# (Astro)Physics

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### Point-free geometry (VII): Intersections and the nearest neighbor

We want to establish some notion of intersection. Each of our regions are independently mushy, so there will be instances where one region starts to take up the same space as another region, so to speak. First, let’s recall the notion of inclusion, I’ll just us a regular arrow. It is our basic morphism (maybe?) between regions:

$A \to B$

Means that $A$ is in $B$, $A$ is included entirely inside of $B$. Ok, let us now add onto this. Let us define, in some clunky notation:

$\mathrm{p}(A) \to B$

Means that part of $A$ is in $B$. Note that $\mathrm{p}(A) \to B \implies \mathrm{p}(B) \to A$; intersection is a mutual thing. So the ordering in these relationships does not really matter. Refer to the crappy picture below.

We have three atomic regions, labeled $A$, $B$, and $C$. I drew them as circles for simplicity. Clearly we have,

$\mathrm{p}(A) \to B, \hspace{0.1in} \&\hspace{0.1in} \mathrm{p}(C) \to B$

And we have chosen these regions such that,

$\mathrm{p}(A)\nrightarrow C$

Which states that $A$ and $C$ do NOT intersect. As an aside, with clever labeling, we can essentially put a coordinate system on our point-free space, but we’ve been able to do that for a while now. It is an affine space; no one region matters. We won’t specify the relationship with the full inclusion, as we feel it a good exercise.

Our notion of distance, using the inclusion map, needs to be changed. We could still go along with the inclusion tree, and we still feel that such an object provides information (whether important or not is not clear to us) about our space. But we likely do not need that anymore.

Specify an atomic region, $R$. We can look at the set of regions which intersect this region, call it:

$I_R = \{ A_n | \mathrm{p}(A_n) \to R\}$

Where $A_n$ are other atomic regions. The CLOSEST region $S \in I_R$ will be, in general, an element of this intersection set. We consider now $I_S$, the intersection set based on region $S$, and we compare it to $I_R$. We count the number of common regions. The region with the MOST elements in common in the two intersection sets is the nearest neighbor to region $R$. Good! This is the equivalent to a point infinitesimally close to another point.

Note: a notion of sameness can be made here. If the intersection set between two regions are the same, then the two regions are the same region:

$I_R = I_S \iff R=S$

We could also think of this intersection relationship as a binary function, too. It could be defined as such:

$\mathrm{p}(A, B) = 1\;\mathrm{ or }\;0$

If $A$ and $B$ intersect or do not, respectively. The inclusion relation can also be seen as such. This translates nicely into set theory formalism. We can almost picture a category defined by atomic regions in a point-free space with these intersections being the morphisms. Associativity and the identity are almost trivial. In fact, this operation is commutative, as stated when we first made this definition. Identity: $\mathrm{p}(A) \to A$ is trivially true: of course a region intersects itself.

Let us think about a quick definition before heading off: a connected set of regions. We can’t make the naive statement that the intersection set for every element of this set of regions must have more than the trivial element (ever region intersects itself). This obviously does not work, we can have two intersecting regions off by themselves. So we will sleep on this.

We will need the notion of both the nearest neighbor and connectedness in order to define lines and curves in our point-free space. This will be our next task, and it may bring us closer to defining the process of moving back from the (possible) continuum limit back down. We can similarly begin to do calculus on this space, and from there physics may result! It has taken a month of on-and-off thinking, but we’re making progress.

Note: we are not sure how robust this definition is with regards to the mushiness of our regions, but it is OK for now, and it allows us to progress in this framework.