tink
Oct24-04, 01:27 PM
Suppose that {Xn}(as n goes to infinity) is a bounded sequence, and that liminfn{Xn} = L. Prove there exists some subsequence of {Xn}, call it {Xnk}(as k goes to infinity) so that lim{Xnk} approaches L as k approaches infinity.
Ok, so here goes what I've got down on this one:
Pick a j,i>=1 so that Xi>=Xj. Make Xi = Xn1 and Xj = Xn2. Continue on this way making Xnk>Xn(k+1). This gives us a new sequence {Xnk}. Since liminfn{Xn} = L, we cannot pick an Xn < L. Since our Xnk are getting smaller as k gets bigger, we now know that we will eventually be picking Xn that are closer and closer to L, but will never be less than L. Therefore, lim{Xnk} approaches L as k approaches infinity.
Is this the right approach to the question? If anybody can give me some suggestions on how to fix this proof (if it is incorrect) I would be sooooo greatful!
Ok, so here goes what I've got down on this one:
Pick a j,i>=1 so that Xi>=Xj. Make Xi = Xn1 and Xj = Xn2. Continue on this way making Xnk>Xn(k+1). This gives us a new sequence {Xnk}. Since liminfn{Xn} = L, we cannot pick an Xn < L. Since our Xnk are getting smaller as k gets bigger, we now know that we will eventually be picking Xn that are closer and closer to L, but will never be less than L. Therefore, lim{Xnk} approaches L as k approaches infinity.
Is this the right approach to the question? If anybody can give me some suggestions on how to fix this proof (if it is incorrect) I would be sooooo greatful!