5

Because $\mathbb{R}$ is homeomorphic to any open interval, and a subset of $\mathbb{R}$ is connected if and only if it is an interval, it has the unusual property of being homeomorphic to any of its proper open connected subsets. In particular, given any point in $\mathbb{R}$, all of its proper connected open neighborhoods are homeomorphic to $\mathbb{R}$ itself.

Additionally, any space with the trivial topology vacuously has this property, since it has no proper open subsets.

Question: Are there any other nontrivial spaces besides $\mathbb{R}$ with this unusual/idiosyncratic property? If so, has anyone ever studied this class of topological spaces and given them a name? If so, what is the name?

Note that $\mathbb{R}^n$ for $n \ge 2$ does not have this property (cf. this question or this one), $\mathbb{R}^2 \setminus \{0 \}$ is a counterexample (and one that shows we have to get algebraic topology involved for $n \ge 2$).

A. Howells
  • 845
hasManyStupidQuestions
  • 1,384
  • 2
    Regarding the one point space example, are you sure that $\emptyset$ should count as a proper subset in this context? Because if so then $\emptyset\subset \mathbb R$ is a proper open connected subset which is not homeomorphic to $\mathbb R$. Once you stop counting $\emptyset$ as a proper subset, every indiscrete topology becomes an example of a space with your property. – A. Howells Mar 07 '22 at 00:46
  • 1
    (I didn't post that as an answer because I figured it wasn't in the spirt of the question and that you'd want to edit to rule it out. If not, let me know and I'll post it as an answer to your first question. I don't know the answer to the second question.) – A. Howells Mar 07 '22 at 00:55
  • 1
    BTW, $\mathbb{R}^2 \setminus {0 }$ not homeomorphic to $\mathbb{R}^2$ does not necessarily involve algebraic topology, see, for instance, my answer to this question. (Same proof applies to higher dimensions.) – Ulli Mar 07 '22 at 10:29
  • @A.Howells Good call/you're right -- feel free to edit the post to make it clearer that that is not intended as an example/coutnerexample. – hasManyStupidQuestions Mar 09 '22 at 22:22
  • 1
    I doubt that this property has a name, as I don't expect that it is a very interesting property in applications (though I am willing to be convinced otherwise). Personally, I might refer to this as a kind of "topological self-similarity". Generally speaking, a metric space is self-similar if it can be decomposed into copies of itself which are geometrically similar (for example, a square can be decomposed into four smaller squares). – Xander Henderson Mar 11 '22 at 01:05
  • 1
    In this case, you are looking for both a stronger condition (in that you are not asking for the mere existence of such a decomposition, but rather that every connected open set is similar to the whole space), but also a weaker condition, in that you aren't asking for geometric similarity, but only topological similarity. So... "connection-strong topological self-similarity"? Can I throw in more adjectives? – Xander Henderson Mar 11 '22 at 01:06
  • @XanderHenderson I agree with you, that's a good interpretation. (E.g. the both stronger and weaker part.) As maybe another example that is not stronger in any sense, $\mathbb{R}^n$ is a manifold in a way that every point possesses some proper connected open neighborhood that is homeomorphic to $\mathbb{R}^n$, even though not all proper open connected neighborhoods have that property. (That being said, that some have this property is more or less a direct consequence of the stronger property of $\mathbb{R}$, i.e. because some neighborhoods are homeomorphic to products of open intervals.) – hasManyStupidQuestions Mar 11 '22 at 04:50

1 Answers1

3

The space of integers with the cofinite topology is connected and has this property. Or the reals with the right ray topology.

Ulli
  • 3,847
  • To clarify the first example (replacing $\mathbb{Z}$ with $\mathbb{N}$), let's say our open set is ${3,4,5,6,...}$, then would a homeomorphism be $\mathbb{N} \to \mathbb{N}$, $n \mapsto n+2$? These are good examples - I wanted to hold off on accepting an answer in case some person turns out to actually know a term or a reference for these kinds of spaces but I'm not sure whether that makes sense – hasManyStupidQuestions Mar 09 '22 at 22:24
  • 1
    @hasManyStupidQuestions: yes, for instance. Note that any injective map from a T1 space into the integers with the cofinite topology is continuous. With this in mind one can also show that it is homeomorphic to each of its infinite subspaces. I'm afraid I also don't name a name for those spaces. – Ulli Mar 10 '22 at 06:04
  • Well it probably doesn't have a name, so I guess I'll just accept the answer for now at least. – hasManyStupidQuestions Mar 10 '22 at 14:49