# Does every sub-lattice of $(2^E, \subseteq)$ have to contain $\emptyset$?

Let $E$ be a finite set, let $2^E$ denote its power set, then it is well-known that $(2^E, \subseteq)$ is not only a poset, but even a poset with the meets equal to intersections and joins equal to unions.

From this it follows that every family of open sets defined on $E$ is a sub-lattice, as well as every family of closed sets, since "arbitrary intersections" corresponds to "finite intersections" since $2^E$ is finite, likewise "arbitrary intersections" corresponds to "finite intersectins" for the same reason.

The question in the title corresponds then to the converse of this fact, namely:

Question: For any finite set $E$, does any sub-lattice of $(2^E, \subseteq)$ define a topology?

(With the topology defined equivalently via open sets or closed sets -- all finite topological spaces are Alexandrov, so the properties of open sets and closed sets are the same.)

Clearly being a sub-lattice implies closure under finite non-empty unions and finite non-empty intersections. But whether a sub-lattice has to be closed under empty unions and empty intersections, hence contain both $\emptyset$ and $E$ and therefore be a topology is what makes my head hurt.

Issue 1: One will often read that the third axiom of open sets (or closed sets), that both $\emptyset$ and $E$ have to be open (respectively closed) sets is supposed to be redundant, since closure under finite intersections (respectively finite unions) implies closure under the empty intersection so that $E$ is open (respectively closure under the empty union so that $\emptyset$ is closed).

However, this logic seems somewhat flawed, depending specifically on how the closure under finite unions/intersections axiom is stated. For definiteness consider the case of closed sets. One will often write that closed sets are closed under finite unions in the following manner:

If $X_1, X_2$ are both closed, then $X_1 \cup X_2$ is also closed.

Clearly, by induction, it follows from this that all non-empty finite unions are again closed sets (the union of one set, case $n=1$, following from taking $X_1 = X_2$).

But $n=0< 1$ and $n=0<2$; so how can closure under an empty union follow by induction from finite unions for $n=1,2$ as base cases?

However, if one simply wrote "Closed sets are closed under finite unions", then it would be clear to me that the empty union should be included, because that is clearly a finite union.

Issue 2: The flats of a matroid with ground set $E$ always form a sub-lattice of $(2^E, \subseteq)$. So if this converse were true, it would mean that every family of flats forms a topology on the ground set.

However, in general matroid axiomatizations and topological space axiomatizations are supposed to be inequivalent. This answer gives an example of how the matroid analog of topological closed sets, flats, need not be closed under unions.

The Kuratowski closure axioms are also evidently different from the matroid closure axioms. Likewise the cyclic set axioms for matroids guarantee closure under finite unions, but not under finite intersections. The hyperplane axioms are similar but not the same as the closed base axioms, and the circuit axioms are similar but not the same as the open base axioms.

Finally, Oxley, in Matroid Theory, only states that the least element of the flat lattice is $cl(\emptyset)$, which seems to imply that the (matroid) closure of the empty set is not always the empty set, hence $\emptyset$ is not always a flat thus not always the least element of the flat sub-lattice of $(2^E, \subseteq)$. In contrast, $cl(\emptyset)=\emptyset$ always for the topological (Kuratowski) closure, since the empty set is always closed, either per fiat or due to closure under finite unions.

The lattice of flats is not a sub-lattice of $(2^E, \subseteq)$, since the join operation is different. The meet operation is the same though.

Also the claim in issue one corresponds to one of the flat axioms being "$E$ is a flat" even though the second flat axiom is "$F_1, F_2$ flats implies $F_1 \cap F_2$ are flats". I.e. the second axiom doesn't seem to imply closure under empty intersection, hence not closure under all finite intersections.

Issue 3: 2: The footnote 6 at the bottom of page 10 here says that [emphasis mine]:

A lattice is a partial order in which any finite non-empty set has an infimum and supremum. Equivalently, any two elements $x$ and $y$ have a supremum $x \lor y$, and an infimum $x \land y$.

In particular, based on this statement it does not seem like any empty set has to have an infimum (it can, but it doesn't have to), which for $(2^E, \subseteq)$ would seem to correspond to sub-lattices not needing to be closed under empty intersections. Likewise, it does not seem like any empty set has to have a supremum, in particular it does not seem like sub-lattices of $(2^E, \subseteq)$ need to contain $\emptyset$.

Moreover, the statement that "the fact that the supremum and infimum only have to exist for non-empty finite sets in the lattice is equivalent to any two elements being closed under meet or join" seems to correspond to my claim that the statement [$X_1, X_2$ closed $\implies$ $X_1 \cup X_2$ closed] can't use induction to prove closure under empty unions.

Issue 4 3: The answers to these questions (1)(2) seem to contradict the statements made in the footnote of the paper referenced above.

## 1 Answer

Of course not. Any collection of subsets of $E$ closed under (binary) union and intersection is a sublattice of $(2^E,\subseteq)$. In particular, $\{A\}$ is a sublattice for any $A\subseteq E$, so you will not have necessarily the empty set and not necessarily have a topology.

• That seems reasonable. But is $\{A\}$ closed under finite intersections and finite unions if it is not closed under the empty intersection and the empty union? Or is closure under finite intersections and finite unions strictly stronger than the conditions to be a sub-lattice? – Chill2Macht Dec 12 '17 at 23:05
• @Chill2Macht A lattice only needs to be closed under binary union and intersection, which extends by induction to nonempty finite union and intersection. A bounded lattice adds a top and bottom explicitly and so is closed under all finite unions and intersections, and a complete lattice is by definition closed under arbitrary union/intersection, from which you derive the existence of a top and bottom. But a sub-lattice need not preserve the top and bottom when they exist - maybe you'd call that a sub-(bounded lattice) but I've never heard of it. – Mario Carneiro Dec 12 '17 at 23:10
• S = {A} has just one element. If K,L in S, then K = L = A so S is closed under intersections and unions. – William Elliot Dec 13 '17 at 0:08
• @WilliamElliot Non-empty finite intersections and unions. Neither $\emptyset$ nor $E$ are in $S=\{A\}$. – Chill2Macht Dec 13 '17 at 0:45
• @Chill2Macht Yes, an Alexandrov topology is the same as a sub-(complete lattice) of $2^E$. (Note that there are complete sublattices of $2^E$ but are not Alexandrov - in fact all topologies are complete sublattices, but the infinite meet operation is not the same as the intersection operator on $2^E$.) – Mario Carneiro Dec 13 '17 at 3:20