Show that the intervals $(a, \infty)$ and $(-\infty, a)$ are are open sets, and intervals $[b, \infty)$ and $(-\infty,b]$ are closed sets
Proof:
consider $(a, \infty)$
By density theorem, $\forall x \in (a, \infty)$, $\exists$ $r_1, r_2$ such that:
$a<r_1<x<r_2<\infty$
$\implies x \in (r_1,r_2) \subset (a, \infty)$ Hence, $(a,\infty)$ is open.
The proof for $(-\infty, a)$:
By density theorem, $\forall x \in (-\infty, a)$, $\exists$ $r_1, r_2$ such that:
$-\infty<r_1<x<r_2<a$
$\implies x \in (r_1,r_2) \subset (-\infty, a)$ Hence, $(-\infty,a)$ is open.
For $[b, \infty)$, the complement of this interval $(-\infty, b)$ is open (as above). Hence, $[b,\infty)$ is closed.
Same reasoning for $(-\infty, b]$
Can anyone please check this and let me if I've done the proof the correct way and if there are any errors or not?
Thank you.
Your proof looks good for me. Like a suggestion, you can say who is explicitly $r_1$ and $r_2$. For example, if $x\in (-\infty,a)$ then if we take $\varepsilon=\frac{|x-a|}{2}$ then $(x-\varepsilon,x+\varepsilon)\subseteq (-\infty,a)$ (why?). The same argument for $(a,\infty)$.