1
$\begingroup$

In the book "Residuated Lattices: An Algebraic Glimpse at Substructural Logics" by Galatos, Jipsen, Kowalski and Ono they have this result (Lemma 3.8(2) page 147)

If $\gamma$ is a closure operator on $\mathcal{P}(A)$ then $\gamma = \gamma_R$ for some relation $R$ with domain $A$.

In this:

  • $A$ is any set, $\mathcal{P}(A)$ is the powerset.
  • The term `closure operator' means that for all $X, Y\subseteq A$: $X \subseteq \gamma (X)$, $\gamma(\gamma(X)) = \gamma (X)$, and if $X \subseteq Y$ then $\gamma(X) \subseteq(Y)$.
  • The closure operator $\gamma_R$ is defined (page 146) when $R\subseteq A \times A$. It comes from the operations on subsets: $X \mapsto X^\triangleright$ and $X \mapsto X^\triangleleft$ where $$X^\triangleright = \{w\in A \mid \forall x \in X\, (x \mathrel{R} w)\}, \text{and } X^\triangleleft = \{w \in A \mid \forall x \in X\, (w \mathrel{R} x)\}.$$ The closure $\gamma_R$ is then defined as $\gamma_R (X) = X^{\triangleright \triangleleft}$.

The proof of the result uses the fact that any closure operator, $\gamma$, on a lattice $\mathcal{L}$ arises from a Galois connection (which in the sense used in the book has a pair of order reversing mappings, not order preserving as sometimes meant). The Galois connection is $\gamma: \mathcal{L} \to \mathcal{C}^{op}$ and $i: \mathcal{C}^{op} \to \mathcal{L}$. Here, $\mathcal{C}^{op}$ is the lattice of closed elements but given the opposite ordering, and $i$ is the (order reversing) inclusion.

Next, they use a result from the exercises (Ex9 page 205). At this point I get confused and have two questions:

  1. The exercise (which itself is clear) is about Galois connections between powersets. But here we have a Galois connection between a powerset and the opposite of a lattice of closed elements. Surely a lattice of closed elements for a closure operator on a powerset does not have to be isomorphic to a powerset? What if $A = \{a,b\}$ and the closure does this: $A \mapsto A, \varnothing \mapsto \varnothing, \{a\} \mapsto \{a\}, \{b\} \mapsto A$. That is a closure, isn't it?
  2. Even if the first question can be sorted (maybe I'm wrong, or maybe you have to generalize the result in the Exercise) there is another thing. The construction of the relation $R$ in the proof is given as $$x \mathrel{R} y \text{ iff } y \in \gamma(x).$$ So that must mean that a closure on a powerset is completely determined by the closures of the singletons. I am sure that cannot be right.
$\endgroup$
2
  • $\begingroup$ I think what it actually means is that the closure operator is completely determined by the closures of singletons and what sets are closed under the operator (and in fact the first is completely determined by the second). This could be shown more directly by the observation that $\gamma(S)$ is the smallest $\gamma$-closed set which contains $S$. $\endgroup$ – Daniel Schepler Feb 23 at 23:16
  • $\begingroup$ @DanielSchepler thank you, and the observation is good. However, I'm still completely puzzled about how the authors construct the relation $R$ from an arbitrary closure on the powerset. $\endgroup$ – John Stell Feb 24 at 10:21

Your Answer

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

Browse other questions tagged or ask your own question.