# Topology of submanifolds of manifolds with boundary

Let $X$ be a compact $n$-manifold with boundary $\partial X$ and let $U$ be a submanifold of $X$ such that $\partial U=\partial X\cap U$. why $X-U-\partial U$ is open in $X$? i know that the topological boundary $\partial A$ of a subset $A$ is always closed since it is the intersection of the closure of $A$ and the closure of $X-A$ but here we are talking about the geometric boundary of a manifold which is different from the topological boundary. I also think from the definition that $\partial X$ is an $(n-1)$-submanifold of $X$ but i don't know if $\partial X$ is always closed? On the other hand i know that an open subset of the manifold $X$ is a submanifold, is the converse true, I mean do we have that each submanifold of $X$ is open in $X$? thank you for your clarifications

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## 1 Answer

The geometric boundary of a manifold is always closed, since it is the complement of the open set of interior points. Also, not every submanifold of $X$ is open, just take an example where the submanifold has smaller dimension.

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I think you are describing the geometric boundary in terms of the definitions related to the topological boundary – palio Oct 7 '12 at 18:06
An interior point of an $n$-dimensional manifold is a point which has a neighborhood homeomorphic to an open ball in $\mathbb{R}^n$. This is obviously an open condition. So the set of non-interior points is closed as its complement. In a manifold with boundary every point is either an interior point or a boundary point, so the boundary is closed. – Lukas Geyer Oct 7 '12 at 19:27
so $X-\partial X$ is open in $X$, what about $X-\partial X-U$ where $U$ is a submanifold of the compact manifold $X$ such that $\partial U=U\cap \partial X$ – palio Oct 7 '12 at 20:02