# Start a new discussion

## Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

## Discussion Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeJan 23rd 2017
• CommentRowNumber2.
• CommentAuthorTobyBartels
• CommentTimeJan 23rd 2017

I decided to add a proof of the corollary (given the theorem) and found it convenient to insert a lemma.

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeJan 23rd 2017

• CommentRowNumber4.
• CommentAuthorTobyBartels
• CommentTimeJan 24th 2017
• (edited Jan 24th 2017)

H'm, when you look at the insertion of the cross references in the document changes, it reverts to the old numbering system (theorem 1, lemma 1; instead of theorem 2.1, lemma 2.2).

ETA: Not in the most recent edit (where I added an anchor name to the corollary), but in Urs's most recent edit (where the text referring to the theorem and lemma numbers was first inserted).

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeJan 24th 2017

Thanks for catching this. This happens for instance when proposition-labels accidentally coincide with section labels or the like.

• CommentRowNumber6.
• CommentAuthorRichard Williamson
• CommentTimeJan 24th 2017
• (edited Jan 24th 2017)

Might be worth noting that it is easy to distinguish $R^1$ from $R^n$ for $n$ not equal 1 using connected components (cut out a point).

1. And if we try to move up a dimension, we see the need for the Jordan curve theorem, if one prefers to think geometrically.

• CommentRowNumber8.
• CommentAuthorTobyBartels
• CommentTimeJan 24th 2017

So inductively, starting with the point $\mathbb{R}^0$ (which is distinguishable from the others already by cardinality), $\mathbb{R}^1$ is distinguished as the only other one that becomes disconnected upon removing a point, $\mathbb{R}^2$ is the only other one that becomes disconnected upon removing an embedding of $S^1$ (which is the one-point compactification of $\mathbb{R}^1$), $\mathbb{R}^3$ is the only other one that becomes disconnected upon removing an embedding of $S^2$ (which is the one-point compactification of $\mathbb{R}^2$), etc? (Assuming that that's all true; the Jordan Curve Theorem seemed obvious but was surprisingly tricky!)

• CommentRowNumber9.
• CommentAuthorTodd_Trimble
• CommentTimeJan 25th 2017

Assuming that that’s all true

Yes; see for instance Proposition 2B.1, page 178 of 559 in Hatcher’s Algebraic Topology. But what the connected components look like can be complicated, as shown in the case of an Alexander horned sphere. See also the generalized Schoenflies theorem, as described in this note by Andy Putman.

• CommentRowNumber10.
• CommentAuthorTobyBartels
• CommentTimeJan 25th 2017

Thanks; and yes, it does work a bit better if it's all spheres.

• CommentRowNumber11.
• CommentAuthorUrs
• CommentTimeJan 25th 2017

Might be worth noting that it is easy to distinguish $R^1$ from $R^n$ for $n$ not equal 1 using connected components (cut out a point).

That point is amplified in the exposition at topology – Introduction. I haven’t moved it over to the entry on invariance of dimension itself.

• CommentRowNumber12.
• CommentAuthorKarol Szumiło
• CommentTimeJan 25th 2017

Proving invariance of dimension by looking at embeddings of spheres is tricky because embeddings of spheres are tricky. The argument of “removing a point from $\mathbb{R}$” generalizes nicely if we think about it homotopically.

If $\mathbb{R}^m$ and $\mathbb{R}^n$ are homeomorphic, then $\mathbb{R}^m \setminus \{ 0 \}$ and $\mathbb{R}^n \setminus \{ 0 \}$ are homeomorphic and so $S^{m - 1}$ and $S^{n - 1}$ are homotopy equivalent. However, the connectivity of $S^{m - 1}$ is $m - 2$ and the connectivity of $S^{n - 1}$ is $n - 2$ and thus $m = n$. The lower bound for the connectivity of the spheres can be proven, e.g. by simplicial approximation and the upper bound follows from Brouwer’s Fixed Point Theorem.

