Solution: Given that $J_n = \left( 0, \frac{1}{n} \right) $. Let \[ J = \bigcap_{n=1}^{\infty} J_n. \] We need to show that the $J$ is empty. Suppose not, let $x \in J$. That means \begin{align*} \forall\ n \in \mathbb{N} \ \left( x \in J_n \right) & \implies \forall\ n \in \mathbb{N} \left( x \in \left( 0,\frac{1}{n} \right) \right) \\ & \implies \forall\ n \in \mathbb{N} \left( 0 < x < \frac{1}{n} \right). \end{align*} Thus for any $n \in \mathbb{N} $, we have $nx < 1$, which contradicts the Archimedean property. Therefore, $J = \emptyset $.

Now we need to show that \[ J\coloneqq \bigcup_{n=1}^{\infty} \left( 0, \frac{1}{n} \right) = (0,1). \] Let $x \in J$. This implies there exists $n \in \mathbb{N} $ such that $x \in \left( 0, \frac{1}{n} \right) \subseteq (0,1) $ and hence $x \in (0,1)$. Therefore, $J \subseteq (0,1)$. On the other hand, let $x \in (0,1)$, then $x \in J_1$ and hence $x \in J$. Thus $J = (0,1)$.