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I am interested in generalizing the fact that the up-sets of a poset always form a Heyting algebra.

Let $P$ be a poset and $H$ a Heyting algebra. $\operatorname{Hom}(P,H)$ can be made a bound lattice using pointwise operations, so that just leaves implication.

I know that if $H$ is complete, we can define implication as a join:

$$ (f \rightarrow g)(x) := \bigvee \lbrace h(x) | h \in \operatorname{Hom}(P,H), h \wedge f \leq g \rbrace. $$

I am curious if the assumption of completeness is necessary.

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  • $\begingroup$ You can prove that $(f\to g)(x) = f(x)\to g(x)$. You might need some equational axioms of Heyting algebra. $\endgroup$
    – amrsa
    Commented Jul 9 at 20:40
  • $\begingroup$ @amrsa But isn't $f(x) \rightarrow g(x)$ not monotone in general? $\endgroup$ Commented Jul 9 at 21:14
  • $\begingroup$ I think you are correct: if $x\leq y$ and $f(x)\leq g(x)$ but $f(y)\nleq g(y)$, then $f(x)\to g(x)=1$ but $f(y)\to g(y)<1$. And it's not difficult to come up with one such example! I'll leave the comment, nevertheless, in case someone else has the same idea. $\endgroup$
    – amrsa
    Commented Jul 9 at 21:31
  • $\begingroup$ Ok, now you have an answer, which seems to me to be correct and I upvoted it, so soon I might come back and delete my comments. All the best! $\endgroup$
    – amrsa
    Commented Jul 10 at 10:25

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I would be interested to hear more about this construction, and your motivation for studying it. However, I believe that in general it should not yield a Heyting algebra. Here is my counterexample idea.

Consider $\omega+1$, the one point compactification of the natural numbers; the dual Boolean algebra of this is the (not complete) finite and cofinite collection of the naturals. Let $H$ be such an algebra. Let $P=\mathbb{N}$ be the natural numbers with their natural order. Let $f$ be the function which enumerates the finite subsets of the \textit{even} numbers, as $f(n)=[0,n]\cap Even$. This is of course a monotone function. Let $\overline{0}$ be the constant function at the empty set. Assume that $h=f\rightarrow 0$ existed. Then by definition of the relative pseudocomplement, we would have $$a\wedge f=0 \iff a\leq h$$ for any $a\in Hom(\mathbb{N},P)$. Now, let $o_{k}$ be a function defined by letting $o_{k}(0)=\{1,3,...,k\}$, where $k$ is some odd number, and then an enumeration of the odds from then on. Then $o_{k}(m)\cap f(m)=\emptyset$, obviously, so $o_{k}\wedge f=\overline{0}$. Hence $o_{k}\leq h$, so in particular $o_{k}(0)\leq h(0)$. Since $o_{k}$ can be made arbitrarily big, and $o_{k}(0)\subseteq h(0)$, then $h(0)$ must be cofinite. But then by monotonicity, this means that $h$ must be eventually constant at a cofinite set. Since $f$ grows arbitrarily large, there must then be some $k$ such that $f(k)\wedge h(k)\neq \emptyset$, which is absurd.

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  • $\begingroup$ Good (+1), but why do you call $o_k$ to a function that doesn't depend on $k$? I guess it actually does, and the idea would be to make $$o_k(0)=\{1,3,5,\ldots,k\},$$ for odd $k$, then $$o_k(1)=o_k(0)\cup\{k+2\}$$ and so on? I think that would make sense. $\endgroup$
    – amrsa
    Commented Jul 10 at 10:23
  • $\begingroup$ You are absolutely right, I edited the answer to reflect this! $\endgroup$ Commented Jul 10 at 10:30
  • $\begingroup$ Nice proof, thanks! The question wasn't really motivated by anything other than curiosity about how weak I could make my assumptions. I am working on a formal proof where it is easier to think of up-sets as monotone functions, which led me to structuring things as above. $\endgroup$ Commented Jul 11 at 1:12

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