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Analysis - What are they doing in this proof ?!?
The thm is "A metric space M is compact iff it is complete and totally bounded".
We have Bolzano-Weierstrass to work with (compact iff sequentially compact).
The "==>" part is direct.
For "<==", they say,
"Let's show that M is sequentially compact. Let {y_k} be a sequence in M and we can assume without loss of generality that all the y_k are distinct."
This is where it gets weird. They show how to construct a Cauchy subsequence that converge but I have no idea what they're doing. They say,
"Given an integer N, cover M with finitely many 1/N-balls, [itex]B_{1/N}(x_{L_1}), ..., B_{1/N}(x_{L_N})[/itex]. An infinite number of the y_k lie in one of these balls. Start with N=1. Write [itex]M=B_{1}(x_{L_1})\cup ...\cup B_{1}(x_{L_N})[/itex], and so we can select a subsequence of {y_k} lying entirely in one of these balls. Repeat for N=2, getting a further subsequence lying in a fixed ball of radius 1/2, and so on. Now choose the "diagonal" subsequence, the first member from the first sequence, the second from the second , ans so on. This sequence is Cauchy and since M is complete, it converge."
Firstly what is up with the L's ? Anyway, my best shot at understanding what he's doing is that he chooses an integer N. Because M is totally bounded, there are finitely many points [itex]\{x_{1},...,x_{L_N}\}[/itex] such that [itex]M=B_{1/N}(x_{1})\cup ...\cup B_{1/N}(x_{L_N})[/itex]. There must be at least one x_j such that [itex]B_{1/N}(x_{j})[/itex] contains infinitely many points. Now, for all n<N, [itex]B_{1/N}(x_{j})\subset B_{1/n}(x_{j})[/itex], so there are also infinitely many points is all those balls. For each n, define a subsequence as just those points in [itex]B_{1/n}(x_{j})[/itex].
But...as indicated in the last bit, the convergeant subsequence we're going to form will consists of the first point of the sequence "spawned" by n=1, the second point will be the second from the sequence spawned by n=2, etc... the Nth will be the Nth term of the sequence spawned by n=N. But that's as far as we go. Nothing tells us that there is even a single point in [itex]B_{1/N+1}(x_{j})[/itex], so our algorithm stops there it seems.
I probably misunderstood the method though, but I've investigated other interpretations and nothing makes sense.
Homework Statement
The thm is "A metric space M is compact iff it is complete and totally bounded".
We have Bolzano-Weierstrass to work with (compact iff sequentially compact).
The "==>" part is direct.
For "<==", they say,
"Let's show that M is sequentially compact. Let {y_k} be a sequence in M and we can assume without loss of generality that all the y_k are distinct."
This is where it gets weird. They show how to construct a Cauchy subsequence that converge but I have no idea what they're doing. They say,
"Given an integer N, cover M with finitely many 1/N-balls, [itex]B_{1/N}(x_{L_1}), ..., B_{1/N}(x_{L_N})[/itex]. An infinite number of the y_k lie in one of these balls. Start with N=1. Write [itex]M=B_{1}(x_{L_1})\cup ...\cup B_{1}(x_{L_N})[/itex], and so we can select a subsequence of {y_k} lying entirely in one of these balls. Repeat for N=2, getting a further subsequence lying in a fixed ball of radius 1/2, and so on. Now choose the "diagonal" subsequence, the first member from the first sequence, the second from the second , ans so on. This sequence is Cauchy and since M is complete, it converge."
Firstly what is up with the L's ? Anyway, my best shot at understanding what he's doing is that he chooses an integer N. Because M is totally bounded, there are finitely many points [itex]\{x_{1},...,x_{L_N}\}[/itex] such that [itex]M=B_{1/N}(x_{1})\cup ...\cup B_{1/N}(x_{L_N})[/itex]. There must be at least one x_j such that [itex]B_{1/N}(x_{j})[/itex] contains infinitely many points. Now, for all n<N, [itex]B_{1/N}(x_{j})\subset B_{1/n}(x_{j})[/itex], so there are also infinitely many points is all those balls. For each n, define a subsequence as just those points in [itex]B_{1/n}(x_{j})[/itex].
But...as indicated in the last bit, the convergeant subsequence we're going to form will consists of the first point of the sequence "spawned" by n=1, the second point will be the second from the sequence spawned by n=2, etc... the Nth will be the Nth term of the sequence spawned by n=N. But that's as far as we go. Nothing tells us that there is even a single point in [itex]B_{1/N+1}(x_{j})[/itex], so our algorithm stops there it seems.
I probably misunderstood the method though, but I've investigated other interpretations and nothing makes sense.
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