# 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 $k\in\mathbb{N}$

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!

• For the direct way the only issue is that you need $n_k\nearrow$ so you have to show that $\{x_n\mid n\ge n_k\}$ is still unbounded (i.e. removing finitely many terms does not change the behaviour). – zwim May 25 '20 at 17:31
• I would like to add that this depends on the axiom of dependent choice. – Yai0Phah May 25 '20 at 18:37

Suppose that the sequence $$(x_n)$$ is not bounded. By definition a sequence is bounded if there exists a positive number $$L$$ such that $$|x_n| for all $$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.
• ok, but how do you ensure the $n_i$ are increasing ? – zwim May 25 '20 at 17:35
• I should choose each $n_i$ greater that the ones already chosen. – Maryam May 25 '20 at 17:36
• Would the issue be resolved by instead of saying remove the element from the sequence, remove all elements from the sequence with index less than or equal to $n_1$. This way the remaining sequence consists of terms from the original sequence, and it is unbounded, but you are guaranteed to have an increasing sequence of $n_i$ – masiewpao May 25 '20 at 18:19