Tell that sequence $(x_n)$ converges if and only if there $n_0\in \Bbb N$ such that $x_n=x_{n_0}$ for all $n\geq n_0.$*

Question

I do not know how to solve the following example so if any of you can help me solve. Please. The example is as follows:

Let $(X,d)$ a discrete metric space and $(x_n)$ is a sequence in $X$. Tell that sequence $(x_n)$ converges if and only if there $n_0\in \Bbb N$ such that $x_n=x_{n_0}$ for all $n\geq n_0.$

Thanky very much. Thanks for your anwers.

score 3 · Answer 1 · answered Nov 12 '13 at 18:51

One direction is easy: a sequence that is eventually constant certainly converges. Suppose now that $\langle x_n:n\in\Bbb N\rangle$ converges to $x$ in the discrete metric space $X$. Since $X$ is discrete, there is an $\epsilon>0$ such that $B(x,\epsilon)=\{x\}$. Since $\langle x_n:n\in\Bbb N\rangle$ converges to $x$, there is an $n_0\in\Bbb N$ such that ... what?

score 1 · Answer 2 · answered Nov 12 '13 at 18:55

1

Let $(x_n)$ converge. Then for all $\epsilon \gt 0$ there exists $n_0$ such that $n\geq n_0 \implies d(x_n, a) \lt \epsilon. \ $ Choose $\epsilon \lt 1$, then since $d$ is discrete, $d(x_n, a) = 0$ for all $n \geq n_0 \implies x_n = a, \ \forall n\geq n_0$.

Now prove the converse.

answered Nov 12 '13 at 18:55

Daniel Donnelly

21,342

Thanks for your answres. Please if you can, Solve the converse direction you please, because I'm not able to do it – Madrit Zhaku Nov 12 '13 at 18:57

Tell that sequence $(x_n)$ converges if and only if there $n_0\in \Bbb N$ such that $x_n=x_{n_0}$ for all $n\geq n_0.$*

2 Answers2