Given topological spaces
and
,
a basis for the product topology on
is given by all sets of the form
,
where
,
and
are open sets.
Note that the sets
form a basis since
so
the sets are in fact closed under finite intersections.
Given topological spaces
and
,
, let
and
be
the projection functions (i.e.
) The product topology on
can
also be described as the topology generated by all sets of the form
,
or
where
,
and
are open sets. By generated we mean "take arbitrary unions of finite
intersections of such sets."
Note that:
and
are
continuous with respect to this topology, essentially by definition.
is continuous if and only if
and
are
continuous.
Due: Thursday 20
Given a topological space
,
a point set
,
and a function
,
show that the collection of sets
,
any
open set
forms a topology on
such that
is
continuous. (see the definition of a Quotient Space)
Given two topological spaces
and
,
a point set
,
and two function
and
, show that the collection of sets
,
any
open
sets
are a basis for a topology on
such that
is
continuous
.
Let
then
is
continuous, where
.
Similarly for
We need only show that the inverse image of basis elements are open. But
or
depending on whether or not
If
and
be
two compact topological spaces then
is
compact.
In general, since any open set is the union of basis elements it suffices to show that any open covering by basis elements admits a finite subcovering. (WHY?)
Let
be an open covering of
Since
is compact and
is
continuous, for each
,
we can select a finite subset
such that
covers
.
Let
.
Note:
Next, since
is compact, there is a finite subset
of
the
that covers
.
Thus
covers
,
as does
since
for
each
.