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It is true that, given any topos $\mathcal{C}$ with a terminal object $1_\mathcal{C}$, $Sub(1_\mathcal{C})$ is a Heyting algebra.

Now suppose that we start with a Heyting algebra $H$. Is it always possible to guarantee the existence of a topos $\mathcal{C}_H$ such that $Sub(1_{\mathcal{C}_H})$ is isomorphic to $H$?

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    $\begingroup$ If you start with a complete Heyting algebra then there is a Grothendieck topos whose lattice of subterminal objects is isomorphic to it; simply take the topos of sheaves on the Heyting algebra with its natural topology, where a sieve $S$ covers $c$ if $c=\bigvee S$. $\endgroup$
    – godelian
    Commented Jan 5, 2017 at 13:38
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    $\begingroup$ @Simon There is an unpublished construction of an elementary topos with given Heyting algebra as $Sub(1)$ by the late D. Pataraia. Unfortunately, although in a sense it is my responsibility, it is still not brought to a publishable form. About half of it is secured in two different ways - Peter Johnstone has an alternative reliable account of that half. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jan 5, 2017 at 14:52
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    $\begingroup$ @მამუკაჯიბლაძე : good to know ! Is this construction an analogue of the construction which to a frame $\mathcal{O}$ attach the category of $\mathcal{O}$-sets, but using only finite $H$-sets so that one does not have to deal with infinite supremum when stating the axiom for $H$-sets and maps between them ? (that was my first idea when I thought about it but I never really try to pursue it and I was wondering if it works...) $\endgroup$
    – Simon Henry
    Commented Jan 5, 2017 at 15:38
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    $\begingroup$ @Simon The finite ones do not suffice in general, as far as I know from Pataraia. E. g. the finite version works for a Boolean algebra essentially because Boolean algebras are unions of their finite Boolean subalgebras. Whereas a Heyting algebra may fail to have enough finite Heyting subalgebras. Pataraia added to $H$ all the higher order quantifiers freely and then factored them back out using a higher order analog of the Pitts construction. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jan 5, 2017 at 18:41
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    $\begingroup$ @Colin Say you want to construct a join map $\operatorname{Sub}(1)^I\to\operatorname{Sub}(1)$ for some infinite set $I$. To approach it somehow you must have some kind of internalization for $\operatorname{Sub}(1)^I$. If $\Omega^I$ is available this is simply $\hom(1,\Omega^I)$. But your topos itself might fail to be complete, so $\Omega^I$ simply does not exist. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jul 29 at 21:09

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