How do we find sub-sequences and limit points?

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gn17
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Hi tsr,

In my lecture notes I'm given the definition of a limit point as:

A real number a is called a limit point of a sequence s subscript n where n is an element of the natural set, if there exists a subsequence such that the limit of the subsequence=a.

This to me makes no sense whatsoever because how are we supposed to know if a subsequence exists and above all how would be find one? I'm given two sequences, 1) (-1)^n. and 2) sin(npi/2). and asked to find their limit points. How do I find sub-sequences of these which give me the limit point. Also can't there be infinite sub-sequences and so infinite limit points so how are we possibly meant to find all of them? So basically how do we find a subsequence to find the limit points and is there any other way to find the limit points of a sequence.


Can I also ask: can an unbounded sequence be convergent? Also, I know a convergent sequence has only 1 limit point but does b imply a, i.e, if a sequence has 1 limit point, is it convergent?




Any help would be much appreciated.
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hassassin04
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(Original post by gn17)
Hi tsr,

In my lecture notes I'm given the definition of a limit point as:

A real number a is called a limit point of a sequence s subscript n where n is an element of the natural set, if there exists a subsequence such that the limit of the subsequence=a.

This to me makes no sense whatsoever because how are we supposed to know if a subsequence exists and above all how would be find one? I'm given two sequences, 1) (-1)^n. and 2) sin(npi/2). and asked to find their limit points. How do I find sub-sequences of these which give me the limit point. Also can't there be infinite sub-sequences and so infinite limit points so how are we possibly meant to find all of them? So basically how do we find a subsequence to find the limit points and is there any other way to find the limit points of a sequence.

Any help would be much appreciated.
As far as I know only by observation. Those 2 examples you gave are very obvious. What are the possible values of (-1)^n and sin(npi/2) ? Can you construct a subsequence with corresponding limits?
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Smaug123
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(Original post by gn17)
Also can't there be infinite sub-sequences and so infinite limit points so how are we possibly meant to find all of them?
Yes, of course - for instance, if we listed the rationals, the list would have every limit point in the reals. You just have to apply common sense, usually.


Can I also ask: can an unbounded sequence be convergent? Also, I know a convergent sequence has only 1 limit point but does b imply a, i.e, if a sequence has 1 limit point, is it convergent?

Consider the sequence 1, 0, 2, 0, 3, 0, ….
It has only one limit point (0), but does not converge.

An unbounded sequence cannot converge. Indeed, suppose it did. Then exists x s.t. for every \epsilon exists N s.t. for all n>N, |x_n - x| < \epsilon. But then x_n < \epsilon + x for all n > N, and if we set \epsilon = 1 then there exists N s.t. for all n>N, x_n<x+1, so convergent sequences are bounded.
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around
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Try writing out the first few elements of the sequences you're giving, and then seeing if you can spot a pattern.
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