Proving T is a topology On R

[Poirot]

Let T consist of the empty set, R and all sets (-a,a) where a>0.

The first axiom is immediate. The second, showing the intersection Of 2 opens sets is open is also easy for me. Now I need to show infinite unions of open sets are open and I am unsure of the best way to proceed.

 

[Opalg]

Let T consist of the empty set, R and all sets (-a,a) where a>0.

The first axiom is immediate. The second, showing the intersection Of 2 opens sets is open is also easy for me. Now I need to show infinite unions of open sets are open and I am unsure of the best way to proceed.

Let $\{(-a_\iota,a_\iota)\}_{\iota\in I}$ be a collection of open sets. If $\{a_\iota\}_{\iota\in I}$ is unbounded then $\bigcup_{\iota\in I}(-a_\iota,a_\iota) = \mathbb{R}.$ Otherwise, let $a = \sup_{\iota\in I}a_\iota$ and show that $\bigcup_{\iota\in I}(-a_\iota,a_\iota) = (-a,a).$

 

[Poirot]

Have you accounted for the case where the empty set or R is included?

 

[Opalg]

Have you accounted for the case where the empty set or R is included?

No, I left that as an exercise for you.

 

[Poirot]

Well, if one of the a(i) is R, then the union is R. If a(i) is empty for all i, the union is empty and finally if a(i) is empty for at least one i then we can remove such sets with the union unchanged

 

[Opalg]

Well, if one of the a(i) is R, then the union is R. If a(i) is empty for all i, the union is empty and finally if a(i) is empty for at least one i then we can remove such sets with the union unchanged

Correct.