AXIOM 11.1.1 THE LEAST UPPER BOUND AXIOMEvery nonempty set of real numbers that has an upper bound has a least upper bound.Some find this axiom obvious; some find it unintelligible. For those of you who find it obvious, note that the axiom is not satisfied by the rational number system; namely, it is not true that every nonempty set of rational numbers that has a rational upper bound has a least rational upper bound. (For a detailed illustration of this, we refer you to Exercise 33.) Those who find the axiom unintelligible will come to understand it by working with it. We indicate the least upper bound ofa set S by writing lub S. As you will see fromthe examples below, the least upper bound idea has wide applicability. (1) lub (??, 0) = 0,(2) lub (?4, ?1) =?1,lub(??, 0] = 0. lub(?4, ?1] =?1.(3) lub {1/2, 2/3, 3/4, , n/(n + 1), } = 1. (4) lub {?1/2,?1/8, ?1/27, , ?13, } = 0. (5) lub {x : x2 0. Since M is an upper bound for S, the condition s ? M is satisfied by all numbers s in S. All we have to show therefore is that there is some number s in S such thatM?

Question 2 In this question you will need to read and understand the following definition and find the error in the proof of the proposition following it. For bonus marks explain how to correct the error. Definition: A function f defined on the interval [a, b] will be said to be locally constant if for every r E [a, b] there is some o > 0 and some c, both depending on r, such that . if x = a then f (=) = c for all z E [a, a + 5) . if x = b then f(=) = c for all z E (b - 6, b] . if a 0 and some c such that . f (=) = c for all z ( [a, a + 6) in the case that w = a . f(=) = c for all z E (w - 6, w + 6) in the case that a a + 6 > a because w is an upper bound for S. This contradicts that a = w. To get a contradiction in the second case, first note that since w is the least upper bound for S and 6 > 0 it follows from Theorem 11.1.2 that there is some s E S such that w-6