Proof
Let
be a function from a set
onto a topological space![]()
The family of subsets of
defined by
is a topology on![]()
Proof
is a topology on
so![]()
Since
is onto,![]()
Also![]()
Hence![]()
Let
represent a family of subsets in
For each![]()
We have![]()
Since
is a topology on![]()
hence![]()
and![]()
Since
is a topology on
and![]()
Hence
and
is a topology on![]()