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March 1st, 2008, 07:35 PM  #1 
Newbie Joined: Oct 2007 From: Baltimore Posts: 4 Thanks: 0  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_n1, then is the limit as n goes to infinity a singleton set, for instance one point on the line segment mentioned above. Thanks, Steve 
March 2nd, 2008, 05:47 AM  #2 
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 
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. 
March 2nd, 2008, 10:41 AM  #3 
Newbie Joined: Oct 2007 From: Baltimore Posts: 4 Thanks: 0 
Yeah I could be left with a cantor set if I removed the middle third at each step.

March 2nd, 2008, 10:54 AM  #4 
Newbie Joined: Oct 2007 From: Baltimore Posts: 4 Thanks: 0 
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.

March 2nd, 2008, 02:15 PM  #5  
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Quote:
I didn't think of having noncontinuous sets. But a nonconvex, continuous set should still reduce towards a single point in some dimension, thus leaving you with a set in R^(n1). 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,....,n1,x), where a(n1) are constant and x is the only thing that changes. In this case, you'll have a convex set in Q^(n1), but that's still not a point. Would the "next" set of these reductions (i.e. to Q^(n2)) 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  
March 3rd, 2008, 12:09 PM  #6 
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 
A friend pointed out, as an example in R^1, the interval (11/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? 

Tags 
chain, convex, markov, sequence, sets 
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