Welcome to our community

Be a part of something great, join today!

Proving T is a topology On R

  • Thread starter
  • Banned
  • #1

Poirot

Banned
Feb 15, 2012
250
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
Feb 7, 2012
2,701
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 $\bigcup_{\iota\in I}(-a_\iota,a_\iota) = (-a,a).$
 
  • Thread starter
  • Banned
  • #3

Poirot

Banned
Feb 15, 2012
250
Have you accounted for the case where the empty set or R is included?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,701
  • Thread starter
  • Banned
  • #5

Poirot

Banned
Feb 15, 2012
250
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
Feb 7, 2012
2,701
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. (Smile)