It is very elementary to show that $\mathbb{R}$ isn't homeomorphic to $\mathbb{R}^m$ for $m>1$: subtract a point and use the fact that connectedness is a homeomorphism invariant.

Along similar lines, you can show that $\mathbb{R^2}$ isn't homeomorphic to $\mathbb{R}^m$ for $m>2$ by subtracting a point and checking if the resulting space is simply connected. Still straightforward, but a good deal less elementary.

However, the general result that $\mathbb{R^n}$ isn't homeomorphic to $\mathbb{R^m}$ for $n\neq m$, though intuitively obvious, is usually proved using sophisticated results from algebraic topology, such as invariance of domain or extensions of the Jordan curve theorem.

Is there a more elementary proof of this fact? If not, is there intuition for why a proof is so difficult?

Arturo Magidin
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    Do you consider the proof using the (n-1)th homology/homotopy group of $\mathbb{R}^n$ minus a point to be too sophisticated? It doesn't seem to me to be much more so than your proof for $\mathbb{R}^2$. – Chris Eagle Mar 03 '11 at 20:46
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    The usual proof is a direct generalization of the proof you gave, using a sequence of homology groups. (That's where the algebraic topology comes in.) $H_0$ measures connectedness, $H_1$ obstructs simple-connectivity, etc. – Cheerful Parsnip Mar 03 '11 at 20:48
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    If you subtract a point doesnt you change the spaces and you show that $\mathbb{R} - \{x\}$ isnt homeomorphic to $\mathbb{R^m} - \{x\}$. What is wrong in my reasoning? – JKnecht Jan 31 '16 at 19:50
  • In fact I do not quite see how R isn't homeomorphic to $R^m$ for $m>1$. Could you please be more explicit? @user7530 – abcdef Mar 07 '17 at 15:20
  • This question is similar to these questions: [Another (non-homological) proof of the invariance of dimension](https://math.stackexchange.com/questions/20308/another-non-homological-proof-of-the-invariance-of-dimension/2316524#2316524) and [Elementary proof of topological invariance of dimension using Brouwer's fixed point and invariance of domain theorems?](https://math.stackexchange.com/questions/1197640/elementary-proof-of-topological-invariance-of-dimension-using-brouwers-fixed-po/2316534#2316534) – IV_ Jul 11 '17 at 20:29
  • @JKnecht I know this response is late, but the property used in that proof is that if $\mathbb{R}^n$ and $\mathbb{R}^m$ are homeomorphic, then they should "behave the same" if you remove a point. But if you remove a point from $\mathbb{R}$, then it is no longer connected, which isn't true for any other $\mathbb{R}^n$. – Calvin Godfrey Jun 18 '19 at 16:03

7 Answers7


There are reasonably accessible proofs that are purely general topology. First one needs to show Brouwer's fixed point theorem (which has an elementary proof, using barycentric subdivion and Sperner's lemma), or some result of similar hardness. Then one defines a topological dimension function (there are 3 that all coincide for separable metric spaces, dim (covering dimension), ind (small inductive dimension), Ind (large inductive dimension)), say we use dim, and then we show (using Brouwer) that $\dim(\mathbb{R}^n) = n$ for all $n$. As homeomorphic spaces have the same dimension (which is quite clear from the definition), this gives the result. This is in essence the approach Brouwer himself took, but he used a now obsolete dimension function called Dimensionsgrad in his paper, which does coincide with dim etc. for locally compact, locally connected separable metric spaces. Lebesgue proposed the covering dimension, but had a false proof for $\dim(\mathbb{R}^n) = n$, which Brouwer corrected.

One can find such proofs in Engelking (general topology), Nagata (dimension theory), or nicely condensed in van Mill's books on infinite dimensional topology. These proofs do not use homology, homotopy etc., although one could say that the Brouwer proof of his fixed point theorem (via barycentric division etc.) was a precursor to such ideas.

Henno Brandsma
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is there intuition for why a proof is so difficult?

Sure: the topological category is horrible. A generic continuous function is bizarre and will violate your geometric intuitions. When we prove that $\mathbb{R}^n$ is not homeomorphic to $\mathbb{R}^m$ by proving that $S^n$ is not homotopy equivalent to $S^m$, much of the work goes into proving that, up to homotopy, we can ignore how bizarre generic continuous functions are. That is, you think that "homeomorphic" is an intuitive condition, but it's not.

This is why the corresponding question in the smooth category is much easier; generic smooth functions are much less bizarre in a way that is quantified by Sard's lemma.

