For boolean algebra, let's take Roman Sikorski's Boolean Algebras as our reference. After giving a set of axioms, he proves (p.9) that the join of A and B is the least element of the algebra such that A and B are its subelements. He also asserts that since that's so, the join of A and B can be defined in terms of the ordering relation.

In the Appendix (p. 198), Sikorski defines a closure algebra as a boolean algebra with an additional operator, a closure operator, and gives axioms for it. He also defines an interior operator in terms of the closure operator.

The assertion about the definability of join in terms of ordering leads one to ask whether the same might be true of the closure and interior operators. Given sufficiently strong axioms for them, one might find that:

  • The closure of A is the least element of which A is a subelement.

  • The interior of A is the greatest element which is a subelement of A.

This idea just occurred to me this evening. I imagine the answer depends on the axioms for the closure operator. Has this topic already been explored? Thanks for any help or references.


Oh sure, this is quite well-known. The closure of an element is the smallest closed element which is greater than or equal to the given element. Dually for the interior operator.

In general, a closure operator $\phi: P \to P$ on a poset $P$ is an order-preserving, inflationary ($x \leq \phi(x)$), idempotent ($\phi(\phi(x)) = \phi(x)$) operation, and if you want a topological closure operator, then you demand $\phi(x \vee y) = \phi(x) \vee \phi(y)$ as well. Alternatively, a closure operator can be specified by an inclusion $i: C \hookrightarrow P$ such that every $p \in P$ has a least upper bound $c \in C$. The assignment $p \mapsto c$ gives an order-preserving mapping $j: P \to C$ with the property

$$j(q) \leq d \Leftrightarrow q \leq i(d)$$

(where the left side is to be read in $C$ and the right in $P$), and the closure operator $\phi$ is the composite $i \circ j$; it is easy to check the order-preserving inflationary idempotent properties. In the context of Boolean algebras $P$, as in Sikorski, you demand that $C$ be closed under joins as well (and that $i$ preserve them), to make $\phi$ a topological closure.

I am writing this answer to bear out a far wider connection with category theory. The posets here are special cases of categories, where there is at most one arrow in any hom-set $\hom(x, y)$, which we write as $x \leq y$. Functors between poset categories amount to order-preserving maps. A closure operator amounts to a monad on a poset. The subcollection $i: C \subseteq P$ of closed elements for that operator is the category of algebras for that monad. The corresponding mapping $j: P \to C$ is the left adjoint to $i$. Any adjoint pair (known under another name as a Galois connection) between posets always induces a closure operator. And so on.

There is an embarrassment of riches of references (i.e., so many that it's hard to think of one that stands out). But a trade secret among category theorists is to understand what general concepts mean in the simplified case of poset categories, and this point of view is explicitly declared in Paul Taylor's Practical Foundations of Mathematics. I think just about any introductory book on category theory will refer to this, though, and you will also find it used heavily in more specialized treatises like Johnstone's Stone Spaces that involve looking at lattices from a categorical point of view.

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  • $\begingroup$ Thanks for this; I knew I could get monads from adjunctions but it didn't occur to me to see what this gave when specialized to posets even though I also knew I could get closure operators from adjunctions between posets...! $\endgroup$ – Qiaochu Yuan Jun 26 '11 at 15:02
  • $\begingroup$ Todd: Thanks for your reply. I will have to learn about category theory. I have one follow-on question: suppose the algebra is such that an element is open if & only if it's closed. Is the closure of A then the least element of which A is a subelement, rather than the least closed element? $\endgroup$ – MikeC Jun 26 '11 at 17:08
  • $\begingroup$ No, it's still the least closed element that is an upper bound of A. (An example of such a space is given by a set equipped with an equivalence relation, and where open sets are unions of equivalence classes.) The element A itself is the least element of which A is a subelement. $\endgroup$ – Todd Trimble Jun 26 '11 at 17:38

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