Topology, complements of finite sets

• Sep 3rd 2010, 06:59 PM
Chris11
Topology, complements of finite sets
Let R be the set of real numbers, then let $\displaystyle \omega$ be the set of all complements of finite sets and the empty set. Prove that (R, $\displaystyle \omega$) is a topological structure.

I was able to prove that it's closed under finite intersections, and I was able to show that the full set is in $\displaystyle \omega$, but I'm having difficulty proving that it's closed under arbitrary unions... Any help would be appreciated. This is not for homework of anykind.
• Sep 4th 2010, 04:46 AM
Plato
Quote:

Originally Posted by Chris11
Let R be the set of real numbers, then let $\displaystyle \omega$ be the set of all complements of finite sets and the empty set. Prove that (R, $\displaystyle \omega$) is a topological structure.

The complement of a union is the intersection of complements: $\displaystyle \left( {\bigcup\limits_\alpha {B_\alpha } } \right)^\prime = \bigcap\limits_\alpha {\left( {B_\alpha } \right)^\prime }$

Surely the any intersection of finite sets is finite.
• Sep 4th 2010, 08:32 AM
Iondor
If M is co-finite and $\displaystyle M\subseteq N$, then N is co-finite too.
Therefore the union of a class of co-finite sets must also be co-finite.