Answer to Question #245998 in Real Analysis for Saqib

Question

State and prove Least Upper Bound (LUB) property of Real Numbers.

Answer 1

Statement;

Let S be a non empty set of real numbers.

A real number x is the least upper bound for S if x is an upper bound for S and x $\le$ y for every upper bound y of S.

Proof;

The least-upper-bound property can be proven using the assumption that every Cauchy sequence of real numbers converges. Let S be a nonempty set of real numbers. So if S has exactly one element, then its only element is a least upper bound. So consider S with more than one element, and suppose that S has an upper bound B₁. Since S is nonempty and has more than one element, there exists a real number A₁ that is not an upper bound for S. Define sequences A₁, A₂, A₃, ... and B₁, B₂, B₃, ... recursively as follows:

(i) Check whether $\frac{(A_n+B_n)}{2}$ is an upper bound for S

(ii) If it is ,let A_n+1=A_n and let B_n+1=$\frac{(A_n+B_n)}{2}$

(iii) Otherwise there must be an element s in S such that $s>\frac{(A_n+B_n)}{2}.$

Let A_n+1=s and B_n+1= B_n.

Then A₁ $\le$ A₂ $\le$ A₃ $\le$ ⋯ $\le$ B₃ $\le$ B₂ $\le$ B₁ and |A_n − B_n| → 0 as n → ∞.

It follows that both sequences are Cauchy and have the same limit L, which must be the least upper bound for S.