Show that $(x_n)$ is unbounded if and only if exists a subsequence $(x_{n_k})$ of $(x_n)$ such that $\|x_{n_k}\|\geq k$ for all $k\in\mathbb{N}$

Question

Let $(x_n)_{n=1}^\infty$ a sequence on $(\mathbb{R},|.|)$

Show that $(x_n)$ is unbounded if and only if exists a subsequence $(x_{n_k})$ of $(x_n)$ such that $\|x_{n_k}\|\geq k$ for all $k\in\mathbb{N}$.

My try:

$\Leftarrow]$ Since the subsequence $(x_{n_k})$ is divergent then is trivial that $(x_n)$ is unbounded.

$\Rightarrow]$ (I was thinking on define a subsequence $x_n > k$ for each $k$ and prove it by induction, but I'm not sure if it's the best way to prove it)

Any suggestions would be great!

Answer 1

Suppose that the sequence $(x_n)$ is not bounded. By definition a sequence is bounded if there exists a positive number $L$ such that $|x_n|<L$ for all $n\in\mathbb{N}$. The negation of this statement is: For any choice of constant $L>0$, there is some $n$ such that $|x_n|>L$.

So in particular if $\{x_n\}$ is not bounded then you can choose $n_1$ such that $|x_{n_1}| >1$.You surely can find such $n_1$, otherwise $\{x_n\}$ would be bounded by the unit interval centered at zero. Then choose $n_2$ such that $|x_{n_2}| >2$ and so on . By induction you get $\{x_{n_k}\}$ such that $|x_{n_k}|>k$ for all $k$ which implies that $x_{n_k}$ is unbounded as well.