1
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ Proof looks fine. $\endgroup$ May 16 '18 at 5:53
  • $\begingroup$ Thank you so much! $\endgroup$
    – Alea
    May 16 '18 at 6:16
  • $\begingroup$ You're very welcome! $\endgroup$ May 16 '18 at 6:42
1
$\begingroup$

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 $(a,\infty)$.

$\endgroup$
3
  • 2
    $\begingroup$ 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! $\endgroup$
    – Alea
    May 16 '18 at 6:05
  • 2
    $\begingroup$ Yes, your reasoning is ok $\endgroup$ May 16 '18 at 6:06
  • 2
    $\begingroup$ Thank you so much! $\endgroup$
    – Alea
    May 16 '18 at 6:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.