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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.

symplectomorphic
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  • If you take a connected open neighborhood (in the subspace topology) of a boundary point of the closed ball, you can retract it to the part of its own boundary contained in itself (the original boundary part of the ball). There are no non-constant retracts off an open ball. – Randall Nov 17 '17 at 19:07
  • I didn't say that well at all. Maybe someone more eloquent can clean it up for me. – Randall Nov 17 '17 at 19:10
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    What @DietrichBurde found looks like a real duplicate; pity the answers there are not very good. – hmakholm left over Monica Nov 17 '17 at 19:20

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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.

symplectomorphic
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    If $X$ is a topological space isn't $\partial X$ (relative to $X$ itself) automatically empty? So $f(\partial X)=f(\varnothing)=\varnothing=\partial Y=\partial f(X)$ doesn't look very interesting here -- at least I cannot see how this implies that $U_x$ and $\mathbb R^m$ can't be homeomorphic. – hmakholm left over Monica Nov 17 '17 at 19:27
  • @HenningMakholm: bah, you're right, of course. Edited. – symplectomorphic Nov 17 '17 at 19:57
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    The problem essentially boils down to showing that Euclidean half-space is not homeomorphic to the whole space; see https://math.stackexchange.com/questions/739962/half-space-is-not-homeomorphic-to-euclidean-space/740016 – symplectomorphic Nov 17 '17 at 20:07