As mentioned in the title, it's well know that boundary of the boundary of a manifold is empty. That is, if $M$ is the boundary of a manifold $N$, i.e. $M=\partial N$, then $M$ is a manifold without boundary, i.e. $\partial M=\varnothing$. For example, the sphere $S^n$ has no boundary because $S^n=\partial B^{n+1}$ where $B^{n+1}$ is the closed unit ball in $\mathbb{R}^{n+1}$. What I would like to ask is that: is there an easy proof or a short proof for this statement?
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What definition of "boundary of a manifold" are you using? – Mariano Suárez-Álvarez Jan 02 '12 at 03:12
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Are there several definitions of "boundary of a manifold"? I didn't know that. But the one I am using is this: http://en.wikipedia.org/wiki/Manifold#Manifold_with_boundary – Paul Jan 02 '12 at 03:16
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1You should probably get a better reference... that blurb in Wikipedia is not exactly a piece of great exposition! – Mariano Suárez-Álvarez Jan 02 '12 at 03:19
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3Of course I know that. However, since you have asked me "What definition of "boundary of a manifold" are you using?", I just quoted the definition which is available from wiki. – Paul Jan 02 '12 at 03:35
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Let us define a (topological) $n$-manifold with boundary to be a (Hausdorff, second-countable) topological space $M$ locally homeomorphic to the closed half space $H$ in $\mathbb R^n$, and the boundary $\partial M$ of $M$ to be the subset of $M$ of points which do not have a neighborhood homeomorphic to an open set in $\mathbb R^n$.
Then:
show that the claim that $\partial\partial M=\emptyset$ follows from the observation that $\partial\partial H=\emptyset$;
show that $\partial\partial H=\emptyset$.
Mariano Suárez-Álvarez
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I am afraid that's what I need. I know how to prove the statement by using definition as you have said. But it's not short if you prove it rigorously by writting down all the details. – Paul Jan 03 '12 at 05:41
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What you have given is just the idea of the proof. If you try to write down all the details of the proof, the proof is not short. – Paul Jan 03 '12 at 05:52
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The first of my two bullets is almost immediate; the second one is a fast computation: first, that $\partial H$ is homeomorphic to $\mathbb R^{n-1}$, and that $\partial \mathbb R^{n-1}$ is empty. – Mariano Suárez-Álvarez Jan 03 '12 at 05:58
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(What is not trivial is that a point is either an interior point or a boundary point) – Mariano Suárez-Álvarez Jan 03 '12 at 06:00
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1Oh really? Ok. Let's have a deal then. If you can write down the proof with all the details with less than, say 15 lines, I will accept your answer as what I required to be a short proof. – Paul Jan 03 '12 at 06:01
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