(Edited Duplicate) Let $(x_{n_n})$ be a sequence of positive real number that has no convergent subsequence. Show lim $x_n$ = +$\infty$ Proof: Suppose $(x_{n})_n$ is a sequence of positive real numbers which has no convergent subsequence. By contradiction we have $(x_{n})_n$ is not bounded, for if it was then it would admit a convergent subsequence.Thus, $\forall M\in\textbf{R}$, $\exists N\in\textbf{N}$  such that $n\geq N$ implies $x_n>M$. Now we must show $x_n>M$ implies $\lim x_{n}=+\infty$. Assume for the sake of contradiction that given $\epsilon>0$, we can choose $N_{1}\in\textbf{N}$ such that $|x_n-L|<\epsilon$. Take $n>\max\{N,N_0\}$ and without loss of generality suppose $M>L$ and $\epsilon =M-L$. 
Thus, $x_n<(M-L)+L =M$, which is a contradiction.
 A: There’s no reason for the nested subscript, so I’ll ignore it. You have a sequence $\langle x_n:n\in\Bbb N\rangle$ of positive real numbers with no convergent subsequence. You’re quite right that if it were bounded, it would have a convergent subsequence, so it must be unbounded. You go a bit astray at the next step, though: this does not say for each $M\in\Bbb R$ there is an $m\in\Bbb N$ such that $x_n>M$ whenever $n\ge m$; it just says that for each $M\in\Bbb R$ there is at least one $n\in\Bbb N$ such that $x_n\ge M$. That is, you can’t conclude simply because that sequence is unbounded that a whole tail of it is at or above $M$: you can only conclude that the sequences rises at least as high as $M$ at some point. Unboundedness simply isn’t enough to guarantee that $\lim_{n\to\infty}x_n=\infty$, as may be seen from the sequence $\langle 0,0,1,0,2,0,3,0,4,0,5,\ldots\rangle$, for example.
You want to show that if $M\in\Bbb R$, there is an $m\in\Bbb N$ such that $x_n\ge M$ for each $n\ge m$, and for this you really will need to use again the fact that no subsequence of $\langle x_n:n\in\Bbb N\rangle$ converges. Suppose that there’s some ‘bad’ $M$ for which it fails. Then for each $m\in\Bbb N$ there is an $n_m\ge m$ such that $x_{n_m}<M$. 
We’d like to argue at this point that $\langle x_{n_m}:m\in\Bbb N\rangle$ is then a subsequence of $\langle x_n:n\in\Bbb N\rangle$ that is clearly bounded below by $0$ and above by $M$, so it must have a convergent subsequence, which will of course then be a convergent subsequence of $\langle x_n:n\in\Bbb N\rangle$, giving us the desired contradiction. There is one problem with this: the indices $n_0,n_1,n_2,\ldots$ might not be strictly increasing, so $\langle x_{n_m}:m\in\Bbb N\rangle$ might not actually be a subsequence of $\langle x_n:n\in\Bbb N\rangle$. This turns out to be only a technical difficulty, however. I’ll leave it to you to show that the last sentence of the previous paragraph can be strengthened to this:

Then for each $m\in\Bbb N$ there is an $n_m\ge m$ such that $x_{n_m}<M$, and moreover, if $m>0$ we may choose $n_m>n_{m-1}$ as well.

Then $\langle x_{n_m}:m\in\Bbb N\rangle$ really is a bounded subsequence of $\langle x_n:n\in\Bbb N\rangle$, and its convergent subsequence gives us the desired contradiction.
