
#1
Oct2710, 05:10 AM

P: 19

a sequence [itex](x_n)[/itex] does not converge to a
means infinitely many elements of [itex]\{x_n:n\in N\}[/itex] not in [itex]B(x,\epsilon)[/itex] why the 2 sentence equaivelent? 



#2
Oct2710, 08:53 AM

HW Helper
P: 1,344

Remember: if a sequence [tex] (x_n)[/tex] does converge to [tex] a [/tex] then, for any [tex] \espilon > 0 [/tex] there is an integer [tex] N [/tex] such that, for all
[tex] n > N [/tex] it is true that [tex] x_n \in B(x,\epsilon)[/tex]. With this in mind, if [tex] (x_n)[/tex] [b] does not converge to [tex] a [/tex], it has to be true that there is no [tex] N [/tex] that satisfies the previous requirement. If saying [tex] x_n \in B(x, \epsilon)[/tex] from some point on is false, it has to be true that [tex] x_n \not \in B(x,\epsilon)[/tex] for infinitely many values of [tex] n [/tex]. 



#3
Oct2710, 09:38 AM

P: 19





#4
Oct2710, 12:18 PM

HW Helper
P: 1,344

a sequence does not converge to a
"Can contains infinitely many in this case????"
In the case of nonconvergence? Sure: consider [tex] (1)^n [/tex]. It doesn't converge to [tex] 1[/tex], but there are infinitely many integers (namely the even ones) for which [tex] (1)^n \in B(1,0.1) [/tex]. 



#5
Oct2910, 03:09 AM

P: 19

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? 


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