By Charles Hayes

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We take an arbitrary non-empty set S of real numbers. We say that S is bounded above if and only if there exists a real number y such that x y whenever x e S. Any number y satisfying this condition is called an upper bound of S. r e S. Any number z satisfying this condition is called a lower bound of S. In case S is bounded both above and below, we say simply that S is bounded. Now we let T= is an upper bound of S}, V = {z z is a lower bound of S}. — Now T 0 if and only ifS is bounded above; and V 0 if and only if Sis bounded below.

Clearly z is rational and 0 < z < t. Also z2 = (12/4) + I + (1/12). By differential calculus, one can show easily that 2 z2. The equality is ruled out since z is rational, hence 2 < z2, and so z e T. This shows that S has no rational least upper bound. is Let us now modify the definitions of Sand T by deleting the requirement that their members must be rational; we let S' and T', respectively, be the corresponding sets. Now, by dropping the references to rationals wherever these occur, the proofs already given can be used to show that (1) S' has no largest member and consequently no member of S' is an upper bound of 5'.

We are using the abbreviation x' z x. Similarly, we shall use the exponential notation to denote positive integral powers of mathematical expressions without further explanation in the future. Concepts of Real Analysis 40 (xiv) If 0