# Is the lattice of topologies a Heyting algebra?

I read that, given $$X$$ a set, if $$Top(X)$$ is the set of all topologies over X, then you can produce a distributive lattice $$(Top(X)< \land, \lor, 0 , 1 )$$. You can achieve this if you interpret $$\land$$ as intersection, $$x \lor y$$ as the topology generated by the sub-basis {x, y} , 0 as the chaotic topology and 1 as the discrete topology. (right?) But why it stops there? Isn't this also a Heyting algebra? There is only one more condition to satisfy once you get a distributive lattice with 0 and 1: it must exist the relative pseudo-complement, i.e., for all $$a$$ and $$b$$, there is a $$x$$ such that

$$a \land x \leq b.$$

(see formal definition https://en.wikipedia.org/wiki/Heyting_algebra)

I can't see how to proof this, or disproof. Are there more requirements to meet?

• Note that while some people have used the picturesque term "chaotic topology", it is much more standard to call it the "indiscrete topology". Commented Jan 29, 2022 at 18:55

The lattice of all topologies on a given set (of at least $$3$$ elements) is not a distributive lattice, so it cannot be a Heyting algebra.