# Open neighborhood of a manifold boundary point

Manifold with boundary:

An $n$-dimensional manifold with boundary is a second countable Hausdorff space in which every point has a neighborhood homeomorphic either to an open subset of $\mathbb{R^n}$, or to an open subset of $\mathbb{H^n}$.

Neighborhood:

If $X$ is a topological space and $p$ is a point in $X$, a neighborhood of $p$ is a subset $V$ of $X$ that includes an open set $U$ containing $p$.

Can someone please exhibit for me the following triple:

• a manifold $M$ with non empty boundary,
• a point $x$ of the boundary $\partial M$ of $M$,
• a neighborhood of $x$ that is not $M$ itself.

I'm probably confused between topological and manifold boundaries, but it seems to me the above does not exist. I'm somewhat convinced the only open subset of $M$ containing $x$ is $M$ itself.

But this also seems absurd as every boundary chart would have to be a global homeomorphism. Hence defying the whole "locally xxx" stuffs of manifolds.

EDIT: Please also be explicit the topology your are using to define a neighborhood, as for in the first answer below I do not believe the defined set to be open...

If $M=D$ the closed unit disk, and $x$ is on the boundary of $D$, then if $U$ is any open disk centered at $x$, then $U\cap D$ is an open neighborhood of $x$ in the sense of $D$. (Neighborhoods like this are homeomorphic to half-spaces.)
• Ok, so a manifold with boundary is always embedded in some ambient space, and the topology is the subspace topology... This was my confusion. In my mind, a "standard" basis for the topology of $D$ would be the set of all open disks $d \subset D$ , hence the set $U \cap D$ was not open... – user2346536 Dec 6 '15 at 22:53
• Right, the manifold need not be a subspace if euclidean space, but $\mathbb H$ of course is. Regarding the open neighborhoods $U\cap D$ or $U\cap\mathbb H$, this is known as the subspace topology. – Samuel Coskey Dec 7 '15 at 2:00