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?
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] 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].
thx! related Question: Can [tex] B(x,\epsilon)[/tex] contains infinitely many[tex]x_n[/tex] in this case????
"Can contains infinitely many in this case????" In the case of non-convergence? 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].
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?