At wikipedia article about Brouwer fixed-point theorem, in the second paragraph, one can read the following:

In its original field, this result is one of the key theorems characterizing the topology of Euclidean spaces, along with the Jordan curve theorem, the hairy ball theorem and the Borsuk–Ulam theorem.

My question is: in what sense does these four theorems characterize the Euclidean space? Is there a theorem that says: "If a space $X$ fulfils the criteria of those four theorems, then $X\simeq A\subseteq \mathbb{R}^n$ " or something along the lines?

Wikipedia has a citation at that point, namely "See page 15 of: D. Leborgne Calcul différentiel et géométrie". Sadly, I don't speak French. An English source or a translation would be apprieciated.

  • $\begingroup$ I'd be highly interested to see an answer of this! +1 for this question $\endgroup$ May 7, 2018 at 7:52
  • $\begingroup$ The Brouwer fixed-point theorem is defined for compact and convex subsets of a Euclidean space to begin with. Or are you simply saying that $X$ should have FPP? Also in the full statement of Jordan curve theorem we talk about bounded subsets. Hence $X$ has to be metric? In that case it can be embedded into a Banach space. So are you asking: can (some properties here) metric space be embedded into a finite-dimensional Banach space? $\endgroup$
    – freakish
    May 7, 2018 at 10:21
  • $\begingroup$ Also I don't think that the wiki implies anything deep (although the connection might exist). It just says that $\mathbb{R}^n$ has these few nice properties. $\endgroup$
    – freakish
    May 7, 2018 at 10:25
  • $\begingroup$ That would be the anwser to my question. Still, the word "characterize" suggest that $\mathbb{R}^n $ is a remarkable space in this context, meaning that no other space has these properties. So that answer would be really unsatisfactory. $\endgroup$
    – mz71
    May 7, 2018 at 10:40
  • $\begingroup$ It might be more of an historical claim, that these particular theorems were major milestones as we developed a topological understanding of $\Bbb R^n$ $\endgroup$ May 7, 2018 at 16:54

1 Answer 1


Considering that the hairy ball theorem and the Borsuk–Ulam theorem are theorems about spheres, not $\mathbb{R}^n$, I cannot see how they can characterize $\mathbb{R}^n$ in a meaningful way.

I do not have access to Leborgne's book, but considering the quote is apparently from page 15, I would assume that this is from the introduction, and that "characterize" is not used in a formal sense.

Rather, I would say that these theorems are nice truths about Euclidean spaces (or spheres...) that were historically important (and remain so). For example the Jordan curve theorem is an intuitive truth about $\mathbb{R}^n$, and the fact that it is so hard to prove is rather crazy and shattered many people's expectations.

  • $\begingroup$ A sphere can be defined in any metric space (in various non-equivalent ways) but the Borsuk-Ulam theorem also requires the notion of antipode. $\endgroup$
    – freakish
    May 7, 2018 at 14:12
  • $\begingroup$ Well, a sphere has a default topology inherited from the Euclidean space. But I understand that this can be the answer to my question. Sadly. Thank you. $\endgroup$
    – mz71
    May 7, 2018 at 14:18
  • $\begingroup$ @mzg147 Yes, but if you want to turn hairy ball theorem into a property of a space, then you have to talk about a sphere in that space. Otherwise it doesn't make much sense. Also note that in the case of the hairy ball theorem you need a notion of calculus to even formulate such property (i.e. a nonvanishing tangent vector field). The statement of the hairy ball theorem probably restricts your $X$ to a differential manifold. And this is trivial since every (connected) manifold embedds into $\mathbb{R}^n$. $\endgroup$
    – freakish
    May 7, 2018 at 14:20
  • $\begingroup$ All in all, I would say don't dig too much into it. The wiki doesn't imply anything deep. $\endgroup$
    – freakish
    May 7, 2018 at 14:24

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