Show that the set of rational numbers is $\textbf{dense}$ in $\mathbb{R}$, meaning that every real number is a limit of rational numbers. Wrote a proof, wondering if it's correct/rigorous enough and if not, where it goes wrong.
So we seek to show that given a $x \in \mathbb{R}$, $\exists$ a sequence of rational numbers $a_n$, $lim_{n\rightarrow \infty} a_n = x$.
By the definition of convergence, we know this means that $\forall \epsilon > 0, \exists N \in \mathbb{N}\ \forall n \in \mathbb{N}, n \geq N \implies |a_n - x| < \epsilon$.
Which we can rewrite as $x - \epsilon < a_n < x + \epsilon$.
By the Archimedean property, we know in $\mathbb{R}, \forall\ x < y \in \mathbb{R}, \exists\ r \in \mathbb{Q}$ such that $x < r < y$. Since $x - \epsilon \in \mathbb{R}$ and $x + \epsilon \in \mathbb{R}$, we know there exists a $r \in \mathbb{Q}, x - \epsilon < r < x + \epsilon$. So we know $\exists$ such an $a_n$.