Theorem
Let
be a metric space. Then
1.
and
- where
is the empty set - are open sets.
2. the intersection of any two open sets is an open set.
3. The union of any family of open sets is an open set.
Proof
1. If
and
then
hence
is an open set.
For each
and
hence
is an open set.
2. Suppose
and
are open subsets of
and
Since
and
are open, open balls
and
exist such that
and![]()
Set
to get
hence
is an open set.
3. Let
be a union of a family of open sets.
Suppose![]()
An open set
exists such that![]()
and since
is open,![]()