Can a Sequence with a Limit of p be Called Infinite?

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Homework Statement


If I have a sequence {Pn} and I know that lim Pn = p, can I call {Pn} infinite? I am trying to use this result in a real analysis proof. I know B(p; r) intersection S is non-empty and I need to show that it has indefinitely many points. I can show that {Pn} is a subset of S and is also a subset of B(p;r). So, if {Pn} is infinite, then B(p;r) intersection S would have indefinitely many points. Our definition of {Pn} is not strictly defined. Just that n is a natural number. I know that sequences can be finite or infinite but I am not sure of the definitions. Thanks!
 
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Do you have a definition for the limit of a sequence?
 
We are using the definition: Let (X, d) be a metric space, {pn} is a subset of C and is a sequence in X and p is an element in X. We say that the sequence converges to p and write lim (as n-> infinity) pn = p provided that for every e>0, there is a real number N so that when n>N the d(p, pn)<e
 
Sorry, {pn} is a subset of X
 
amanda_ou812 said:
We are using the definition: Let (X, d) be a metric space, {pn} is a subset of C and is a sequence in X and p is an element in X. We say that the sequence converges to p and write lim (as n-> infinity) pn = p provided that for every e>0, there is a real number N so that when n>N the d(p, pn)<e
You wrote, "lim (as n-> infinity) pn = p". Doesn't that imply that if a sequence does converge to a limit, then it must be an infinite sequence ?

Added in Edit:

Of course, just because a sequence is infinite, it doesn't follow that all of its terms are distinct.
 
hm...I see your point. So, {pn} infinite and {pn} as a subset of B(p; r) intersection S does not imply that B(p;r) intersection S has indefinitly many points. Is there another way I can get to that conclusion?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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