[ Home ] [ Up ] [ Info ] [ Mail ]

Prove. The measure of the set R of all rational numbers in the interval [0, 1] is zero.

The proof of the theorem utilizes the following theorem.

Theorem 1.

Proof. The derivation of 1) is as follows: The progression

is a geometric progression a, ar, ar2, ar3 , .... where a = 1 and r = 1/2.. The sum of 2) over n terms

is given by the formula

The sum of the progression 3) for n = is then given by

Thus, from 3),

Proof of main theorem: The measure of the set R of all rational numbers in the interval [0, 1] is zero.

We have already shown that the set R is countable. We now arrange the points of R in a sequence

r1, r2, ... , rn, ... .

Next, for a given ε > 0, we enclose the point rn by an open interval Cn of length ε/2n. The sum C = Σ Cn is an open covering of set R. The open intervals Cn may intersect so the measure mC of C is given by

mC = m(Σ Cn) < Σ mCn = Σ (ε /2n) = ε

(where in the last equality we employ Theorem 1). Since ε can be chosen arbitrarily small, we have mR = 0, which is what we set out to prove.

[ Home ] [ Up ] [ Info ] [ Mail ]