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?

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ $0$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.