Let be $m,n\in\mathbb{N}$, $p\in(0,1)$, $S_n$ is a binomially distributed random variable and $ \begin{align*} &A_{n,m}:=\left\{\omega\in\Omega\mid \left|\frac{S_n(\omega)}{n}-p\right|\geq \frac{1}{m}\right\}. \end{align*} $
Here, $\Omega=\{0,1\}^{\mathbb{N}}$ and we can assume that there exists a probability space $(\Omega,\mathcal{F},P)$ such that $A_{n,m}\in \mathcal{F}$.
Show that $ \begin{align*} \bigcup\limits_{n\mid n\geq N}A_{n,m}=\left\{\omega\in\Omega\mid\sup\limits_{n\mid n\geq N}\left|\frac{S_n(\omega)}{n}-p\right|\geq\frac{1}{m} \right\}. \end{align*} $
The relation $\subset$ is not a problem but I am not sure why $\supset$ holds.
I tried to prove it with a contraposition $(\lnot B\implies \lnot A)\iff (A\implies B)$ but it gets me nowhere:
We choose an $\omega\in\Omega$ with $\sup\limits_{n\mid n\geq N}\left|\frac{S_n(\omega)}{n}-p\right|\geq\frac{1}{m}$ and fix it. If we assume that there exists no $n\geq N$ such that $\left|\frac{S_n(\omega)}{n}-p\right|\geq \frac{1}{m}$, then there must be a $n\geq N$ with $\left|\frac{S_n(\omega)}{n}-p\right|< \frac{1}{m}$. But this doesn't lead to a contradiction...
Any suggestions? Am I missing someting?