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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?

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

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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.

See Thm 1.6 in The Lattice of Topologies: A Survey by R.E. Larson and S.J. Andima: https://projecteuclid.org/download/pdf_1/euclid.rmjm/1250130634

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    $\begingroup$ Are you sure that the exercise is not: prove that the open sets of any topology form a Heyting algebra? (Which is a very different claim.) Or perhaps they are only talking about the two-element set? $\endgroup$
    – Pilcrow
    Commented Jan 29, 2022 at 19:11

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