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Let $(x_n)$ a sequence such that $$\forall M\in\mathbb N, \exists n_M: x_{n_M}>M.$$ show that $\lim_{M\to \infty }n_M=+\infty $

So, I want to prove that $$\forall K>0, \exists p\in \mathbb N : \forall t\in\mathbb N, t\geq p\implies n_t\geq K.$$

Let $K>0$. I wanted to set $p=\sup\{t\mid x_{n_t}??\}$ but I don't see... I'm thinking about this for few hours, but I can't solve it... I also tried by contradiction, but it doesn't work.

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  • $\begingroup$ That is actually the definition of a sequence that approaches infinity $\endgroup$ – Heisenberg Mar 12 '17 at 16:07
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What you wrote is $$\limsup_{n\to \infty }x_n=+\infty.$$ It looks a little bit weird to consider $\{n_k\}_{k}$ as a sequence since $(x_{n_k})$ is not necessarily a subsequence of $(x_n)$. But anyway, it's indeed a sequence of $\mathbb N$ that diverge to $+\infty $. Let $K>0$. Then, $x_K\geq K$. Now, let $$M=\max\{|x_0|,...,|x_K|\}+1.$$ In particular, there is $n_M$ s.t. $x_{n_M}\geq M$ and $n_M> K$. So, if you set $p=n_M$ you will have that $n_t\geq K$ for all $t\geq p$. The claim follow.

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Suppose that $n_1,n_2,n_3,\dots$ is not tending to $+\infty$.

Then $(n_M)_M$ will have a bounded subsequence $(n_{M_k})_k$.

Then set $\{n_{M_k}\mid k=1,2,\dots\}$ is finite and consequently set $\{x_{n_{M_k}}\mid k=1,2,\dots\}$ is finite.

Let $y:=\max\{x_{n_{M_k}}\mid k=1,2,\dots\}$

For $k=1,2,\dots$ we have: $$x_{n_{M_k}}>M_k$$ This leads to a contradiction if we choose $k$ in such a way that $M_k\geq y$.

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