• CommentRowNumber13.
• CommentAuthorRichard Williamson
• CommentTimeJan 25th 2017
• (edited Jan 25th 2017)

Here is what seems to me to be a fun and very simple (certainly much simpler than using the Jordan curve theorem or homology, etc) way to prove that $\mathbb{R}^{3}$ is not homeomorphic to $\mathbb{R}^{n}$ for any $n \geq 2$ and $n$ not equal to $3$: if $\mathbb{R}^{3}$ were homeomorphic to $\mathbb{R}^{n}$ for such an $n$, then all knots would be trivial (ambient isotopic to the unknot), which is not the case.

So long as we restrict to $n \gt m$, this proof actually works for any $m$ in place of $3$, replacing knots by $(m-2)$-knots (embeddings of $S^{m-2}$ in $\mathbb{R}^{m}$). The proof that there are no knots in codimension $\geq 3$ in this general case is more involved, though (a famous theorem of Zeeman, but certainly simpler than higher dimensional analogues of the Jordan curve theorem). This leaves the codimension 1 cases for $m \geq 4$, where a (not simple) Schönflies theorem implies that there are no knots in all cases except for the case $m=4$, which is open, I believe.

• CommentRowNumber14.
• CommentAuthorTodd_Trimble
• CommentTimeJan 25th 2017

The Brouwer invariance of domain theorem is saying more than that $\mathbb{R}^m \ncong \mathbb{R}^n$ if $m \neq n$, so I’m not clear on why we’re focused so much on that particular corollary in this discussion.

But FWIW, I like Karol’s proof of that statement. :-)

• CommentRowNumber15.
• CommentAuthorKarol Szumiło
• CommentTimeJan 25th 2017

Well, this is a topic about invariance of dimension not invariance of domain, isn’t it?

What I really like about the argument I sketched is that it can be made very low tech. Last year I taught a point-set topology course which was not supposed to go anywhere near algebraic topology, but I still wanted to finish it off with invariance of dimension. So I had to think hard to simplify it as much as possible. In the end I managed to give a proof that doesn’t even mention homotopies at all, most of the effort goes into combinatorics of barycentric subdivisions and Sperner’s Lemma.

• CommentRowNumber16.
• CommentAuthorTodd_Trimble
• CommentTimeJan 25th 2017

Karol: oh you’re so right. I misread “domain” for “dimension”. My eyes are used to certain things (and invariance of domain was also in the article).

• CommentRowNumber17.
• CommentAuthorUrs
• CommentTimeJan 26th 2017

Karol, do you have notes prepared for this course you gave?

• CommentRowNumber18.
• CommentAuthorKarol Szumiło
• CommentTimeJan 26th 2017
• (edited Jan 26th 2017)

I only have handwritten notes that are a little too messy to make public (but it might be a good idea to clean them up one day). I don’t mind sharing them privately though.

• CommentRowNumber19.
• CommentAuthorUrs
• CommentTimeJan 26th 2017
• (edited Jan 26th 2017)

Thanks. I’d be grateful if you would indeed send me a scan by private email!

Or better, for a +500 bonus of karma points: how do you feel about adding the outline of your proof to the $n$Lab entry?

• CommentRowNumber20.
• CommentAuthorKeithEPeterson
• CommentTimeJan 28th 2017
• (edited Jan 28th 2017)

Instead of showing that $R^n$ is disconnected, why not just start with $\mathbb{R}^1\setminus\left \{ * \right \}$ as a base case, since as Richard points out it’s simple to show it’s disconnected, and then take the product with a cartesian space $\mathbb{R}^{(n-1)}$, ie $(\mathbb{R}^1\setminus\left \{ * \right \})\times\mathbb{R}^{(n-1)}$? Then if $R^n \cong (\mathbb{R}^1\setminus\left \{ * \right \})\times\mathbb{R}^{(n-1)}$, then $R^n$ is disconnected.

• CommentRowNumber21.
• CommentAuthorTodd_Trimble
• CommentTimeJan 28th 2017

$\mathbb{R}^n$ is not disconnected. The isomorphism in the last line doesn’t exist.

