Union of Closures of Singleton Rationals is Rational Space

Theorem
Let $B_\alpha$ be the singleton containing the rational number $\alpha$.

Then:
 * $\displaystyle \bigcup_{\alpha \mathop \in \Q} B_\alpha^- = \Q$

where $B_\alpha^-$ denotes the closure of $B_\alpha$ in $\R$.

Proof
Let $\alpha \in \Q$.

The open intervals:
 * $\left({-\infty \,.\,.\, \alpha}\right)$ and $\left({\alpha \,.\,.\, +\infty}\right)$

are open in $\R$.

Thus $S := \left({-\infty \,.\,.\, \alpha}\right) \cup \left({\alpha \,.\,.\, +\infty}\right)$ is open in $\R$.

But:
 * $\R \setminus S = \left\{{\alpha}\right\}$

Thus by definition $B_\alpha = \left\{{\alpha}\right\}$ is closed in $\R$.

From Closed Set Equals its Closure, it follows that:
 * $B_\alpha = B_\alpha^-$

Hence the result.