Prove that if sup($a_n$) = $\infty \Rightarrow$ $(a_n)_k \to \infty$ So i've looked and searched for some resource that might shed some light on this situation that I can actually understand, but I am sure there is something I missed. I've been on this question for hours:
Prove that if sup($a_n$) = $\infty \Rightarrow$ $\exists b_k = a_{n_k} \to \infty$
So given $sup(a_n) = \infty $ then we know that $a_n$ has no upper bound.
Next we take some sub-sequence of $a_n$, such as $b_k = a_{n_k}$
Next we know the following (I am pretty sure I need to get something like this, but am iffy on my logic of getting the following)
$$a_{n_1} > n_1; a_{n_2} > n_2; \dots; a_{n_k} > n_k$$
$$\therefore b_k = a_{n_k} > k => b_k > k$$
Finally we show that $b_k \to \infty$ little unsure on this logic now to and am pretty sure this isn't quite right.
$$b_k > k => k \to \infty, b_k \to \infty$$
Just some remarks. I know that is proof is very likely wrong, especially at the points where I say $a_{n_1} > n_1$ (Im sure I missing some statement for this to be true)
and when I am proving $b_k \to \infty$ (Not following the def of a limit)
But this is the best I can think of.
 A: We want to show that there exists a subsequence $a_{n_k} \to +\infty$. We can do this easily. Note that $\sup_n a_n > 1$. Hence, there exists $n_1$ such that $a_{n_1} > 1$. Suppose that $a_{n_k} > k$ for each $k \leq N$. Then since $\sup_{n > n_N} a_n > N+ 1$, we have that there is some $n_{N+1} > n_N$ for which $a_{n_{N+1}} > N$. Thus, in this way, we construct inductively a sequence $a_{n_{k}} > k$ for all $k$, such that $n_1 < n_2 < n_3 < \cdots$. Hence $a_{n_k}$ is a subsequence and $\liminf_k a_{n_k} \geq +\infty$, so $a_{n_k} \to +\infty$. 
A: Your approach seems OK to me.
0) $\sup a_n=\infty$:  There is no upper bound $L \in \mathbb{R}$, $L >0$ for $a_n$.
Need to show that there is a subsequence $a_{n_k} \rightarrow \infty$. 
1) $a_n$ is not bounded above:
For every  $k \in \mathbb{N}$ there is an index $n_{k}$ s.t.
$a_{n_k} >k$.
We show:
The subsequence $a_{n_k}$ converges to $\infty$;
A) For every $L \in \mathbb{R}$, $L >0$, there is, by the Archimedean principle, a 
$k_0 \in \mathbb{Z^+}$, s.t. $k_0 >L$.
B) For $k \ge k_0$ :  $n_{k} \ge n_{k_0}$, and
$a_{n_k} \gt k_0> L$, and we are done.
A: I think it is trivial, but to prove it ;
$\sup{a_n}=\infty$ means $(\not\exists M)\ a_n<M$. This implies $(\forall\epsilon,\exists n)\ a_n>\epsilon$, so we can write $$\limsup_{n\to\infty}a_n=\infty$$
now the proof is complete.
