Naive Topology Intuitions and Logic

While playing around a bit with CAD programs lately, I realize I have two separate — and potentially contradictory! — bundles of intuitions about shapes in space ought to undergo unions and intersections.

Shapes to a Topologist

One I'll call the "topologist" idea of shape. A shape is a subset of $\R^n$, and unions are set-theoretic unions, and intersections are set-theoretic intersections.

Naive Shapes

The other I'll call the "naive" idea of shape, but I don't mean that as an insult. It actually strikes me as quite natural. It's what I expect and get from CAD software that manipulates 3-d shapes. According to this intuition, if I take two equal-sized squares side-by-side,

If I compute their union, I should get a rectangle twice as wide as it is tall.
If I compute their intersection, I should get the empty shape.

The Problem

Here's why this is at odds with the topologist's intuition: ask yourself, is the boundary of the square in the square? That is, are we dealing with an open square $(0,1) \x (0,1)$ or a closed square $[0,1] \x [0,1]$? If open, then the union of the two squares $(0,1) \x (0,1) \cup (1,2) \x (0,1)$ is not the rectangle $(0,2) \x (0,1)$. If closed, then the intersection of the two squares is not empty!

A Solution I Don't Like

We could say that the squares are products of half-open intervals like $[0,1) \x [0,1)$, and that would make this particular example work, (in the sense that conditions (1) and (2) above are satisfied) but it's ugly and asymmetric, and I'm not sure if there's any sensible story to tell about rotation. Surely at least saying the rotation of a shape is the pointwise rotation of the points belonging to it can no longer be the case, since this would break the half-open-in-the-correct-direction invariant that we're tacitly introducing.

A Solution I Like

My goal is to give some formal account of what the "Naive Shapes" idea is saying. I feel like a core part of it is believing that

If I try to ask whether the boundary of these shapes "belongs to" the shapes or not, this is a meaningless question

So I want to do something like quotienting out by the presence or absence of boundary points. Alternatively, I can think of it as picking out a subset of the possible shapes that are "well-behaved", which excludes lower-dimensional shapes like the single face that is the intersection of two squares sitting next to each other. Then the operation of taking their intersection is modified to instead be finding the "best approximation to their intersection by a well-behaved shape"

Well-Behaved Shapes

After some rumination, I decided that a definition I like is:

A subset of $\R^n$ is well-behaved if it is equal to the interior of its closure.

because this means the set itself is open, and also doesn't have any weird "holes" that closure would have repaired.

Cut off by LLM

I had a longer bit of explanation and proof planned for this post, but then I popped

Is there a standard name for the property of a subset X of a topological space where X is equal to the interior of its closure?

into ChatGPT, and it turns out this is in fact a well-known property of subsets of a topological space: the property is called being regular open.

The interesting fact is that these do form a complete boolean algebra. Being regular open is preserved by intersection, and although it isn't preserved by union or complementation , you can define a "corrected union" \[ A \sqcup B = \mathrm{Int}(\mathrm{Clo}(A \cup B))\] and a "corrected complement" \[ {\sim}A = \mathrm{Int}(\mathrm{Clo}(X \setminus A))\] and these are the appropriate operations for the boolean algebra.

The Logical View

I notice the definition of regular open is the same as something that smells a lot like (although I suppose it's strictly stronger than) the axiom that characterizes the modal logic S5 : if we interpret Int as $\square$ and Clo as $\dia$ in a classical modal logic interpreted into a topological space, then we're requiring \[ A \dashv\vdash \square \dia A \] So, $A$ being true here is the same as A always eventually being true.

Or to see it another way, we're basically picking out the double-negation-stable propositions from an intuitionistic point of view, where (intuitionistic) negation is interpreted into modal logic as $\square \lnot$, and asking for \[ A \dashv\vdash \square \lnot \square\lnot A \] is just asking that $A$ is equivalent to its intuitionistic double-negation.