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 December 26th, 2016, 12:21 AM #1 Newbie   Joined: Oct 2016 From: Netherlands Posts: 3 Thanks: 0 k-dim manifold, measure 0 Could someone explain to me why: if M is a k-dim manifold in R^n and k
 December 26th, 2016, 11:58 AM #2 Senior Member   Joined: Aug 2012 Posts: 2,324 Thanks: 715 Step 1. Can you see that the unit circle has measure zero in the plane?
 December 28th, 2016, 03:14 PM #3 Senior Member     Joined: Oct 2016 From: Arizona Posts: 207 Thanks: 37 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems. I can't wait until I'm at this level of math! I can't wait to learn measure theory! The end of Rudin's book starts it, but I'm only on Chapter 2 so far.
December 28th, 2016, 04:36 PM   #4
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Quote:
 Originally Posted by kNiEsSoKk Could someone explain to me why: if M is a k-dim manifold in R^n and k
The proof of this is not exactly easy. However, the intuition is pretty straight forward. It follows these ideas.

1. One sufficient condition for a set to have measure (with respect to $n$ dimensional Lebesgue measure) zero is that you can cover it by a set of arbitrarily small measure. Convince yourself that any $k$ dimesional affine subset of $\mathbb{R}^n$ must have measure zero.

2. An intuitive way of thinking about a manifold is as "locally" Euclidean subset. This means that for any point there is a neighborhood on which the topology is locally homeomorphic to a ball in $\mathbb{R}^n$.

Combining the above 2 ideas should hopefully give you an intuitive idea about why the theorem must be true.

 February 4th, 2017, 04:43 PM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 The area of a line in 2 dimensional space is 0. The volume of a curve or plane in 3 dimensional space is 0.

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