Suppose a set
is 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 closure
of
must satisfy these conditions
![]()
![]()
![]()
![]()
With the closure defined as above, we can write down the theorem:
Let
represent a set and let
represent a function such that
1.![]()
2. for each![]()
3.
for![]()
4.
for each![]()
Then the family
is a topology on
where
is the discrete topology on
and
for each![]()
Proof
If
then
(1)
In fact if
then
and![]()
If
then since
we have
hence
Also since![]()
If
then
and any finite intersection is in![]()
Let
then
for each
By (1) we obtain
for each![]()
Hence
and
with![]()
The union of any subset of
must be in
By using
as the topology on
we show that
In fact
because each
is closed and
so
Similarly![]()
Since
is open in
for some![]()
Since
we obtain![]()