Theorem. (Nested Interval Property). For each n , assume we are given a closed interval I_n = [a_n, b_n], where I_n = { x : a_n < x < b_n }. Assume also that I_n contains I_n+1. Then by the Axiom of Completeness, the resulting sequence of closed intervals

I₁ I₂ I₃ ... has a nonempty intersection.

I_n .

Proof. Let A = { a_n : n}. Notice that for any n, b_n is an upper bound for the set A. Let x = sup A. We know that such an x exists because for all n, x > a_n and x < b_n. Now consider any I_n. xI_n. Therefore,

I_n .

Theorem. (Density of in ). Between any x, y a rational number a can be found such that x < a < y. Likewise, an irrational number b can be found such that x < b < y.

Proof. Follows through the use of the Archimedian Property.

Interestingly enough, since both a, b , we can find more rational and irrational numbers between them. Continuing this, we can find yet more rational and irrational numbers between those. Therefore, between any two real numbers, there exists an uncountable number of real numbers. This will be discussed more in the Topology of .