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QUESTION: Let $M$ be a $k$-manifold-without-boundary in $\mathbb R^n$ and $N$ be another manifold-without-boundary in $\mathbb R^n$ such that $M\subseteq N$.

Assume that there exists a point $\mathbf p\in M$ such that for each open set $U$ in $\mathbb R^n$ which contains $\mathbf p$, there is a point $\mathbf q\in U$ such that $\mathbf q\in N\setminus M$.

Then can $N$ possibly be a $k$-manifold?

Intuitively it seems obvious that the dimension of $N$ should be greater than $k$ but I am unable to make it into a proof.

Can somebody help?

Thanks.

caffeinemachine
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  • For inclusion, dimensions could be equal, e.g., a surface and any open set in the surface. Do you also want $M$ to be a submanifold of $N$? – user99680 Mar 01 '14 at 05:11
  • i realize that inclusion doesn't alone force the dimension of $N$ to be greater than that of $M$. But there's another condition in my OP which I think is the key condition. I do not require $M$ to be a submanifold of $N$ but if you can prove something interesting using this extra condition then that is welcome. :) – caffeinemachine Mar 01 '14 at 05:24
  • The thing is that a chart for $M$ is a map from a subset of $M$ into $\mathbb R^n$. And any p would be contained in a chart, and the chart is strictly in $M$, so I can't see how that could happen. – user99680 Mar 01 '14 at 05:27
  • Do you have an example of the situation you're describing? – user99680 Mar 01 '14 at 05:33
  • I don't know if you are familiar with mechanical linkages or tensegrity frameworks. The question makes sense in that context. I am not sure what do you mean by an 'exmaple of the situation'?? – caffeinemachine Mar 01 '14 at 06:10
  • Sorry, I'm not familiar with them ; please see my reply below, I changed it. – user99680 Mar 01 '14 at 06:16
  • Thank you for the heads up. I am not familiar with abstract manifolds yet. That is, I am not acquainted with charts, atlases, etc. Can you answer the question in terms of coordinate patches, like the way it is done in advanced calculus? (Sorry if this is too much too ask). – caffeinemachine Mar 01 '14 at 07:33
  • No problem; please give me some time and I'll do it. – user99680 Mar 01 '14 at 07:40

1 Answers1

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I think the answer is no; if $M$ is a stand-alone manifold and $ M \subset N$ , then if $p$ is a point in $M \subset N$ , then $p$ is in a chart $(U, \phi )$ for $M$ , and $U \subset M$ (inclusion is strict unless the manifold has just 1 chart.), and all open sets containing $p$ are intersections of charts, each of which is contained in $M$.

EDIT: Let $p$ be in the manifold. Then there is a coordinate patch $(U_p, \phi_p)$containing $p$. Now, $U_p$ is open in the manifold, so that there is an open ( in the manifold topology) set $V_p$ containing $p$, and ( by openness of $V_p$ ) $V_p$ is strictly contained in $U_p$ . So, in a manifold, the condition you described is not possible.

user99680
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