Countable Non-Differentiable Points on Convex Curve Boundaries

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I am having trouble proving the following:

Suppose that E is a convex region in the plane bounded by a curve C. Show that C has a tangent line except at a countable number of points.

E is convex iff for every x, y \in E, and for every \lambda \in [0,1], (1-\lambda) x + \lambda y \in E.

I am considering an approach where I parametrize C in a fixed orientation and then look at the places where it is not differentiable, showing somehow that corners with some angular measure a \in [0,\pi) are the only flavor of non-differentiable parts on this curve, and then showing that the number of corners is bounded by \frac{2\pi}{\pi - a} for the largest a.

Any thoughts?
 
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I think you are on the right general track. But there can be a countably infinite number of nondifferentiable points, yes? Can you show that an uncountable sum of positive numbers must be infinite?
 
Ah, right, we could have a polygon(?) with angles ##\pi, \frac{3\pi}{2}, \frac{7\pi}{4}...##

How do you take an uncountable sum? That sounds quite exotic! I am trying to imagine a proof by contradiction and pigeon-hole.
 
alanlu said:
Ah, right, we could have a polygon(?) with angles ##\pi, \frac{3\pi}{2}, \frac{7\pi}{4}...##

How do you take an uncountable sum? That sounds quite exotic! I am trying to imagine a proof by contradiction and pigeon-hole.

It's not that exotic. Suppose you are summing c_i over an uncountable index i belonging to a set I. Define I_n to be the set of all i such that c_i>1/n for n a positive integer. Then if the sum is finite, I_n must be finite for all n, right? What's the union of all of the I_n? Forgive me for not TeXing this.
 
It would be all the positive values in I, so I has at most a countable subset of nonzero values when the value of the sum is finite, as the union of countably many finite sets is at most countable. Thanks for the help!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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