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March 1st, 2008, 07:35 PM   #1
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Sequence of convex sets, markov chain

I made a wacky claim in some notes I have posted in my personal webspace and now I'm filled with doubt.

Starting with the set of all convex linear combinations of the usual basis of R^n, call it C_0.

For instance in R^2 this would be the line segment connecting (1,0) to (0,1).

Suppose I had a sequence of convex sets C_n such that C_n is a proper subset of C_n-1, then is the limit as n goes to infinity a singleton set, for instance one point on the line segment mentioned above.

Thanks,
Steve
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March 2nd, 2008, 05:47 AM   #2
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It's in R^n, right? So, each C_n has at least one point not in C_(n+1), which means C_0 has at least an infinite number of points not in C_inf.

This would imply what you said, except the reals are not countably infinite. I really don't think you can take it for granted that C_inf is a single point or empty.

I don't even think the convexity plays a role in this, although convexity might limit which subsets can be chosen in such a way that you'll see the result you want.

My two cents.*


*Do not take my word as Gospel.
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March 2nd, 2008, 10:41 AM   #3
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Yeah I could be left with a cantor set if I removed the middle third at each step.
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March 2nd, 2008, 10:54 AM   #4
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Oh but a cantor set is not convex, I think convexity does come into play. Any two points in a convex set must be connected by a line segment within the set.
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March 2nd, 2008, 02:15 PM   #5
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Quote:
Originally Posted by StephenKauffman
Oh but a cantor set is not convex, I think convexity does come into play. Any two points in a convex set must be connected by a line segment within the set.
It's not continuous though, is it?
I didn't think of having non-continuous sets.

But a non-convex, continuous set should still reduce towards a single point in some dimension, thus leaving you with a set in R^(n-1). Of course, you could take an infinite number of subsets such that it is reduced only in one dimension.

The biggest problem I'm still running into is that if you're working in R^n, there's still the fact that the interval (0,2) has an uncountable number of elements that are not in (0,1). Thus, you could take every subset to be a superset of a smaller uncountably infinite set, and still be working in R^n.

If you limit it to Q^d, then you'll run into the previous problem of removing an infinite number of points of the form (a,b,c,....,n-1,x), where a-(n-1) are constant and x is the only thing that changes. In this case, you'll have a convex set in Q^(n-1), but that's still not a point.

Would the "next" set of these reductions (i.e. to Q^(n-2)) make the number of operations uncountable? I'm not sure... this is where my intuition of infinite fails me.
Wouldn't you have to do n*inf=inf reductions of this sort? So would it work in Q^n?

I don't know about Q^n. However, I'm almost positive your original problem doesn't hold in R^n
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March 3rd, 2008, 12:09 PM   #6
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A friend pointed out, as an example in R^1, the interval (-1-1/n, 1+1/n), as n goes from 1 to infinity. Each set is contained within the previous.

Is that right? Or is it something on the right track?
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