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 .