Here is a specific example of what I mean. The reason we can distinguish $\mathbb{R}$ from $\mathbb{R}^m, m > 1$ by removing a point is because continuous functions send points to points. It is tempting to argue as follows: we can distinguish $\mathbb{R}^2$ from $\mathbb{R}^m, m > 2$ by removing a line, since the result is not connected for $\mathbb{R}^2$ but is connected for $\mathbb{R}^m$. But of course this argument doesn't work because continuous functions need not send lines to lines; the image of a line can be much bigger, e.g. all of $\mathbb{R}^m$.

This is weird. For $\mathbb{R}^2$, as you say, we can rescue this proof by removing a point because we know about simple connectedness and because, again, continuous functions send points to points. But for $\mathbb{R}^3$ we are stuck: removing a plane doesn't work, and removing a line doesn't even work, so if we want to stick to our "removing a point" strategy we had better figure out what the analogue of simple connectedness is in higher dimensions. This naturally leads to homotopy and homology, which happen to be strong enough tools to deal with the fact that continuous functions are bizarre, but they don't change the fact that continuous functions are bizarre.

Qiaochu Yuan
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    $\mathbb{R}^n$ is homotopy equivalent to $\mathbb{R}^m$ (both contractible). – Chris Eagle Mar 03 '11 at 22:29
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    He meant those spaces minus a point. – aaron Mar 03 '11 at 22:34
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    Ha, whoops; I switched "homotopy equivalent" and "homeomorphic." – Qiaochu Yuan Mar 03 '11 at 23:10
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    I'm reasonably sure that the "removing a line" strategy can be made to work, in that, if $m \ge n+2$, then cutting an embedded copy of $\mathbb{R}^n$ out of $\mathbb{R}^m$ can't leave you with a disconnected space. Certainly the analogous result for $S^n$ is true (it's in Hatcher), and that's enough to tell you that $\mathbb{R}^n$ and $\mathbb{R}^m$ are not homeomorphic since $S^n$ is the one-point compactification of $\mathbb{R}^n$. Of course, this strategy requires substantially more work than the standard homology/homotopy approach. – Chris Eagle Mar 03 '11 at 23:58
  • @ChrisEagle You might run into subtleties involving the difference between connectedness and path connectedness. I'm pretty sure you can embed a line in $\Bbb R^3$ such that its complement is connected but not path-connected. – Akiva Weinberger May 09 '16 at 23:22
  • Hi Qiaochu Yuan, can u please guide us with the next: https://math.stackexchange.com/questions/2440926/measurable-functions-and-existence-decreasing-function – mathreda Sep 25 '17 at 01:31

Well, I might recast your proofs of the first two cases as follows:

Suppose that $\mathbb{R}^n$ and $\mathbb{R}^m$ are homeomorphic. Then for any $P \in \mathbb{R}^n$, there must exist a point $Q \in \mathbb{R}^m$ such that $\mathbb{R}^n \setminus \{P\}$ and $\mathbb{R}^m \setminus \{Q\}$ are homeomorphic. (Since the homeomorphism group of $\mathbb{R}^m$ acts transitively, really any point $Q$ is okay, but I'm trying to be both simple and rigorous.)

Now your proof when $n = 1$ is equivalent to the observation that $\pi_0(\mathbb{R}^1 \setminus \{P\})$ is nontrivial, while $\pi_0(\mathbb{R}^n \setminus \{Q\})$ is zero for all $n > 1$. (Note that $\pi_0(X)$ is in bijection with the set of path-components of $X$, so really we are using that $\mathbb{R}^1$ minus a point is path-connected and $\mathbb{R}^n$ minus a point is not, for $n > 1$.)

When $n =2$, your proof is literally that $\pi_1(\mathbb{R}^2 \setminus \{P\})$ is nonzero whereas $\pi_1(\mathbb{R}^n \setminus \{Q\})$ is zero for all $n > 2$. It is a little questionable to me whether the fundamental group counts as "elementary" -- you certainly have to learn about homotopies and prove some basic results in order to get this.

If you are okay with such things, then it seems to me like you might as well also admit the higher homotopy groups: the point here is that

for all $m \in \mathbb{Z}^+$, $\pi_{m-1}(\mathbb{R}^m \setminus \{P\}) \cong \mathbb{Z}$ whereas for $n > m$, $\pi_{m-1}(\mathbb{R}^n \setminus \{Q\}) = 0$.

If I am remembering correctly, the higher homotopy groups are introduced very early on in J.P. May's A Concise Course in Algebraic Topology and applied to essentially this problem, among others.

[By the way, if we are okay with homotopy, then we probably want to replace $\mathbb{R}^n \setminus \{P\}$ with its homotopy equivalent subspace $S^{n-1}$ throughout. For some reason I decided to avoid mentioning this in the arguments above. If I had, it would have saved me a fair amount of typing...]

