# Proof Verification:Show that sets are open or closed in $\mathbb{R}$

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.

• Proof looks fine. May 16 '18 at 5:53
• Thank you so much!
– Alea
May 16 '18 at 6:16
• You're very welcome! May 16 '18 at 6:42

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 $(a,\infty)$.
• Yes, since we have $-\infty$ to the left, we need not worry as any $\epsilon$ can not go beyond it. However, for $a$, we have to worry about it so no matter how close your $x$ is to $a$, it'll never become $a$ and hence, you can always take $\epsilon$ to be half the distance (or even a third and so on). Is my reasoning ok? Also, many thanks!