Suppose a setis given. To promote it to a topological space
we must describe a family of subsets of
which satisfy certain conditions. Of the many ways which define such a family of subsets, one is using a function
which sends every subset
of
to its closure
The closureof
must satisfy these conditions
With the closure defined as above, we can write down the theorem:
Letrepresent a set and let
represent a function such that
1.
2. for each
3.for
4.for each
Then the familyis a topology on
where
is the discrete topology on
and
for each
Proof
Ifthen
(1)
In fact ifthen
and
Ifthen since
we have
hence
Also since
Ifthen
and any finite intersection is in
Letthen
for each
By (1) we obtain
for each
Henceand
with
The union of any subset ofmust be in
By using
as the topology on
we show that
In fact
because each
is closed and
so
Similarly
Sinceis open in
for some
Sincewe obtain