An open set with end point real as a union of open sets with rational end point.

Question

Suppose that we have the open interval of $\overline{\mathbb{R}}$ $(a,+\infty]$ and suppose that we want to write this as a union of open intervals of the same type but with rational endpoint.

My attempt. For the density of $\mathbb{Q}$ in $\overline{\mathbb{R}}$, exists an decreasing sequence of rational $\{q_n\}$ such that $q_n\to a$, then $$(a,+\infty]=\bigcup_{n=1}^{+\infty}(q_n,+\infty].$$

Here's my test:

if $x\in\bigcup_{n=1}^{+\infty} (q_n,+\infty]$, then $x\in (q_n,+\infty]$ for same $n\in\mathbb{N}$, since $q_n\ge a$ we have that $(q_n,+\infty]\subseteq (a,+\infty].$

Edit for the vice versa

Let $x\in (a,+\infty]$, since the sequence $\{q_n\}$ is decreasing exists $n_0\in\mathbb{N}$ such that $a<q_n<x$ for all $n>n_0$, therefore $x\in (q_n,+\infty]$ for all $n>n_0$.

Question. It's correct?

Thanks!

Answer 1

(a,oo] = $\cup${ (r,oo] : a < r, r in Q }.