A Manifold Contained in Another 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.
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.
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.