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What topological properties uniquely characterize the local topology of $\mathbb{R}^n$?

I know $\mathbb{R}$ is the unique complete ordered field, but that's partly algebraic.

Better if the characterization avoids sneaking in the topology of $\mathbb{R}$ through the side door by saying it has a metric which is continuous, for example. And certainly we don't want to sneak it in by assuming the notion of a manifold.

A start: a similar question on MathStack, and a 1965 paper giving an axiomatic characterization of $S^n$. (Chapter II of the paper is the meat.)

Eric Auld
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  • every open set in $\mathbb{R}^n$ can be written as a countable disjoint union of open cubes – Nick Castillo Aug 23 '21 at 20:09
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    What's wrong with '$S^n$ minus a point'? Or, 'the one-point compactification of the space is $S^n$'. – Berci Aug 23 '21 at 20:19
  • @NickCastillo: Do you have a reference for this claim? I believe it without "disjoint", but I'm having trouble seeing how to write a ball as such a union. Supposing we can do it for some ball, projecting to a single coordinate axis gives us a non-trivial decomposition of $(-1,1)$ into disjoint intervals. (Non-trivial means more than one interval is used in the decomposition). I don't think such a decomposition exists by a limit argument using the left and right end points of the intervals used. That said, I haven't written down a proof, so maybe I am missing something. – Jason DeVito - on hiatus Aug 23 '21 at 20:34
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    @Berci Because I'm interested in local topological properties of $\mathbb{R}^n$, and $S^n$, as a manifold, is built out of $\mathbb{R}^n$, so that is a bit circular. – Eric Auld Aug 23 '21 at 21:58
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    For $\Bbb R$: https://mathoverflow.net/questions/76134/topological-characterisation-of-the-real-line – Alessandro Codenotti Aug 24 '21 at 04:26
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    Note that your title question is different from the question in the first line. Which one are you actually asking? Also, do you regard the notion of a manifold as given? For instance, one can characterize $R^n$ as a contractible manifold simply connected at infinity (except maybe in dimension 4, not sure about that). But if you are asking about a characterization of open subsets of $R^n$, then it is rather hopeless, I think, even in dimension 3. – Moishe Kohan Aug 24 '21 at 10:39
  • @MoisheKohan No, the notion of manifold is surely not given...its local topology is based on $\mathbb{R}^n$. Alessandro's response is the sort of thing I'm looking for, but only treats $\mathbb{R}^1$. – Eric Auld Aug 24 '21 at 19:12
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    @AlessandroCodenotti That's a good start, thank you. This is just the kind of thing I'm looking for. – Eric Auld Aug 24 '21 at 19:13
  • @JasonDeVito What Nick wrote is simply false, apart from dimension 1. For instance an open annulus is not homeomorphic to a discount union of open cubes. – Moishe Kohan Sep 02 '21 at 16:35
  • @EricAuld Your question remains unclear as written. References given in comments give you a topological characterization of manifolds. Are you interested in a characterization of open subsets (as the body of your question suggests) or the entire $R^n$ (as in the title and the 2nd line)? Using what tools? Algebraic topology? – Moishe Kohan Sep 03 '21 at 17:32
  • @MoisheKohan Thanks for your feedback. I'm interested in the local topology, not anything global. I'd be open to changing the question title if you have a suggestion to make it more clear. Which reference gives a topological characterization of manifolds? Re algebraic topology, all of these are contractible, so not interesting from a homotopy theoretic perspective. – Eric Auld Sep 03 '21 at 18:46
  • @EricAuld: For instance, read the paper by Harrold: The paper does not take the notion of a manifold for granted, as is made clear even by the title of the paper. The description is quite complicated and is by induction on $n$. As for "all these are contractible", this is simply false (for $n>1$) if you mean "open connected subsets of $R^n$," which is what the first sentence of your question is asking about. – Moishe Kohan Sep 03 '21 at 21:20
  • @MoisheKohan You’re quite right re contractible, excuse me. Local only, no algebraic topology. I agree Harrold is a good example. – Eric Auld Sep 04 '21 at 01:09

2 Answers2

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You are actually quite mistaken that algebraic topology is irrelevant in regards to your question. But the math involved in it is quite deep, here are some glimpses.

