prod_{i\in n}U_i)$ generated by the parallelepipeds $\prod_{i\in n}X_i$. Namely, the multi-join-homomorphism $\Pi:\prod_{i\in n}\mathscr PU_i\to\mathscr P_{\vee\Pi}(\prod_{i\in n}U_i)$ sending $X\in\prod_{i\in n}\mathscr PU_i$ to $\prod_{i\in n}X_i$ seems to be universal: for a multi-join-homomorphism $m:\prod_{i\in n}\mathscr PU_i\to S$ a join-homomorphism $m_\Pi:\mathscr P_{\vee\Pi}(\prod_{i\in n}U_i)\to S$ with $m=m_\Pi\circ\Pi$ is uniquely determined by $m_\Pi(\prod_{i\in n}X_i)=m(X)$. It just remains to find out whether this $m_\Pi$ is correctly defined, i. e. whether there are "extra" relations between joins of parallelepipeds in $\mathscr P(\prod_{i\in n}U_i)$...


Note also that there is a map from $\mathscr P(\prod_{i\in n}U_i)$ to ideals of $\mathscr P_{\vee\Pi}(\prod_{i\in n}U_i)$ sending a subset of $\prod_{i\in n}U_i$ to the set of all finite joins of parallelepipeds contained in it. Thus ideals of $\mathscr P(U^n)$ correspond to (possibly all) staroids.

Realized later:

In fact coproducts of distributive lattices are given by the tensor product, so under Stone/Priestley duality they correspond to cartesian products of spectral/Priestley spaces. Thus $(\mathscr PU)^{\otimes n}$ is the $n$-th copower of $\mathscr PU$ in the category of distributive lattices, hence its dual is the $n$-th power of the dual of $\mathscr PU$, i. e. of $\beta U$. Thus staroids correspond to open/closed sets of the Stone space $(\beta U)^n$.

Explicitly, to an open set $W$ of $\prod_{i\in n}\beta U_i$ corresponds the staroid $$ \{\ (X_i)_{i\in n}\ |\ \exists (X_i\in\chi_i\in\beta U_i)_{i\in n}\textrm{ s. t. }(\chi_i)_{i\in n}\notin W\ \} $$ whereas to a staroid $\Sigma\subseteq\prod_{i\in n}\mathscr PU_i$ corresponds the open set $$ \{\ (\chi_i)_{i\in n}\in\prod_{i\in n}\beta U_i\ |\ \exists (X_i\in\chi_i)_{i\in n}\textrm{ s. t. }\ (X_i)_{i\in n}\notin\Sigma\ \}. $$

This now works for infinite $n$ (and also for non-constant families $(U_i)_{i\in n}$).

  • $\begingroup$ A reference text (preferably cheap) about multi-join-homomorphism? $\endgroup$
    – porton
    Dec 23, 2014 at 22:06
  • $\begingroup$ In the case $n=2$ staroids bijectively correspond to filters (or ideals) on the boolean lattice of finite unions of binary cartesian products on $U$. There is (again if $n=2$) a "natural" injection (which is not a surjection) from $2$-staroids to filters on $U^2$. See mathematics21.org/algebraic-general-topology.html - Let us relate this with your suppositions ("PS" and "PPS" which I haven't yet read carefully enough). $\endgroup$
    – porton
    Dec 23, 2014 at 22:11
  • $\begingroup$ If you google for tensor product of (semi)lattices you will find several relevant texts. Just to give an example - probably the most relevant one I found is "The Boolean Algebra of Cubical Areas as a Tensor Product in the Category of Semilattices with Zero" $\endgroup$ Dec 23, 2014 at 22:18
  • 1
    $\begingroup$ @მამუკაჯიბლაძე I think I follow what you're saying, but there might be a slight sticking point in accounting for Victor's axiom 3. I'm going to reformulate some of it in a separate answer; I'd be obliged if you'd have a look. $\endgroup$
    – Todd Trimble
    Dec 24, 2014 at 22:01
  • 1
    $\begingroup$ I just realized (pretty belatedly) that the infinite tensor product that gives copowers of distributive lattices or Boolean algebras (a filtered colimit of finite tensor powers) is not the same as the universal recipient of join-multilinear maps $PU^n \to PU^{\otimes n}$. Related: mathoverflow.net/questions/11767/infinite-tensor-products In fact infinite tensor products in the latter sense seem difficult to understand and work with. $\endgroup$
    – Todd Trimble
    Jan 1, 2015 at 14:44

What follows is another point of view on the answer by მამუკა ჯიბლაძე (which I wouldn't immediately know how to pronounce if I weren't already familiar with the transliteration "Mamuka Jibladze").

Lemma 1: If $L$ is a (bounded) distributive lattice, then the poset of join-preserving maps $\phi: L \to \mathbf{2}$ under the pointwise order is a co-frame.

Proof: Evidently for such $\phi$ the set $\phi^{-1}(0)$ is an ideal in $L$, so the poset $[L, \mathbf{2}]$ is dual to the poset of ideals under inclusion. It is reasonably well-known that the poset of ideals in a distributive lattice $L$ (or if you prefer, the poset of filters in $L^{op}$, also a distributive lattice) is a frame. QED

Lemma 2: If $L_1, \ldots, L_n$ are (finitely many) distributive lattices, then their tensor product $L_1 \otimes \ldots \otimes L_n$ (as commutative monoids under their join operations) is also a distributive lattice.

Proof: Here we are viewing distributive lattices as rigs; all that really needs to be checked is that the multiplication on the tensor product rig is indeed the same as conjunction on its underlying poset. But that is an easy calculation using the fact that $\top_1 \otimes \ldots \otimes \top_n$ is the top element of the tensor product, together with the idempotence and order-preservation of multiplication. QED

Proposition: If $L_1, \ldots, L_n$ are distributive lattices, then the poset of maps $L_1 \times \ldots \times L_n \to \mathbf{2}$ that preserve finite joins in separate arguments is a co-frame.

Proof: Because the poset of such maps is identified with the poset of join-preserving maps $L_1 \otimes \ldots \otimes L_n \to \mathbf{2}$, where the domain is a distributive lattice by Lemma 2; this poset is a co-frame by Lemma 1. QED

Applying this proposition to the special case $L_i = P(X_i)$, we have that the poset of $n$-"staroids" for finite $n$ is a co-frame.

However, I do not see that an $n$-"staroid" for infinite cardinals $n$ can be viewed as equivalent to maps out of the infinite tensor product (an appropriate filtered colimit of finite tensor products), because Porton's axiom 3. imposes an extra condition that cannot be deduced from separate join-preservation, as far as I can see. It might however be arguable that condition 3. is not well-motivated.

  • $\begingroup$ The condition 3 is motivated by the fact that "principal staroids" (staroids corresponding to $n$-ary relations in a certain way) are always upper sets. But I am interested in both cases: with and without axiom 3. $\endgroup$
    – porton
    Dec 24, 2014 at 23:51
  • $\begingroup$ Thank you for the accurate answer! I will try to add to mine more details about what does this approach give in the infinite case. $\endgroup$ Dec 25, 2014 at 8:02

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