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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!

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    $\begingroup$ 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). $\endgroup$ – zwim May 25 '20 at 17:31
  • $\begingroup$ I would like to add that this depends on the axiom of dependent choice. $\endgroup$ – Yai0Phah May 25 '20 at 18:37
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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.

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    $\begingroup$ ok, but how do you ensure the $n_i$ are increasing ? $\endgroup$ – zwim May 25 '20 at 17:35
  • $\begingroup$ I should choose each $n_i$ greater that the ones already chosen. $\endgroup$ – Maryam May 25 '20 at 17:36
  • $\begingroup$ why can you do so ? $\endgroup$ – zwim May 25 '20 at 17:36
  • $\begingroup$ because, when I have chosen an element in the subsequence, I remove that element from the sequence and what remains is still unbounded, I suppose $\endgroup$ – Maryam May 25 '20 at 17:37
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    $\begingroup$ 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 $n_i$ $\endgroup$ – masiewpao May 25 '20 at 18:19

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