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I'm reading Michael Henle's A Combinatorial Introduction to Topology where he defines topological spaces and nearness thusly:

A topological space is a set 𝒥 together with the choice of a class of subsets N of 𝒥 (each of which is called a neighborhood of its points) such that

(a) Every point is in some neighborhood.

(b) The intersection of any two neighborhoods of a point contains a neighborhood of that point.

[...]

Let 𝒥 be a topological space. Let A be a subset of 𝒥 and P be a point of 𝒥. P is near A, written P ← A, if every neighborhood of P contains a point of A.

I'm more familiar with category theory than topology, so I tried to extend nearness to a binary relation between sets so I could use it to define morphisms between subsets of 𝒥:

Given A, B ⊆ 𝒥, say A ← B if ∀ P ∈ A, P ← B

But then after rereading the normal definition of nearness, I realized that there's no requirement that this binary relation is reflexive or transitive, so this definition is useless for defining a category.

For example, I could define a topological space on ℝ2 where a neighborhoods of a point are open disks around the reflection of that point about the diagonal x=y; that is a neighborhood of (a,b) is { (x,y) | (x - b)2 + (y - a)2 < δ2 }.

In this topological space, my extension on nearness is neither reflexive nor transitive. (a,b) ← { (b,a) }, and (b,a) ← { (a,b) }, but (a,b) ← { (a,b) } only if a = b.

My question is, how often do topological spaces like this come up? Is there a name for topological spaces where the extension of nearness to a binary relationship between sets is non-reflexive and/or non-transitive?

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    $\begingroup$ Of possible relevance is the paragraph beginning with "Note that this notion of being close to $x$ has more structure to it than" in my answer to Confusion Regarding Munkres's Definition of Basis for a Topology (see also the paragraph beginning with "One the things that (b) in Theorem 4 does"). $\endgroup$ – Dave L. Renfro Jan 7 at 18:50
  • $\begingroup$ "I'm more familiar with category theory than topology, so I tried to extend nearness to" --- This sounds like a "hammer in search of a nail" approach. However, a lot of interesting results in math have been obtained in this way! $\endgroup$ – Dave L. Renfro Jan 7 at 19:45
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I talked about this at lunch with a topologist, and I think I was wrong.

I misread the definition - I missed the fact that every neighborhood is a neighborhood of each of its points. That is, for any topological space 𝒥, a neighborhood A is a neighborhood of P if and only if P ∈ A.

In the usual plane topology, I'd been thinking of the open disc centered at a point as a neighborhood belonging to only the center point, not a neighborhood belonging to all of its points.

My example using ℝ2 with neighborhoods reflected about x=y is then not a valid topological space.

You can extend nearness to a reflexive and transitive binary relation between subsets of 𝒥.

Let A, B ⊆ 𝒥. Say A is near B if for all P ∈ A, P is near B.

  • Reflexivity: ∀ P ∈ A, ∀ neighborhoods X of P, P ∈ X ⇒ ∃ some point of A in X, so P is near A.

    Therefore A is near A.

  • Transitivity: Suppose ∃ A, B, C ⊆ 𝒥.

    Consider neighborhood X of P ∈ A.

    If A is near B, then P is near B, so ∃ Q ∈ B s.t. Q ∈ X. This means X is also a neighborhood of Q ∈ B.

    If B is near C, then Q is near C, so ∃ R ∈ C s.t. R ∈ X.

    Therefore P ← C, so A ← B and B ← C implies A ← C.

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  • $\begingroup$ $0$ is near $(0,1)$ and $1$ is near $(0,1)$, but $0$ is not near $1$. Nearness in the usual sense is of course not transitive. There's a chain of near places all the way to a far place. Of course this is a different concept but it doesn't seem to me like it should be transitive. $\endgroup$ – Matt Samuel Jan 8 at 4:40
  • $\begingroup$ @MattSamuel nearness of subsets, using my definition above, forms a Transitive Relation. Your example does prove that nearness is not symmetric, however. $\endgroup$ – rampion Jan 8 at 16:21
  • $\begingroup$ I don't think it's true that nearness is antisymmetric (the irrationals and the rationals being a good counterexample), so I suspect nearness is a preorder. $\endgroup$ – rampion Jan 8 at 16:25

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