Show that whenever m≠n. It should then be obvious that cannot have any convergent subsequence.
S is the set of all bounded, infinite sequences (x_k)=(x_1,x_2,...) of real numbers where (x_k) is bounded if its supremum is less than infinite for every pair of such sequences x=(x_k) and y=(y_k) where d(x,y)=sup_k abs (x_k-y_k). the question says denote by B(0,x)- the closed unit ball in (S,d) centered at 0 B(o,x)-={x: d(0,x)<=1}.
1. observe by construction B(0,x)- is closed and bounded.
2. for each n in the naturals, define x^(n)=(x_n,k) by x_n,k=1 if x=n and 0 otherwise. prove that sequence x^1,x^2,x^3,... of elements in (S,d) lies in B(0,x)- and has no convergent subsequences. conclude that B(0,x)- is not sequentially compact hence by bolzano-weierstrauss thm cannot be compact.
thanks for any help.
i dont get what they mean when they say "by construction," and also, i'm not seeing how there's no convergent subsequence.. why does distance between and m and n not equaling 1 contribute to the nonexistence of convergent subsequence? yes i know... i'm lost
I'm also uncertain what they mean by "by construction", which is why I didn't comment on that part of the question. For a start, I don't know why the notation B(0,x)- is used. Shouldn't it be B(0,1), the ball centred at 0 with radius 1? It's certainly obvious that this set is bounded (because everything in it is within distance 1 of the point 0). It's not immediately obvious that it's closed. That's presumably something you're meant to prove.
Have you come across the concept of a Cauchy sequence? If a sequence converges then it has to be Cauchy, which means that the distance between any two terms tends to 0 if you go far enough along the sequence. But in this sequence the distance bwteen any two terms is 1. So it's impossible for any subsequence to be Cauchy. Therefore no subsequence converges.
It was given in the statement that B(o,x)-={x: d(0,x)<=1} or .
However, I did try to comment because frankly I did not understand the notation “define x^(n)=(x_n,k) by x_n,k=1 if x=n and 0 otherwise.”
Did simple sequences become sequences of pairs?
It just seems to me to be a confused question.
That phrase "by construction" is baffling. I don't know what to make of it. But there is still something to prove. The set {x: d(0,x)*#8804;1} is certainly bounded, and its closure is certainly closed. But when you take the closure, is it still bounded? The answer is Yes, and in fact the set is already closed, so you don't actually need to take the closure. But that fact does need some explanation.
The “construct” is really construction and it goes with part (a) and has nothing to do with part b).
For part b) look at this description of the sequence: “That x^(n) is supposed to be the element of S (the space of all bounded, infinite sequences) which has a 1 in the n-th coordinate and zeros everywhere else.” Now is should be clear to you that any two terms of that sequence are exactly 1 unit apart. Therefore, the sequence can have no limit point. Because if a sequence has a limit then almost all the terms of the sequence are ‘close’ to the limit. But these are all 1 unit apart, none is close another.
ok so the terms of the sequences go like
x^(1)=1,0,0,0,...
x^(2)=0,1,0,0,...
x^(3)=0,0,1,0,...
and so on. it would seem to me that there would be a convergent subsequence if we took out the 1 from each term... i think i'm missing something to this problem...