# A sequence does not converge to a

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?

Homework Helper
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 $$n$$.

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 many$$x_n$$ in this case????

Homework Helper
"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 $$(-1)^n \in B(1,0.1)$$.

"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.