In a topological space a collection is a base for if and only if it consists of open sets and for each point and open neighborhood of there is a set such that .
Proof:
We need to show that a subset of is open if and only if it is a union of elements in . However, the if part is obvious, from the facts that the elements in are open, and that so are arbitrary unions of open sets. Thus, we only have to prove, that any open set indeed is such a union.
Let be any open set. Consider any element . By assumption, there is at least one element in , which both contains and is a subset of . By the axiom of choice, we may simultaneously for each choose such an element . The union of all of them indeed is . Thus, any open set can be formed as a union of sets within .
Show that the collection of all open intervals in is a base for a topology on .
Show that the collection of all closed intervals in is not a base for a topology on .
Show that the collection of half open intervals is a base for a topology on .
Show that the collection of half open intervals is a base for a topology on .
Let . A Partition over the closed interval is defined as the ordered n-tuple ; the norm of a partition is defined as For every , define the set . If is the set of all partitions on , prove that the collection of all is a Base over the Topology on .