How would you rigorously show
$\displaystyle \cap_{n=1}^{\infty}[a-\frac{1}{n},\infty)= (a,\infty) $ for each $\displaystyle a\in\mathbb{R}$
Clearly $\displaystyle [a,\infty)\subseteq \left[a-\tfrac{1}{n},\infty\right),\text{ }\forall n\in\mathbb{N}$. Now, suppose that $\displaystyle b<a\in\bigcap_{n=1}^{\infty}\left[a-\tfrac{1}{n},\infty\right)$. We clearly have that $\displaystyle b<a\implies a-b>0$ and thus by the Archimedean principle there exists some $\displaystyle m\in\mathbb{N}$ such that $\displaystyle \frac{1}{m}<a-b\implies b<a-\frac{1}{m}\implies b\notin\left[a-\tfrac{1}{m},\infty\right)$. It follows that $\displaystyle b\notin\bigcap_{n=1}^{\infty}\left[a-\tfrac{1}{n},\infty\right)$