Added: Of course one could also use homology groups instead, as others have suggested in the comments. One might argue that homotopy groups are easier to define whereas homology groups are easier to compute. But this one computation of the "lower" homotopy groups of spheres is not very hard, and my guess is that if you want to start from scratch and prove everything, then for this problem homotopy groups will give the shorter approach.

As to why the problem is hard to solve in an elementary way: the point is that the two spaces $\mathbb{R}^m \setminus \{P\}$ and $\mathbb{R}^n \setminus \{Q\}$ look the same when viewed from the lens of general topology. [An exception: one can develop topological dimension theory to tell them apart. For this I recommend the classic text of Hurewicz and Wallman. Whether that's "more elementary", I couldn't say.] In order to distinguish them for $m,n$ not too small, it seems that you need to develop various notions of the "higher connectivities" of a space, which leads inevitably to homotopy groups and co/homology groups.

Another alternative is to throw out homeomorphism and look instead at diffeomorphism. This puts a reasonable array of tools from differentiable topology and manifold theory at your disposal rather quickly (see e.g. Milnor's book Topology from a differentiable viewpoint). It is not hard to show that the dimension of a manifold is a diffeomorphism invariant! So maybe the subtlety comes from insisting on working in the topological category, which often turns out to be more difficult than working with topological spaces with nice additional structures.

Pete L. Clark
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    Since you're making revisions, you write that "$\pi_0(X)$ is a free $\mathbb{Z}$-module on the set of path-components of X" - this is not true. The set $\pi_0(X)$ is in bijection with the path-components. I believe you're thinking about $H_0(X)$. – Jason DeVito May 16 '12 at 17:10

I second Brandsma's answer. Still such proofs tend to be 10 pages long if everything is spelled out. (Engelking spends 7 on proving Brouwer's fixed point theorem using Sperner's lemma, then several pages on dimension theory.)

I wrote down a fairly simple self-contained proof of invariance of dimension (for cubes, not ${\mathbb R^n}$) in four pages. See http://arxiv.org/abs/1310.8090

The proof consists of (i) proving a cubical version of Sperner's lemma, (ii) using that to prove that the $n$-cube $I^n$ is $n$-connected in the following sense:

Let $I_i^-=\{ x\in I^n\ | \ x_i=0\},\ I_i^+=\{ x\in I^n\ | \ x_i=1\}$ be the faces of $I^n$. For $i=1,\ldots,n$, let $H_i^+, H_i^-\subset I^n$ be closed sets such that for all $i$ one has $I_i^\pm\subset H_i^\pm$ and $H_i^-\cup H_i^+=I^n$. Then $\bigcap_i(H_i^-\cap H_i^+)\ne\emptyset$.

In step (iii) one shows that this implies dim$(I^n)\ge n$ for a somewhat ad-hoc, but convenient, notion of dimension (defined in the same spirit as the above definition of $n$-connectedness), and (iv) one invokes a simple but clever proposition of van Mill showing dim$(I^n)\le n$. Now the claim follows from the (obvious) fact that dim is homeomorphism invariant.

Note: the proofs of (i) and (ii) are due to Kulpa and replace barycentric subdivision by a simple compactness argument.

Daniel Fischer
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M. Mueger
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One possible approach is the following:

  1. Show Borsuk–Ulam theorem.
  2. Deduce that $S^n$ cannot be embedded to $\mathbb{R}^n$.
  3. Let us consider $\mathbb{R}^n$ and $\mathbb{R}^m$, where $m > n$. Now $S^n$ cannot be embedded to $\mathbb{R}^n$, but it can be embedded to $\mathbb{R}^{n+1} \subseteq \mathbb{R}^m$. Therefore $\mathbb{R}^n$ and $\mathbb{R}^m$ cannot be homeomorphic (they allow different embeddings).

In fact the dual argument (cannot be embedded to the same class of topological spaces) of 3. shows, that $S^n$ is not homeomorphic to $S^m$. This might not be strictly the easiest (the most direct) but at least for me this seems to be the nicest argument: I never claimed that Borsuk-Ulam theorem is easy.

Jesse Railo
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Consider the one point compactifications, $S^n$ and $S^m$, respectively. If $\mathbb R^n$ is homeomorphic to $R^m$, their one-point compactifications would be, as well. But $H_n(S^n)=\mathbb Z$, whereas $H_n(S^m)=0$, for $n\ne m,0$.


Sperner showed in the article below, invariance of open sets, invariance of domain and invariance of dimension can be proved already with elementary combinatorial methods alone.

[Sperner 1928] Sperner, Emanuel: Neuer Beweis für die Invarianz der Dimensionszahl und des Gebietes. In: Abh. Math. Sem. Univ. Hamburg. Band 6, 1928, 265–272

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  • Bringing some of the ideas here would help in case the linked content becomes unavailable. Furthermore, simply copying answers is not a good practice. – robjohn Jul 10 '17 at 23:10