Spivak & Dimension of Manifold

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1. Jan 10, 2016

NihilTico

1. The problem statement, all variables and given/known data

I'm taking a swing at Spivak's Differential Geometry, and a question that Spivak asks his reader to show is that if $x\in M$ for $M$ a manifold and there is a neighborhood (Note that Spivak requires neighborhoods to be sets which contain an open set containing the point of interest) of $x$ such that this neighborhood is homeomorphic to $\mathbb{R}^n$, then this $n$ is fixed for any other such neighborhood of $x$.

If I could prove this, then I could prove his other assertion that if $U\subseteq M$ is a neighborhood of some $x\in M$, and is homeomorphic to $\mathbb{R}^n$ (for any $n$), then $U$ must be open; via the Invariance of Domain, the proof follows almost immediately from what I'm proving here.

2. The attempt at a solution

I'm a little far removed from topology so I'm not entirely confident in my proof. In particular, when I tried to find some sort of verification online, I found this which doesn't look anything like my argument.

I more or less mimic my proof that $\mathbb{R}^n$ is not homeomorphic to $\mathbb{R}^m$ if $n\ne m$ by the invariance of domain in my proof:

Spivak has shown that one can choose, for any neighborhood $U$ of $x\in M$, a subset $V\subseteq U$ such that $V$ is open with respect to the topology of $M$ and homeomorphic to the same $\mathbb{R}^n$ as $U$. This is straight-forward.

So if I let $x\in M$, and $N_1$, $N_2$ be two neighborhoods of $x$ homeomorphic, respectively, to $\mathbb{R}^n$ and $\mathbb{R}^m$. Let's denote by $f\colon N_1\to\mathbb{R}^n$ and $g\colon N_{2}\to\mathbb{R}^m$ the homeomorphisms. WLOG we can assume $n<m$, attempting to obtain a contradiction.

I can widdle $N_1$ and $N_2$ down into subsets $V_1$ and $V_2$ open in $M$ as I noted above. Let $V=V_1\cap V_2$. Then, of course, $V\subseteq N_{1}$ and $V\subseteq N_2$; moreover $V$ is still open in $M$. Since $g$ and $f$ are homeomorphisms, the image of $V$ is open in $\mathbb{R}^n$ under $f$ and the image of $V$ under $g$ is open in $\mathbb{R}^m$. Denote by $\pi$ the embedding of $\mathbb{R}^n$ into $\mathbb{R}^m$. Then the map $h:=\pi\circ f\circ g^{-1}$ passes $g(V)$ to $f(V)\subseteq\mathbb{R}^n$ viewed as a subset of $\mathbb{R}^m$. But in the metric topology of $\mathbb{R}^m$ (where, again, $n<m$), any subset of $\mathbb{R}^n$ is closed. So this contradicts the invariance of domain, since $h$ defines a continuous injection of an open subset of $\mathbb{R}^m$ into $\mathbb{R}^m$.

Does this look alright? Any pointers?

Last edited: Jan 10, 2016
2. Jan 11, 2016

Samy_A

I don't know where the statement I colored in red comes from. I think it is wrong.
Remove it, and your proof looks correct, as $f(V) \subseteq\mathbb R^n$ and $g(V) \subseteq \mathbb R^m$ would be homeomorphic non-empty open sets, and that is not possible if
$m \neq n$.

3. Jan 11, 2016

Fredrik

Staff Emeritus
By "subset of $\mathbb R^n$", I assume that he means "subset of $\{x\in\mathbb R^m|x_{n+1}=x_{n+2}=\cdots =x_m=0\}$". This set is closed because the limit of a convergent sequence in that set is in that set.

4. Jan 11, 2016

micromass

Staff Emeritus
In that interpretation, it's not wrong that $\mathbb{R}^n$ is closed in $\mathbb{R}^m$. But it is definitely not true in that interpretation that any subset of $\mathbb{R}^n$ is closed in $\mathbb{R}^m$.

5. Jan 11, 2016

Samy_A

Thought so too.
If by "any subset" he meant "any closed subset" (in the $\mathbb R^n$ topology), then I agree.

Anyway, it is a red herring, as he doesn't really use the statement.

6. Jan 11, 2016

Fredrik

Staff Emeritus
Ah, of course. Didn't really think about that part. Thanks.

7. Jan 13, 2016

NihilTico

Hey all, thanks for the replies and comments and I apologize for getting back a little late.

Not really sure why I threw that statement in there, there are plenty of counter-examples.

I appreciate the help, thanks!