Spivak & Dimension of Manifold

[NihilTico]

1. Homework Statement
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?

 

[Samy_A]

Homework Statement

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?

(color added)
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
$m \neq n$.

 

[Fredrik]

I don't know where the statement I colored in red comes from. I think it is wrong.

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.

 

[micromass]

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.

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 $\mathbb{R}^m$.

 

[Samy_A]

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.

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 $\mathbb{R}^m$.

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.

 

[Fredrik]

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 $\mathbb{R}^m$.

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

 

[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!