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?
Boundary of the boundary of a manifold is empty
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0What definition of "boundary of a manifold" are you using? – 2012-01-02
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0Are 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 – 2012-01-02
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1You should probably get a better reference... that blurb in Wikipedia is not exactly a piece of great exposition! – 2012-01-02
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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. – 2012-01-02
1 Answers
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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 $\partial\partial H=\emptyset$.
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0I 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. – 2012-01-03
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0*What* is not short? – 2012-01-03
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0What 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. – 2012-01-03
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0The 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. – 2012-01-03
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0(What is not trivial is that a point is *either* an interior point *or* a boundary point) – 2012-01-03
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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. – 2012-01-03
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1Yes, really. But I think it is better if *you* do it. – 2012-01-03
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1@Mariano: Can you give a hint as to how to proceed further? – 2015-10-24