Proof that unbounded sequence has a subsequence whose reciprocal converges to zero

Question

I am trying to solve Exercise 6.6.3. in Tao's analysis textbook.

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 $(a_n)_{n=0}^{\infty}$ such that $\lim\limits_{n \to \infty} \frac{1}{b_n}$ exists and is equal to zero. (Hint: for each natural number $j$, recursively introduce the quantity $n_j := \min \{n \in \mathbb{N}: |a_n| \geq j; n > n_{j-1} \}$ (omitting the condition when $n > n_{j-1}$ when $j = 0$), first explaining why the set $\{n \in \mathbb{N}: |a_n| \geq j; n > n_{j-1} \}$ is non-empty. Then set $b_j := a_{n_j}$.

I am having difficulty using the hint. As a first note, the set $\{n \in \mathbb{N}: |a_n| \geq j; n > n_{j-1} \}$ is certainly nonempty because $(a_n)$ is unbounded, so if we can find a term that is larger than any given natural number. The 'minimum' operation is surely defined by the Well-Ordering principle, and the outcome is unique, and thus the sequence well-defined. By iterating this process, it seems that we will construct a sequence, $(b_j)_{j=0}^{\infty}$ such that $b_j \geq j$ for all $j \in \mathbb{N}$, which is also unbounded and diverges to $+\infty$. The reciprocal of something converging to $+\infty$ is surely $0$.

I am unsure on whether this argument is formal enough or contains any holes.

Any help would be greatly appreciated.

Answer 1

You forgot $| \cdot|$ !

Since $b_j = a_{n_j}$ we have $|b_j|=|a_{n_j}| \ge j$ for all $j$, hence

$$ \frac{1}{|b_j|} \le \frac{1}{j}$$

for all $j$. This gives $\lim\limits_{j \to \infty} \frac{1}{b_j}=0.$

Answer 2

Since $a_n$ is unbounded there is an index $j_1$ such that $|a_{j_1}|>1$. Since $\{a_1,...a_{j_1}\}$ is bounded $\{a_{j_1+1},...\}$ is unbounded so there is an index $j_2$ in $\{a_{j_1+1},...\}$ such that $|a_{j_2}|>2$.

By continuing like this we can define sequence $0<b_k={\dfrac {1}{|a_{j_k}|}<\dfrac{1}{k}}$.

Squeeze and obtain the result.