Proving some number is a subsequential limit

Question

Let $X_n$ be a sequence of real numbers. Suppose that for every $\epsilon>0$ and for every $m\in{N}$, there exists $n\geq m$ with $|x_n|<\epsilon$. Prove that 0 is a subsequential limit of the sequence $x_n$.

Now I have attacked this a couple different ways, getting stuck on each of them. My one big question would be, should you approach this as a subsequence, or try going to the base sequence, and attempting it so that if the base sequence converges, so will the subsequence and etc.

Answer 1

Let us assume that $0$ is not a subsequential limit. Then, no subsequence of $(x_n)$ converges to $0$. Thus, given $\epsilon>0$:

For every $N\in \mathbb{N}$, there exists $k\geq N$ such that $|x_k|>\epsilon$.

Why is this true?

(If this were NOT true, then for some $M \in \mathbb{N}$, all terms after $M$ would be less than $\epsilon$ implying that we can form a subsequence which converges to $0$.)

But, from the given statement:

For every $\epsilon>0$ and for every $m \in \mathbb{N}$, there exists $n \geq m$ with $|x_n|<\epsilon$.

Contradiction!

Answer 2

The plan is to use your condition to pick a subsequence which goes to $0$. Define the subsequence $x_{n_k}$ as follows: choose $x_{n_k}$ so that $x_{n_k}<\frac1k$. We are allowed to do this by hypothesis, and it's clear that $x_{n_k}\to0$.

Answer 3

For every $\epsilon>0$ and for every $m\in{N}$, there exists $n\geq m$ with $|x_n|<\epsilon$, Set for that $n$, $a_{n_m}=:x_n$. Now for every $\varepsilon>0$, there exists a $m$ such that for any $n\geq m$, $|a_{n_m}|<\varepsilon$. So $a_{n_m}\to 0$ when $m\to \infty$.