let $\Omega$ be an open subset of $\mathbb{R}^n$ and $f$ be a smooth real valued function on $\Omega$. Let $M=\{p\in\Omega:f(p)\ge 0 \}$, and $\Gamma=\{p\in\Omega :f(p)=0\}$. Suppose $M\neq \phi$ and that $df_p\neq 0\forall p\in\Gamma$ Then how do we give a manifold structure on $M$ and how to admit its dimension $n$ and $\Gamma$ is of $n-1$ dimensional manifold? Is it true that If M is a compact manifold in $\mathbb{R}^n$ then $\partial M=\Gamma$ is also compact?and If $dim M=n$ then $\partial M=\text{boundary }M$? I have no rigorous idea on manifold with and with out boundary, will be pleased if anyone can give motivation for that also and intuitive idea.
Construction of manifold
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differential-geometry
manifolds
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0If $M$ is compact then $\partial M \subset M$ and therefore $\partial M$ is also compact! – 2012-06-23
1 Answers
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Under your conditions $M$ isn't necessarily an $n$-dimensional manifold. Here is a counter example
For $0
Since $S_r^1$ is a $(n-1)$-dimensional manifold and $A(r,R)$ an $n$-dimensional one, therefore $M=S_r^1\cup A(r,R)$ is not a manifold.