Is the dual/opposite of a Heyting Algebra still a Heyting Algebra?

Question

Is the dual/opposite of a Heyting Algebra still a Heyting Algebra? What about locales? I don't see any obvious way to prove it but don't know enough examples either to test for counterexamples.

Answer 1

Let $H = \mathbb N \times \{0,1\} \cup \{\top\}$, and define $a=(0,1)$, $b=(0,0)$, and let us see that $a \to b$ doesn't exist, that is, there doesn't exist $$a \to b = \max\{c : a \wedge c \leq b\}.$$ If $a \wedge c \leq b$, that is, $a \wedge b = (0,0)$, then necessarily $c = (n,0)$ for some $n \in \mathbb N$, but there is no greatest such $n$. Moreover, $$\bigvee\{c : a \wedge c \leq b\} = \top,$$ but $\top$ is not one of those. So $H$ is not a Heyting algebra.

Can you see that $H^{\partial}$ (the order-dual of $H$) is a Heyting algebra (it's certainly a distributive lattice)?