Let
be the set of all real numbers. A set
is open if for each
there exists
such that![]()
is a metric space where
Let the family of all open intervals in
be called
then
is a topology on![]()
Proof
Obviously
and
are open in
so
and![]()
Suppose
are open sets, then![]()
In fact
such that
where![]()
Suppose each
is open then
is also open (
may be infinite here), since![]()