Let $A$ be a set of real numbers.
Prove that a function $f: A \rightarrow \mathbb{R}$ is not bounded by any number if and only if there is a sequence $x_n \in A$ so that $|f(x_n)|> n, \forall n$
$\rightarrow$ Since $f(x)$ is not bounded by any number then we can't have $1 \geq |f(x)|$ for all $x \in A$. Thus there exists an $x_1 \in A$ where $|f(x_1)|>1$. Hence there is a $x_n \in A$ where $|f(x_n)|>n ,\forall n$
$\leftarrow$ Now if there is a sequence $x_n \in A$ such that $|f(x_n)| > n, \forall n$, Then by definition the function is not bounded.
It seems pretty straight forward just wanted to see if this was an ok approach or to see any other cleaner approaches.
$f $ bounded at $A \iff $
$$\exists M\ge 0 \;\;:\; \forall x\in A \;\;|f (x)|\le M $$
thus the negation is
$f $ not bounded at $A \iff $ $$\forall r\ge 0 \;\; \exists x\in A \;:\; |f (x)|>r $$
$x $ depends on $r $. we should write $x_r $.
this is true if we replace $r $ by $n $
so $\exists x_n\in A \;:\: |f (x_n)|>n $.