Show that the class of closed intervals [a, b], where a is rational and b is irrational and a < b, is a base for a topology on the real line r.

See proof below.

Step-by-step explanation:

Remember that a collection of subsets B of a set X is a base for a topology of X if the following conditions hold

The union of B is equal to XIf T,S∈B and p∈T∩S then there exists some Q∈B such that x∈Q and Q⊆T∩S.

In this case, X=R and B={[a,b]: a is rational and b is irrational}. Let's prove the statements above

The inclusion ∪B⊆R is true for any collection of subsets of the real line. To prove that R⊆∪B., let x∈R. If x is rational choose an irrational number y>x. (this can be done because irrational numbers are not bounded above) Then x∈[x,y] and [x,y]∈B, thus x∈∪B. If x is irrational, choose some rational number z<x (it can be done because rationals are not bounded below). Then x∈[z,x] for some [z,x]∈B, thus by definition of union, x∈∪B. In any case, x∈∪B therefore R=∪B.Let T=[a,b], S=[c,d]∈B be arbitrary elements of B. Suppose that p∈T∩S. Define Q=T∩S. Q is a closed interval, its starting point is either a or c (the greatest of these), which are rational, and its endpoint is either b or d (the smallest of these), which are irrational. In any case Q∈B and we have that p∈Q⊆T∩S.

This is the graph x=1