Final answer:
The class of closed intervals [a, b], where a and b are rational numbers, does not satisfy the necessary conditions for a base of a topology on the real line. Closed intervals don't produce open sets upon union and cannot represent arbitrarily small neighborhoods around points, both of which are required for a base.
Step-by-step explanation:
In topology, a basis (or base) for a topology on a set X is a collection of open sets such that every open set in the topology can be written as a union of sets from this collection. To prove that the class of closed intervals [a, b], where a and b are rational numbers and a < b, is not a base for a topology on the real line, we must use the definition of a base and the properties of the real line.
Firstly, the union of any collection of closed intervals [a, b] is also a closed set, not an open set. This is because if you take any two points in the union, the closed intervals contain all the points between them, and so does the union. A base for a topology, however, must consist of open sets, since open sets are the building blocks of a topology.
Secondly, one requirement for a collection of sets to be a base is that for every point x in the space and for every open set O containing x, there must be a base set B containing x such that B is a subset of O. However, since our intervals are closed, any single point x would be an element of [x, x], and yet this set cannot be contained in any open set around x as it includes the endpoints which are not interior points.
Therefore, because the class of closed intervals cannot satisfy these necessary conditions to create the open sets of a topology through unions, it fails to be a base for a topology on the real line.