TITLE: How to prove that the closed unit ball in $\mathbb{R}^n$, with the subspace topology, is not a manifold?
QUESTION [5 upvotes]: Our definition of a manifold $M$ is a Hausdorff topological space such that for every $x \in M$, there exists a neighborhood $U_x$ that is homemorphic to $\mathbb{R}^m$ for some $m$. We define the closed unit ball in $\mathbb{R}^n$ to be the set $\{x \in \mathbb{R}^n \colon \|x\| \leq 1\}$.
The claim is that the closed unit ball is not a manifold. The open unit ball is clearly a manifold, so I assume that for every point $x$ on the boundary of the closed unit ball, none of the neighborhoods of $x$ are homemorphic to $\mathbb{R}$. However, I am having trouble doing so. I have tried proving this by contradiction by supposing such a homemorphism exists and showing there exists a topological property of $\mathbb{R}^n$ that the closed unit ball doesn't have.

REPLY [5 votes]: I assume you mean the closed unit ball $\mathbb{B}$ with the subspace topology.
Take a point satisfying $\|x\|=1$. Every open set $U_x$ around $x$ can be written $U_x=U\cap\mathbb{B}$ for some open set $U$ of $\mathbb{R}^n$.
Suppose we had a homeomorphism $f$ between $U_x=U\cap\mathbb{B}$ and $\mathbb{R}^m$ for some $m$.
Now delete the point $x$ from $U_x$ and the point $f(x)$ from $\mathbb{R}^m$. Then $f$ restricts to a homeomorphism between $U_x-x$, which has trivial homotopy type, and $\mathbb{R}^m$ minus a point, which doesn't.