Theorem
Let
represent a set and
a family of subsets of
such that
1.![]()
2. The union of any elements of
is a member of![]()
3. The intersection of any elements of
is a member of![]()
Then if
is a family of subsets of
such that
if and only if
then
is a topology on![]()
Proof
Since
and![]()
![]()
![]()
Suppose
then![]()
Also
so![]()
This can be extended to any intersection of sets.
Suppose
is a family of sets of![]()
is a family of subsets of
and![]()
Hence
is a topology on
Elements of
are called open sets.