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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.

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If $M$ is compact then $\partial M \subset M$ and therefore $\partial M$ is also compact! – Mercy King Jun 23 '12 at 10:47
up vote 1 down vote accepted

Under your conditions $M$ isn't necessarily an $n$-dimensional manifold. Here is a counter example

For $0<r<R$ let $\Omega=B_R(0):=\{x \in \mathbb{R}^n:\ |x|<R\}$, and $f:\Omega \to \mathbb{R}, \ f(x)=|x|^2-r^2$. Then $$ M=\{x \in \Omega:\ f(x) \ge 0\}=S_r^1\cup A(r,R), \ \partial M=\Gamma=S_r^1\cup S_R^1, $$ with $$ S_a^1=\{x \in \mathbb{R}^n:\ |x|=a\} \ \forall a>0, \ \text{ and } A(r,R)=\{x \in \mathbb{R}^2:\ r<|x|<R\}. $$ We have $df_x(v)=2\langle x,v\rangle$ for every $x \in \Omega, \ v \in \mathbb{R}^n$, in particular $df_x \not\equiv 0$ for every $x \in \Gamma$.

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.

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