2

Let $M\subset \mathbb{R}^n$ be a $k$-dimensional manifold and $X\subset M$ a subset. The boundary of $X$ in $M$, denoted by $\partial_M X$, is the set of all $x\in X$ such that each neighborhoud of $x$ contains points in $X$ and $M\setminus X$.

I want to show that the `smooth boundary' of $X$, as defined here: Definition 6.6.2, is a smooth $(k-1)$-dimensional manifold.

By implicit function theorem I can see that in a small neighborhood $V$ of $x\in \partial_M X$, the locus $Y\subset V$ defined by $f=0, g=0$ is a $(k-1)$-dimensional manifold. But I can't see why each such point in the locus has to be a boundary point of $X$ with respect to $M$.

Any thoughts would be appreciated.

student
  • 21
  • $M$ has dimension $k$ So $X$ donot need to have dimension $k$. Why smooth boundary of $X$ has dimension $(k-1)$ ? – HK Lee Oct 13 '13 at 02:14
  • $X$ is not an arbitrary subset of $M$. Part 2 of the definition 6.6.2 above is strong enough to ensure that smooth boundary of $X$ is a $(k-1)$-dimensional manifold. – student Oct 13 '13 at 02:52

0 Answers0