• CommentRowNumber22.
• CommentAuthorKeithEPeterson
• CommentTimeJan 28th 2017

I never claimed it was.

• CommentRowNumber23.
• CommentAuthorTodd_Trimble
• CommentTimeJan 28th 2017

Then I’m not following the argument.

• CommentRowNumber24.
• CommentAuthorKeithEPeterson
• CommentTimeJan 28th 2017

If a Real line without a point is disconnected, then the product of such a thing with a Real line produces a plane with a line disconnecting it.

In fact, for any Cartesian space, the product with it and the Real line without a point will also be a disconnected space. Any space $R^n$ that is isomorphic/topologically deformable to such a space will also be disconnected.

• CommentRowNumber25.
• CommentAuthorTodd_Trimble
• CommentTimeJan 28th 2017
• (edited Jan 28th 2017)

But how is this showing $\mathbb{R}^m$ is not homeomorphic to $\mathbb{R}^n$ if $m \neq n$? The first two sentences in the last post are trivially correct, as is the assertion that any space that is homeomorphic (or even just homotopy equivalent) to a disconnected space is disconnected.

• CommentRowNumber26.
• CommentAuthorRichard Williamson
• CommentTimeJan 29th 2017
• (edited Jan 29th 2017)

Keith, I think that morally speaking the argument that Toby gave in 8 is the way to prove invariance of dimension (all the ways to prove it discussed in this thread are based on this idea). The principal difficulty is that in dimensions higher than 1, we have to be able to control the image of a line, plane, …, and this is very tricky (it is exactly what the Jordan curve theorem and generalisations give us control over). The proof that $\mathbb{R}^{2}$ is not isomorphic to $\mathbb{R}^{n}$ for $n \gt 2$ for instance can go as follows: take some nice circle in $\mathbb{R}^{n}$, for which it is easy to show that cutting it out does not disconnect $\mathbb{R}^{n}$; then note that if $\mathbb{R}^{2}$ were isomorphic to $\mathbb{R}^{n}$, then cutting out in $\mathbb{R}^{2}$ the image of this circle under the homeomorphism must not disconnect it, whereas cutting out any circle in $\mathbb{R}^{2}$ will in fact disconnect it. The latter statement is obvious intuitively, but is difficult to prove rigorously: it is exactly what the Jordan curve theorem establishes.

One might argue that the need for the Jordan curve theorem and its ilk to prove something that is intuitively obvious is a pathology, indicating that something is amiss with founding topology on analysis. And I would agree with this, as would many others, Grothendieck for example. It is not so easy to find a better foundations, though (there is an approach using logic that Todd knows a lot about and which I am somewhat familiar with, but I’ve never been much sold on it).

• CommentRowNumber27.
• CommentAuthorTodd_Trimble
• CommentTimeJan 29th 2017

I think the thing I am supposed to know a lot about in Richard’s #26 is what is described in van den Dries’s book Tame Topology and O-minimal Structures, which is a development coming from model theory. A good example is the class of semi-algebraic sets, where the typical pathological examples that Richard is alluding to (like Alexander’s horned sphere, or space-filling curves, etc.) simply can’t be described in that framework. Grothendieck was advocating looking at a stratification of such theories of “Tame Topology” in section 5 of his Esquisse d’un Programme, and sometime around the early 90’s or so, the model theorists managed to describe one possible realization of his vision via their notion of O-minimal structures.

• CommentRowNumber28.
• CommentAuthorUrs
• CommentTimeApr 26th 2017

I have re-arranged and expanded a little bit at topological invariance of dimension.

In a Statement section I collection the statement of topological invariance of dimension and of its strengthening to the “invariance of domain”, and tried to make clearer that the proof that used to be there is just that the latter implies the former.

Then I started a new section Proofs and filled in the proof via ordinary cohomology, and a pointer to the proof via K-theory. Further proofs should be added here.