# Do the monotone maps from a poset into a Heyting algebra form a Heyting algebra?

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.

• You can prove that $(f\to g)(x) = f(x)\to g(x)$. You might need some equational axioms of Heyting algebra. Commented Jul 9 at 20:40
• @amrsa But isn't $f(x) \rightarrow g(x)$ not monotone in general? Commented Jul 9 at 21:14
• 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. Commented Jul 9 at 21:31
• 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! Commented Jul 10 at 10:25

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.
• 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. Commented Jul 10 at 10:23