Let a seqence be in a metric space, can it has more than one accumulation point?
My answer is no, because if it has more than one, then it would converge to both points, implies they are the same. Would that be the right start?
Let a seqence be in a metric space, can it has more than one accumulation point?
My answer is no, because if it has more than one, then it would converge to both points, implies they are the same. Would that be the right start?
I actually need to know the same answer. I was thinking that a sequence could have more than one accumulation point. For example the sequence {(-1)^n}? I may be wrong though. If there is more than one accumulation point, how would I go about proving that?
An accumulation point of a squence is not the same thing as the limit of the sequence. An accumlation point may be taken to be defined as a limit of a sub-sequence, or alternativly as point such that every open interval containing the point also contains an element of the sequence.
See Plato's post for an example.
oops sorry.
Could this adapted somehow to prove that a sequence can have more than one accumulation point?
I am aware that the proof in the textbook proves a sequence cannot have more that one limit, so I am doubtful, but it would be great if it could provide a starting point.
If you deal with a bounded sequence which does not have limit.
Then this means that this sequence have more than one convergent subsequences.
Refer the example given by Plato (see also Bolzano?Weierstrass theorem - Wikipedia, the free encyclopedia).