The unit Ball is a manifold with boundary

Question

I try to prove that $\overline{\Bbb B}:=\{x \in \Bbb R^n: \|x\| \leq 1\}$ is a manifold with boundary. For this I have to show that for all $x \in \Bbb B$ there exist an open subset $U$ s.t. $U \subset \Bbb B$ and there exist and open subset $\tilde{U} \subset \Bbb B$, s.t. $U$ and $\tilde{U}$ are homeomorphic. But I can't find how to define this homeomorphismus.. Can someone helps me?

Answer 1

I am not sure what to make of $\tilde U$ since with no further conditions, once we have $U$, we can take $U = \tilde U$.

To find such a $U$, note that $\mathbb{B} := \{ x \in \mathbb{R}^n : \|x\| < 1 \}$ so given such an $x \in \mathbb{B}$, $\|x\| < 1$ implies that $$\epsilon :=\min_{\|y\|=1} \|y-x\|$$ is well-defined, since distance from $x$ along the boundary is a continuous function on a compact domain, and greater than $0$, since if $\epsilon =0$, then $\|x\| =1$, a contradiction. Then take the ball $U = B\left(x,\frac{\epsilon}{2}\right)$ the open ball centered at $x$ with radius $\frac{\epsilon}{2}$.

Clearly, $U \subseteq \mathbb{B}$.

Answer 2

First of all, you should notice that every open set $A$ of $\mathbb{R}^n$ is a manifold $\textit{without}$ boundary because for every point your manifold itself is "the open set of $A$ homeomorphic to some open set of $\mathbb{R}^n$ you require in the definition of manifold w/o boundary.

Now I want to give you an intuitive idea of what is a manifold with boundary and why it has the definition it has: when you talk of manifold with boundary, you need to change a little bit the previous definition because you want to have a way to include boundary points: if you think about the example of the closed ball you can see the difference between interior points and boundary points as "having some manifold all around you in every direction" if you are in an interior points and "having some manifold only below your feet" if you happen to be "walking" on the boundary of the ball.

How do you translate this into mathematical definitions? You take the same definition you had for manifolds and require that your open set is homeomorphic (actually diffeomorphic) to some open set in a $\textit{half-space}$ of $\mathbb{R}^n$ instead of just $\mathbb{R}^n$. A half space is for example the set of all point whose first (or last o third or whatever) coordinate is greater or equal than zero.

Why do we give such a definition? This definition allows you to have two substantially different kind of open subset to be playing with: standard open balls-like sets (the one you already had in the definition of manifold) and "cut-in-half" open balls (think of them as the intersection of a standard unit ball with the sets of all point with first coordinate greater or equal than zero). Now, if you are a point on the boundary of the manifold, it's kind of like if you have the manifold only on one side and in this sense it's like if you are staying on the cut edge of the half ball.

For all the points that lies in the interior of the manifold, this new definition doesn't really change anything because you can still take the open set you had with the previous definition and see them as open sets in the half space.

While for the points that form the boundary you clearly shouldn't be able to find any open set homeomorphic to any open set of $\mathbb{R}^n$, because those kinds of open sets are "all around you" and if you have manifold "all around you" you should be an interior point, right? So for this kind of points we use the "cut in half" open sets to parametrize their neighborhood, meaning that: a point is said to be boundary if the open set with which you are parametrizing (the homeomorphism [ugh, again it should be a diffeo] is the parametrization) its neighborhood is of the "cut ball" type and the point correspond under this parametrization to a point on the "cut-edge" of the ball (meaning it has, for example, first coordinate equal to zero).

With this in mind it should be intuitively reasonable that the closed ball is a manifold with boundary because, for all the point in the interior you can find an open ball centered to that point which lies inside the ball: that is an open set of $\mathbb{R}^n$, so the homeomorphism is just the inclusion of that open set into the ball. Now, the points on the boundary (those of the sphere) don't have any open neighbourhood (in $\mathbb{R}^n$) contained in the ball but if you take any open ball in $\mathbb{R}^n$ around a point on the boundary and take the intersection with the ball, what you get is something homeomorphic (more, diffeomorphic) to a cut-ball set of the half-space (it's like if you deformed it to give it a round shape, I believe you can construct this homeomorphism using the stereographic projection). There's no hope in trying to find an open set (in the sense of the subspace topology on the ball) around boundary point which is homeomorphic to some open set of $\mathbb{R}^n$.

Sorry for the length of my answer, I hope this was useful at least!