Manifolds Question
Let M be a compact nmanifold with boundary. Show that there is an open set U containing the boundary that is diffeomorphic to (the boundary) x [0,1)
 By definition of the boundary of a manifold, every point on the boundary has an open neighborhood that is diffeomorphic to R^{n1} x [0,infinity), which of course I can shrink to R^{n1} x [0,1). Also, by compactness, I can cover the boundary with finitely many such neighborhoods. The hint here is that I'm supposed to use the theorem on existence of partitions of unity. Supposedly, that allows one to stitch together local information into global. Unfortunately, the book doesn't have any examples of that. I'd actually prefer to work this problem out on my own, so... Any other examples of how to use partitions of unity to stitch together local information into global? Thanks 
[QUOTE=jinydu;341099]Let M be a compact nmanifold with boundary. Show that there is an open set U containing the boundary that is diffeomorphic to (the boundary) x [0,1)
 By definition of the boundary of a manifold, every point on the boundary has an open neighborhood that is diffeomorphic to R^{n1} x [0,infinity), which of course I can shrink to R^{n1} x [0,1). Also, by compactness, I can cover the boundary with finitely many such neighborhoods. [/QUOTE] Looks good to me. [QUOTE] The hint here is that I'm supposed to use the theorem on existence of partitions of unity. Supposedly, that allows one to stitch together local information into global. Unfortunately, the book doesn't have any examples of that. [/QUOTE] I don't understand the hint either. 
I suspect your "finitely many such neighborhoods" constitute a POU.

[QUOTE=ewmayer;341156]I suspect your "finitely many such neighborhoods" constitute a POU.[/QUOTE]
'POU'??? Acronym? Compact sets have finite covers by open sets. 
Here's the partition of unity theorem:
Let X be a manifold. For any covering of X by open sets {U_alpha}, there is a sequence of smooth functions {theta_i} on X such that: (a) 0 <= theta_i <= 1 (b) Each x\in X has a neighborhood on which all but finitely many of the theta_i are zero (c) Each theta_i is supported on one of the U_alpha (d) \sum_i theta_i = 1 identically I understand the statement of the theorem. Just not seeing how to apply it... Here's another problem that has partition of unity as a hint, by the way: Let X be a manifold with boundary. Then there is a smooth nonnegative function f:X>R with regular value at 0 such that (boundary)X = f^1(0) Thanks 
How about a concrete example? Take M to be the closed unit disc in the complex plane. Then its boundary is the unit circle. Taking U = {re^it: r>1/2}, we get a diffeomorphism of U with S^1 x [0,1) via re^it > (e^it, 2(1r)).
The problem with that is it is too specific to the example and doesn't really shed light on the general case. So... Here's another construction that I think captures the difficulty of the general case better: Let E = {e^it + (1)s: pi/3 < t < pi/3, 0 <= s < 1}, N = iE, W = E, S = iE. Clearly, E is diffeomorphic to {e^it: pi/3 < t < pi/3} x [0, 1) and similarly for the other three; and the four together cover S^1. But is it possible to stitch the four diffeomorphisms together to get a single diffeomorphism of E U N U W U S with S^1 x [0,1)? 
[QUOTE=R.D. Silverman;341104]Looks good to me.
I don't understand the hint either. Compact sets have finite covers by open sets [/QUOTE] If you don't understand something, then ask for explanation. Any subset of a topological space has finite cover by open sets ( for example whole space ), not only compact subsets. We all know what you meant and all know that you used irony again writing something different than it should be. So, 3 worthless sentences in 2 short posts. 
[QUOTE=jinydu;341099]By definition of the boundary of a manifold, every point on the boundary has an open neighborhood that is diffeomorphic to R^{n1} x [0,infinity), which of course I can shrink to R^{n1} x [0,1).[/QUOTE]
(Thinking aloud here) Can you do said shrinking via a mapping which satisfies the same restrictions as the theta_i "basis functions" from the partition of unity theorem? 
The professor said this was an "easy" result... Guess I should feel good about not being the only one who doesn't see it right away...

[QUOTE=jinydu;341512]The professor said this was an "easy" result... Guess I should feel good about not being the only one who doesn't see it right away...[/QUOTE]
Don't jump into this conclusion so fast. Solution looks to be rather technical, but it seems to me that to write it with full details requires some work and not many people here are interested in manifolds. For sure the fact that nobody answered shouldn't make you feel good. If this is a step for something more serious, then I wish you good luck. 
Perhaps partition [0,1) { measure of unity } finitely,
then take union of inverses of sets in partition. 
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