# Proving that the pointful and the pointless forms of the well-inside relation coincide

In Stone Spaces by Johnstone, a pointless version of the well-inside order is given:

Definition 1. Let $$L$$ be a locale and let $$x, y \in L$$. $$x$$ is said to be well-inside $$y$$ iff $$\begin{equation*} x \eqslantless y \quad:=\quad \exists z \in L.\; x \wedge z = \bot \;\text{and}\; y \vee z = \top. \end{equation*}$$

My understanding of Johstone is that this is a pointless manifestation of the following relation we can define on topological spaces (as pointed by Johnstone on pg. 80 of Stone Spaces).

Definition 2. Let $$X$$ be a topological space and let $$U, V \in \Omega(X)$$. $$U$$ is well-inside $$V$$ iff $$\mathsf{Clos}(U) \subseteq V$$ (where $$\mathsf{Clos}$$ denotes the usual closure of a set).

Proposition. Let $$X$$ be a topological space and let $$U, V \in \Omega(X)$$. $$U$$ is well-inside $$V$$ (by Defn. 2) iff $$U \eqslantless V$$ with respect to the locale $$\Omega(X)$$ of open sets of $$X$$ (Defn. 1).

Johnstone says that this is a direct consequence of the fact that $$x \eqslantless y$$ iff $$\neg x \vee y = \top$$ in any Heyting algebra and that $$\neg U$$ is the interior of the complement of $$U$$. I don't understand this and the validity of this proposition is not apparent to me. Can someone please a provide a proof for this claim?

Suppose $$U$$ is well-inside $$V$$ by Definition 2. Let $$W = X\setminus\text{Clos}(U)$$, the complement of the closure of $$U$$. We have $$U\subseteq \text{Clos}(U)\subseteq V$$, so $$\emptyset \subseteq U\cap W \subseteq \text{Clos}(U) \cap W = \emptyset$$, and $$X\supseteq V\cup W \supseteq \text{Clos}(U)\cup W = X$$. Thus $$U$$ is well-inside $$V$$ by Definition 1.

Conversely, suppose $$U$$ is well-inside $$V$$ by Definition 1. Then there is some open set $$W$$ such that $$U\cap W = \emptyset$$ and $$V\cup W = X$$. Let $$C = X\setminus W$$, and note that $$C$$ is closed. Since $$U\cap W = \emptyset$$ and $$V\cup W = X$$, we have $$U\subseteq C\subseteq V$$. Since $$C$$ is a closed set containing $$U$$, $$\text{Clos}(U)\subseteq C\subseteq V$$, so $$U$$ is well-inside $$V$$ by Definition 2.

That was a direct proof - it's worthwhile understanding Johnstone's comment, though. It breaks down into the following pieces:

1. In any Heyting algebra (and thus in any locale), $$x$$ is well-inside $$y$$ (Definition 1) if and only if $$\lnot x\vee y = \top$$.
2. In the locale of open sets in a topological space, $$\lnot U$$ is the interior of the complement of $$U$$ (which is equal to the complement of the closure of $$U$$).
3. Thus, $$U$$ is well-inside $$V$$ by Definition 1 iff $$\lnot U \cup V = X$$ iff $$\text{Clos}(U)\subseteq V$$ iff $$U$$ is well-inside $$V$$ by Definition 2.

The first two points are rather important facts - which are worth spending some time to internalize if you haven't already - and the third point is an easy verification.

• many thanks for explaining Johnstone's remark! Oct 8 '20 at 9:20