Help with a proof of Analysis (subsequences) I have a bad time with the next problem. I'd appreciate any help. 
Problem:  Let $(a_n)_{n=0}^{\infty}$ be a sequence which is not bounded. Show that there exists a subsequence $(b_n)_{n=0}^{\infty}$ of $(a_n)$ such that $(1/b_n)\rightarrow0$.
Scratchwork: 
Claim 1: The set  $A_j: =\{\, n \in \mathbb{N}: |a_n|\ge j\,\}$ is not empty. Let $j\in \mathbb{N}$ be given. Since $(a_n)$ is not bounded, then there is some $N (j)$ so that $|a_n|\ge j$, in particular  $|a_N|\ge j$. Thus $N \in A_j$ which shows that is non-empty.
Claim 2: The function $n: \mathbb{N}\rightarrow \mathbb{N}$, $j\mapsto \text{min} (A_j)$ is well define. This follows immediately from (1), because each $A_j$ is not empty, and the uniqueness of the minimum. 
Claim 3: The function $n$, is an ordering preserving map, i.e., $n(j)< n(j+1)$. 

Here is where I'm stuck. I can show $\,n(j) \le n(j+1)$ but  not $n(j) \not= n(j+1)$ 

Claim 4: $|a_{n(j)}|\le |a_{n(j+1)}|$. We argue for contradiction, suppose $|a_{n(j)}|>|a_{n(j+1)}|$. So, we have $|a_{n(j)}|>j+1$, and then $n(j)\in A_{j+1}$, which contradicts that $n(j+1)$ is the minimum of $A_{j+1}$, since $n(j) < n(j+1)$ by (2). By induction we can show that $|a_{n(j)}|\le |a_{n(\ell)}|$, whenever $n(j)< n(\ell)$.
We set $b_j:= a_{n(j)}$
Claim 5: $(1/b_j)\rightarrow 0$. Let $\epsilon>0$ be given.  Then there is some $N$ such that $|a_n|\ge \left\lfloor{1/\epsilon}\right\rfloor+1$ for infinitely many $n\ge N$. So, if $j \ge N$ then $n(j)\ge N$, which means $|a_{n(j)}|\ge \left\lfloor{1/\epsilon}\right\rfloor+1$ and since $|a_{n(\ell)}|\ge|a_{n(j)}|$ for each $n(j)<n(\ell)$. Thus $|b_j|\ge \left\lfloor{1/\epsilon}\right\rfloor+1$, for each $j\ge N$ and hence $1/|b_j| \le \epsilon$, which proves that $(1/b_j)\rightarrow 0$, as desired.
Using this definition of the function $n$ is possible to show that conserve the strict inequality (because this is the hint which the author gives)? Thanks. 

Here are some changes: 

We define $n$ recursively as follows: Let $n_1:= \text{min} (A_1)$ and $n_{j+1}:= \text{min } (A_{j+1}-\{n_{k}\})$. 
We claim that $n(j) < n(j+1)$. Suppose for the sake of contradiction that $n(j) > n(j+1)$ (notice that $n(j) \not= n(j+1)$ otherwise we reached a contradiction ). Since,  $n(j+1) \in A_{j}$ (this is because $|a_{n(j+1)}|\ge j+1>j$), $n(j+1)< n(j)$ contradicts the minimality of $n(j)$. 
We thus have $|a_{n(j)}|\ge j$, which implies $|1/a_{n(j)}|\le 1/j$ and so $(1/a_{n(j)})\rightarrow 0$. Setting, $b_j:= a_{n(j)}$ we're done.
 A: You are doing too much work.
Let $n(1) = \min A_1$, $n(k) = \min A_k \cap \{n(k-1)+1,...\}$. Then you always have $n(k+1) > n(k)$, and $|a_{n(k)}| \ge k$.
Then you have  $| \frac{1}{a_{n(k)}} | \le \frac{1}{k}$, and so $\lim_{k \to \infty} \frac{1}{a_{n(k)}} = 0$.
Let $b_k = a_{n(k)}$.
A: I think you're on the right track. I'm going to assume that the sequence is unbounded above without loss of generality. What we'll do here is construct a sequence $\{b_n\}$ which is bounded below by $n$ (i.e. for every $n\in N$, $b_n > n$) which implies that $$0 < \frac{1}{b_n} < \frac{1}{n}$$ which is our desired result.
First, argue that for any $j >0$, the sequence $\{b^j_n\}$ defined by $b^j_n = a_{n+j}$ is unbounded if $a_n$ is unbounded. In other words, we can take out the first $j$ elements of our original sequence, and it will remain an unbounded sequence. To see this, assume it is bounded above. There is some $M_1$ so that $a_1,\dots,a_{j} < M_1$. Since by assumption, $b^j_n$ is bounded, there is some $M_2$ so that for all $i > j$, $a_i < M_2$. Let $m = \max(M_1,M_2)$. Then we have produced a bound for $a_n$, and we have a contradiction.
We'll define our desired sequence $\{b_n\}$ inductively. We need to go through a bit of a complicated song and dance to ensure that we are picking our subsequence in such a way that if $b_{n+1} = a_i$ and $b_{n} = a_j$ then $i>j$. In other words, we want to ensure that we are actually choosing a subsequence.
Let $a^1_n = a_n$ (you'll see why we've defined it this way in a moment). By definition of being unbounded, there is some $a^1_{j_1}$ so that $a^1_{j_1} > 1$. Let $b_1 = a^1_{j_1}$. Then define $\{a^2_n\}$ with $a^2_n = a^1_{n+j_1}$. In other words, $a^2_n$ will be the unbounded sequence which is obtained by starting $a^1_n$ from the element after $b_1$.
Now assume we have defined $b_i = a^i_{j_i} > i$. Then, define the next sequence where we can pick $b_{i+1}$ from by $a^{i+1}_n = a^{i}_{n+j_i}$. This sequence is still unbounded and we can argue then that there is some $j_{i+1}$ for which $a^{i+1}_{j_{i+1}} > i+1$ (sorry for the complicated notation). 
So let $\{b_n\}$ be the sequence so inductively defined. It is easy to see that it satisfies that $b_n > n$ for every $n\in N$. So then, we have that $$0 \leq \lim_{n\rightarrow \infty} \frac{1}{b_n} < \lim_{n\rightarrow \infty} \frac{1}{n} = 0$$ and we have proved the desired result. Let me know if you need any clarification on this method.
