Proof
Let
be a set and let
be a family of subsets of
so that
the discrete topology. There is a unique smallest topology
such that
given by![]()
is said to be generated by A.
Proof

Let
and let
be the intersection of all topologies
containing![]()
is a topology since
1.
for each
so ![]()
2. If
for each
then
for each
so ![]()
2. If
for each
then
for each
so![]()
Since
is the intersection of all the topologies containing
it must be the smallest topology containing
If there is any other smallest topology
then
is a smaller topology than either
or
containing A - a contradiction - so
is unique.
Also since
and
is a topology, it must contain all intersections and unions of elements of![]()