# Topological boundary of a specific manifold.

Let $\emptyset\neq X\subset\mathbb{R}^n$, $n\ge2$, be an $(n-1)$-dimensional differentiable submanifold, i.e. for every $p\in X$ there is an open $U_p\subset\mathbb{R}^n$ with $p\in U_p$, and a differentiable map $f:U_p\to\mathbb{R}$ such that $f^{-1}(0)=X\cap U_p$ and $rk(Df(p))=1$. Suppose that $X$ has the following property:

There exists an open $U\subset\mathbb{R}^n$ with $X\subset U$, and a differentiable function $g:U\to\mathbb{R}$ s.t. $X=g^{-1}(0)$ and $Dg(p)\neq0$ for all $\in X$.

Now the questions asks to show that $X$ is the topological boundary in $U$ of $A:=\{p\in U\,|\,g(p)<0\}$.

My first take is always to show both inclusions...but:

When starting with $p\in\partial A$, we must show that $p\in X$, i.e. $f(p)=0$. We know that for all $N\in\mathcal{N}_p$ it holds that $N\cap A\neq\emptyset$ and $N\cap(U\setminus A)\neq\emptyset$. So, for any open $U$ around $p$ there exist $p_1,p_2\in U$ such that $f(p_1)<0$ and $f(p_2)>0$. I guess that $f$ has to switch signs somewhere because of this, but I'm stuck on how to prove this, and also the other inclusion. Any help is appreciated.

Let's attack this more conceptionally, less point-wise.

Your map $g$ is continuous, hence it divides your space $U$ into 3 components, namely $U_\pm = g^{-1}(\mathbb R_\pm)$ which are open and $X$ which is closed, i.e. we write $U$ as disjoint union

$$U= U_+ \sqcup X \sqcup U_-.$$

By this we have that both $U_\pm \sqcup X = U _ \mp^c$ are closed with intersection contained in $X$ (by disjointness). Hence $\partial U_+ \subset X$.

Now is the point where the differentiability comes in! It tells you that $X$ is 1-codimensional, which means that every open ball around a point in $X$ hits $U_+$, hence $X \subset U_+$.

Another way to argue is for $p\in X$ it is $Dg(p) \neq 0$, hence there is $\epsilon_0$, such that

$\forall \epsilon < \epsilon_0 :g(B_\epsilon(p))$ is open hence $$\forall \epsilon < \epsilon_0 :g(B_\epsilon(p))\cap \mathbb R_+ \neq 0,$$ which means $X\subset U_+$.

• I guess this things is symmetric in $U_+$ and $U_-$, so if $\partial U_+\subset X$ then also by the same reasoning $\partial U_-$. But why does it follow that $\partial U_+\subset X$ from what you write? And further on you show $X\subset U_+$, why is this sufficient to conclude $\partial U_+=X$? – B. Pasternak May 9 '16 at 16:29
• If you show both inclusions, the equality follows. And yes it is symmetric. However you definitely need that every point in $X$ is a boundary point of $U_\pm$ for the other inclusion. – Daniel Valenzuela May 10 '16 at 10:29

With the help of the answer by @Daniel Valenzuela and some more work, I have come up with the following solution, which doesn't seem to use the fact that $f$ is differentiable (and I think this makes sense since this is a topological property we are proving) (OP appreciates any comments).

Clearly, $X=g^{-1}(0)$ is closed in $U$, and $A=g^{-1}((-\infty,0))$ and $B:=g^{-1}((0,\infty))$ are both open in $U$, since $f$ is continuous. It is clear that $U=X\sqcup A\sqcup B$, and also that $(X\sqcup B)^\circ=B$. Then, since $(U\setminus A)^\circ=U\setminus\overline{A}$, it follows that $U\setminus\overline{A}=(X\sqcup B)^\circ=B$, so $\overline{A}=X\sqcup A$. As $A$ is open, $A=A^\circ$, and it follows that $\partial A=\overline{A}\setminus A^\circ=(X\sqcup A)\setminus A=X$. Am I missing something?

• You are on the right track, but you will definitely need the differentiability! Because only then $g^{-1}0$ will be a nice closed set with no $n$-dimensional interior, namely an $(n-1)$-dimensional manifold. That's why $(X\sqcup B)^\circ = B$ does not need to hold in general. Consider e.g. the continous function $\mathbb R^n \to \{0\} \hookrightarrow \mathbb R$. – Daniel Valenzuela May 10 '16 at 10:26
• You are absolutely right, thank you for your help. – B. Pasternak May 10 '16 at 13:40
• Glad it helped! – Daniel Valenzuela May 10 '16 at 20:21