Definition 1. A finite-dimensional locally compact metrizable topological space $X$ is called ANR (absolute neighborhood retract) if it is locally contractible.

Definition 2. An (integer) $n$-dimensional homology manifold is a metrizable 2nd countable topological space $X$ satisfying $$ H_*(X, X-\{x\}) \cong H_*({\mathbb R}^n, {\mathbb R}^n \setminus \{0\}) $$ for every $x\in X$. (Due to the excision, this condition is actually local.)

Definition 3. A generalized manifold is an ANR which is also a homology manifold.

All topological $n$-dimensional manifolds are generalized manifolds. It is a well-known open problem to find extra topological conditions under which the converse is also true. For instance, if $X$ is a 2-dimensional generalized manifold, then it is also a topological surface. This is a nontrivial theorem due to Moore (1920s). From this, with a bit more work, one gets a characterization of 2-dimensional disks. (One needs to define generalized manifolds "with boundary." Then the 2-disk is the compact contractible 2-dimensional generalized manifold with boundary.)

In higher dimensions ($n\ge 3$), Moore's result fails and one needs extra conditions. One such condition (which kicks in once $n\ge 5$) is known as DDP (the Disjoint Disk Property):

Definition 4. A metrizable topological space is said to satisfy the DDP if any two (continuous) maps of the 2-disk into $X$ admit arbitrarily small perturbations, which have disjoint images.

DDP fails for all manifolds of dimension $\le 4$ but holds for all manifolds of dimension $\ge 5$. For a long time, it was believed that every generalized manifold of dimension $\ge 5$, that also satisfies DDP, is, in fact, a topological manifold. This, turned out to be false:

Bryant, J.; Ferry, S.; Mio, W.; Weinberger, S., Topology of homology manifolds, Ann. Math. (2) 143, No. 3, 435-467 (1996). ZBL0867.57016.

Nevertheless, there is an extra invariant, introduced by Frank Quinn (of local algebro-topological nature) of generalized manifolds $X$, whose vanishing (plus DDP) is equivalent to the the condition that $X$ is a topological manifold provided that $n\ge 5$ (see the reference in the above paper). But I stop here.

Moishe Kohan
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  • Great answer, thanks! Should have thought about local topology defined that way. Do you know any good examples of generalized manifolds which aren’t topological manifolds? – Eric Auld Sep 04 '21 at 15:51
  • @EricAuld: Here's Bing's example. Take a wild arc $\alpha$ in $R^3$ whose complement is not simply connected. Contract $\alpha$ to a point. The quotient will be a generalized 3-manifold which is not a manifold (the space is simply-connected but the complement to a point is not simply connected). – Moishe Kohan Sep 04 '21 at 16:59
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This is a difficult question, and I don’t think it’s solved. An easier problem is to characterize the topological spaces that embed as a connected closed subset of some Euclidean space.

The necessary and sufficient topological properties are: $(1)$ connected; $(2)$ compact; $(3)$ second countable; and $(4)$ of finite dimension (large inductive, small inductive, covering, they’re all the same in this setting). If you fix $n=1$ then this amounts to conditions $(1,2,3)$, plus having at most two noncut points (so you’ve got to be either degenerate or an arc). In the case $n=2$, there are all kinds of partial characterizations of continua that live in the plane.

A very elegant theorem is due to Kuratowski: A connected topological graph is planar if it does not contain either the complete graph on five vertices or the $3–3$ bipartite graph. As for some necessary conditions vis a vis your question: If $ X$ is a connected open subset of $\mathbb{R}^n$ then $X$ is: $(1)$ path connected; $(2)$ locally path-connected; $(3)$ second countable; and $(4)$ of dimension $\le n$ .

There may be lots of other interesting topological properties I’ve forgotten; one interesting feature of our space $X$ is that if $f:X→\mathbb{R}^n$ is continuous and one-one, then $f$ is a homeomorphism onto $f(X)$ and $f(X)$ is also open in $\mathbb{R}^n$. (This is from Brouwer’s invariance of domain theorem, $1912$.)

Sorry if this doesn't answer your question but I hope it helps.

Jakeup
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