# Proving T is a topology On R

#### Poirot

##### Banned
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

##### MHB Oldtimer
Staff member
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

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

#### Opalg

##### MHB Oldtimer
Staff member
Have you accounted for the case where the empty set or R is included?
No, I left that as an exercise for you. #### Poirot

##### Banned
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

##### MHB Oldtimer
Staff member
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. 