A sequence does not converge to a

[eileen6a]

a sequence (x_n) does not converge to a
means
infinitely many elements of \{x_n:n\in N\} not in B(x,\epsilon)

why the 2 sentence equaivelent?

 

[statdad]

Remember: if a sequence (x_n) does converge to a then, for any \espilon > 0 there is an integer N such that, for all
n > N it is true that x_n \in B(x,\epsilon).

With this in mind, if (x_n) does not converge to a, it has to be true that there is no N that satisfies the previous requirement. If saying x_n \in B(x, \epsilon) from some point on is false, it has to be true that x_n \not \in B(x,\epsilon) for infinitely many values of n.

 

[eileen6a]

Remember: if a sequence (x_n) does converge to a then, for any \espilon > 0 there is an integer N such that, for all
n > N it is true that x_n \in B(x,\epsilon).

With this in mind, if (x_n) does not converge to a, it has to be true that there is no N that satisfies the previous requirement. If saying x_n \in B(x, \epsilon) from some point on is false, it has to be true that x_n \not \in B(x,\epsilon) for infinitely many values of n.

thx! related Question: Can B(x,\epsilon) contains infinitely manyx_n in this case?

 

[statdad]

"Can contains infinitely many in this case?"

In the case of non-convergence? Sure: consider (-1)^n. It doesn't converge
to 1, but there are infinitely many integers (namely the even ones) for which (-1)^n \in B(1,0.1).

 

[eileen6a]

"Can contains infinitely many in this case?"

In the case of non-convergence? Sure: consider (-1)^n. It doesn't converge
to 1, but there are infinitely many integers (namely the even ones) for which (-1)^n \in B(1,0.1).

thx great example.
how about this case?
(x_n) converge to b.
Can a ball centered at a contains infinitely many x_n, while a is not equal to b?