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Suppose $X$ is a topological space. What are the properties such that if $X$ satisfies them, then $X$ is homeomorphic to $\mathbb{R}^{n}$ for some non-negative integer $n$?

There are answers to this for the real line such as here and here. I am wondering if there are similar answers to the case of general Euclidean spaces?

This type of question leaves more than one answer possible, so any suggestions are appreciated.

MaximusIdeal
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    the problem is that completeness is not a topological notion, so any "topological characterization" of $\mathbb{R}^n$ would work for $\mathbb{Q}^n$ –  Oct 07 '21 at 19:02
  • @Masacroso That's a fair point. We can consider metric spaces if that helps (i.e. we can assume $X$ is a Cauchy complete metric space). – MaximusIdeal Oct 07 '21 at 19:06
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    @Masacroso I'm pretty sure $\mathbb{R}^{n}$ is locally compact whereas $\mathbb{Q}^{n}$ is not, so if it comes down to those two, local compactness can discriminate between the two. – MaximusIdeal Oct 07 '21 at 19:11
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    @Masacroso Connectedness is another purely topological property distinguishing $\mathbb{R}^n$ and $\mathbb{Q}^n$. – Noah Schweber Oct 07 '21 at 19:13
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    @Maximal yes. Locally compact spaces are generally used in topology as a substitution for completeness. Maybe there is a characterization using this –  Oct 07 '21 at 19:13
  • @Noah alright, my "example" was bad. After all the topology of $\mathbb{Q}$ is defined from the topology in $\mathbb{R}$. What I wanted to say is that if you can characterize just with topological notions $\mathbb{R}$ then you can describe completeness using just topological notions, and I dont think that this can be done. –  Oct 07 '21 at 19:16
  • A first stab might be: contractible, path connected, finite topological dimension, and can be the underlying space of a topological group. Does that give any "false positives"? – Noah Schweber Oct 07 '21 at 19:19
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    @Masacroso Actually you can characterize $\mathbb{R}$ topologically despite completeness not being a topological property. You just can't use completeness to do so. For example (following the links in the OP): "connected, locally connected separable metrizable space, such that every point is a strong cut point" (note that metrizability, and complete metrizability for that matter, is a topological property). – Noah Schweber Oct 07 '21 at 19:23
  • @MaximalIdeal Note that the characterization of $\mathbb{R}$ does yield a characterization of $\mathbb{R}^n$, namely "a finite topological power of $\mathbb{R}$." So you might want to add further conditions to the question to rule this sort of thing out. – Noah Schweber Oct 07 '21 at 19:25
  • @Noah mmm... ok, very interesting, I didnt knew that something like this could be possible. I will read about it, seems very interesting –  Oct 07 '21 at 19:50
  • @NoahSchweber complete can be replaced by topologically complete (which has an internal topological characterisation). – Henno Brandsma Oct 07 '21 at 22:15
  • This was asked and answered in the past at MSE. It depends on what kind of a characterization you are interested in and how much algebraic topology do you know. – Moishe Kohan Oct 07 '21 at 23:52
  • As far as I am concerned, this is a duplicate of this. – Moishe Kohan Oct 08 '21 at 14:38

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Following up on one of my comments above, here is a weak (see below) positive answer: responding to a question on MO, Taras Banakh observed that a result of Gleason/Palais implies that every Hausdorff contractible topological space which is the underlying space of some topological group and has finite covering dimension is some $\mathbb{R}^n$.

This characterization has two suboptimal features in my opinion. First is the introduction of algebraic ideas, namely the property of being a "groupizable" space. This could be viewed as a positive, though. Second, and much more seriously in my opinion, is the use of the word "finite" which feels a bit like cheating. Note that if we're allowed to refer to finiteness we could just say "some finite topological power of $\mathbb{R}$" (via an appropriate characterization of $\mathbb{R}$).

Noah Schweber
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There is a simpler description, which I discovered after investigating ideas presented by our StackExchange users: Noah Schweber and Taras Banakh. Namely "A compact contractible topological group is trivial" paper by Burkhard Hoffman states:

Corollary 1. Let $G$ be a contractible locally compact topological group. Then $G$ is homeomorphic to a finite product of real lines.

With this we at least get rid of "finite dimension" condition. Although the algebraic aspect is still there.

freakish
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    contractible uses $\Bbb R$ (or $[0,1]$) in its definition. It's also not "pure" for that reason... – Henno Brandsma Oct 08 '21 at 10:26
  • @HennoBrandsma depends on your definition of "pure". Note that $[0,1]$ interval can be defined purely axiomatically, for example as the terminal object in the category of bipointed spaces: https://mathoverflow.net/questions/92206/what-properties-make-0-1-a-good-candidate-for-defining-fundamental-groups – freakish Oct 08 '21 at 10:46
  • Does the category proof also show its existence ? – Henno Brandsma Oct 08 '21 at 10:47
  • I've been a bit naive thinking there'd be a simple answer. However, this seems like the best candidate, and I actually like this result. – MaximusIdeal Oct 29 '21 at 13:12
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The Wikipedia article on inductive dimension notes that a theorem of Urysohn states that when $X$ is a normal space with a countable basis then

LDim X = IDim X = iDim X

(These are in turn, the Lesbegue dimension and the large and small inductive dimension).

And go on to say that the Nobeling-Pontryagin theorem then states if this dimension is finite then they are subspaces of Euclidean spaces.

This gives a purely topological characterisation of subspaces of Euclidean spaces as the real line is not used anywhere here and no algebra is used.

Mozibur Ullah
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