2
$\begingroup$

Let $M$ a $n$-manifold whit boundary, i.e., for each $x\in M$, there exist $U_x\subseteq M$ open in the topology of $M$ such that $U_x$ is homeomorphic to $\mathbb{R}^n$ or homeomorphic to $\mathbb{H}^n$, where

$$ \mathbb{H}^n = \{ (x_1,\ldots,x_n) \in \mathbb{R}^n \;:\; x_n \geq 0\}. $$

Denote by $\partial M $ the boundary of $M$, i.e., $\partial M = \{x \in M\;:\;U_x \cong \mathbb{H}^n \}$.

Suppose that $M$ is embedding in a topological space $X$ and denote by $\partial_T M$ the topological boundary of $M$, i.e., $\partial_T M = X \setminus (Ext(M) \cup Int(M))$.

I conjecture that $\partial M \subseteq \partial_T M$. Is it true? This make sense? If yeah, you can give me a good argument? If not, you can show me a counter-example?

$\endgroup$
3
$\begingroup$

You probably had some assumptions in mind, about $M$ or $X$ or the embedding, that you didn't state in the question. What you actually wrote allows the possibility that $X=M$, and then $\partial_TM=\varnothing$.

$\endgroup$
3
$\begingroup$

Think about the circle in the plane. What is its topological boundary? What condition on $M\subset X$ is necessary and sufficient for $M$ not to be the topological boundary, and hence for $\partial M$ to be the topological boundary?

$\endgroup$
1
$\begingroup$

Suppose $X$ is a manifold. Since the assertion is local at every point in the boundary, it is enough to consider the case $X=\mathbb R^n$. The dimension $n$ is relevant. If $n-1\ge\dim(M)$, then the topological boundary is the whole $M$, and the assertion holds trivially. If $n=\dim(M)$, then $\text{Int}(M)=M\setminus\partial M$ is an open set in $\mathbb R^n$ and $\partial M$ a hypersurface in $\mathbb R^n$. This hypersurface locally splits $\mathbb R^n$ into two components, one in $M$ and the other in $\mathbb R^n\setminus M$. Consequently the hypersurface $\partial M$ is exactly the topological boundary of $M$. Here the assertion holds more to the point.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.