A sequence doesn't converges $\iff\exists\epsilon _0 >0$ and a subsequence $(x_{n_k})$ of $(x_n)$ such that $d(x_{n_k},x^*) \geq \epsilon _0$

Question

Prove that sequence doesn't converges $\iff\exists\epsilon _0 >0$ and a subsequence $(x_{n_k})$ of $(x_n)$ such that $d(x_{n_k},x^*) \geq \epsilon _0$

My try:

Definition of not convergence: there exists $\epsilon _0 >0$ such that $\forall N=N(\epsilon_ 0)\in \mathbb{N}$ exists $n\geq N$ such that $d(x_n,x^*)\geq \epsilon _0$.

1. For $N=1$ exits $n_1 \geq 1$ such that $d(x_{n_1},x^*)\geq \epsilon _0$.
2. For $N= n_1 +1, n_2 \geq N >n_1$ such that $d(x_{n_2},x^*)\geq \epsilon _0$
3. If $n_1<n_2<...<n_k$ are such that $d(x_{n_j},x^*)\geq \epsilon _0$, we choose $N=n_k+1, n_k+1 \geq N$ such that $(d(x_{k+1},x^*) \geq \epsilon _0$.
4. Finally, we obtain the subsequence $(x_{n_k})$ of $(x_n)$ such that $d(x_{n_k},x^*) \geq \epsilon _0$

I don't know if this is enough to prove both inclusions. Any suggestions would be great!

Answer 1

For the direction of $\Longleftarrow$:

Suppose on the contrary the sequence $(x_n)$ converges. Then for any $\epsilon>0$, there is some $M\in\mathbb{N}$ such that whenever $n\geq M$, one has $d(x_n,x)<\epsilon$. But by assumption, there is a subsequence $(x_{n_k})$ such that there exists $\epsilon_0>0$, $ N\in\mathbb{N}$ such that $d(x_{n_k},x)\geq\epsilon_0$ (for all $n_k\geq N$). Choose $\epsilon=\epsilon_0$. Then we get a contradiction.

Answer 2

A sequence $(x_n)_{n\in \Bbb N}$ converges to $x^*$ in a metric space $(X,d)$ iff $\{n\in \Bbb N:x_n\not \in B_d(x^*,r)\}$ is finite whenever $r>0,$ ... where $B_d(x^*,r)=\{y\in X:d(y,x)<r\}.$ This comes directly from the def'n of $\lim_{n\to \infty}d(x_n,x^*)=0.$

So suppose $(x_n)_{n\in \Bbb N}$ does not converge to $x^*.$ Take $\epsilon_0=r>0$ such that the set$S=\{n\in \Bbb N:x_n\not \in B_d(x^*,r)\}$ is infinite. Then $(x_n)_{n\in S}$ is the desired sub-sequence.