Real Analysis: Unbounded Sequences Currently, I'm working on the following problem about unbounded sequences and their subsequences. Though, I really don't understand how to prove the following, it appears to be a direct result of the definition of an unbounded sequence. Does anyone know the method for how to do this: 
If $\{x_n\}$ is any unbounded sequence of real numbers, then there is a subsequence with $v_1 < v_2 < v_3 < \cdots$ diverging to $\infty$ or $v_1 >v_2 >v_3 >\cdots$ diverging to $−\infty$. 
Any help would be greatly appreciated. 
Thanks :-)
 A: Let's say that $\{x_n\}$ is not bounded from above. It follows that $1$ is not an upper bound of the sequence and therefore there are infinitely many subscripts $k\in\mathbb N$ such that $x_k>1$ (could have be finitely many?). In particular there is a subscript $k_1\in\mathbb N$ such that $x_{k_1}>1$.
Since $\max\{2,x_{k_1}\}$ is not an upper bound of $\{x_n\}$ it follows that there are infinitely many subscripts $k\in\mathbb N$ such that $x_k>\max\{2,x_{k_1}\}$. In particular there is a subscript $k_2\in\mathbb N$ with $k_2>k_1$ such that $x_{k_2}>\max\{2,x_{k_1}\}$.
Since $\max\{3,x_{k_2}\}$ is not an upper bound of $\{x_n\}$ it follows that there are infinitely many subscripts $k\in\mathbb N$ such that $x_k>\max\{3,x_{k_2}\}$. In particular there is a subscript $k_3\in\mathbb N$ with $k_3>k_2$ such that $x_{k_3}>\max\{3,x_{k_2}\}$.
Do you see what I am doing here? Finish it using induction (i.e. prove that there is a sequence of subscripts $k_1<k_2<\ldots<k_n<\ldots$ such that $x_{k_n}<x_{k_{n+1}}$ and $x_{k_n}>n$ for all $n\in\mathbb N$).
