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

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 $$n_1 are such that $$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!

• Your proof only works for $\Longrightarrow$. But can be shortened to: leong the subsequent to be just the original sequence $(x_n)$. May 14, 2020 at 1:42
• I know this is a silly thing to say, but this is just the negation of "a sequence converges iff all its subsequences have the same limit", so you can use that proof (which is in every textbook) and conclude this statement is true by contrapositive.
– user762914
May 14, 2020 at 2:04

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.
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 $$B_d(x^*,r)=\{y\in X:d(y,x) This comes directly from the def'n of $$\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.