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If $M \subset \mathbb R^n$ is a compact smooth manifold with boundary, and ${M_\varepsilon }$ is the closed $\varepsilon$-neighborhood of $M$ in $\mathbb R^n$, then whether for sufficiently small $\varepsilon$, ${M_\varepsilon }$ is a smooth manifold?

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1 Answer 1

Yes. This follows from the tubular neighborhood theorem, which you may find in many differential geometry/topology books. See e.g. http://www.google.com/search?q=tubular%20neighborhood%20theorem&um=1&ie=UTF-8&hl=en&tbo=u&tbm=bks

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Wouldn't this be rather a comment than an answer? –  t.b. Nov 13 '11 at 0:46
    
@t.b.: what's the difference? –  Damian Sobota Nov 13 '11 at 1:04
    
@Damian: look at an answer e.g. by Arturo Magidin and look at this which basically says: it is a consequence of the tubular neighborhood theorem, look, here's what Google gives you. Then you should see a fundamental difference (in quality, content, helpfulness, etc...) –  t.b. Nov 13 '11 at 1:11
    
@t.b: you changed my link-- I linked to google books instead of google, which I think is more useful. I also made this community wiki in case the subtext of your comment was accusing me of being a "karma whore." –  Eric O. Korman Nov 13 '11 at 1:21
    
I wouldn't have used such a strong word. The link didn't work for me, (it still doesn't), so I stripped out the 's' from https and the rest of the Google noise, sorry that I overlooked the &tbm=bks modifier. –  t.b. Nov 13 '11 at 1:27

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