p.42 5c - The absolute values are zero at 1 and -1. If x<-1, |x-1|=1-x and |x+1|=-x-1; since 1-x > -x-1, true for x<-1 If -1 *LT* x *LT* 1, |x-1|= -x+1, |x+1|=x+1 ; since x+1 *LT* -x+1 for x *LT* 0, true for -1 *LT* x *LT* 0 If x>1, |x-1|=x-1, |x+1|=x+1; since x-1 < x+1, false for x>1. True for x=-1, false for x=1 Therefore true for x<0 11 - Let epsilon = |a-b|/3. If c is in both the epsilon neighborhood of a and b, The triangle inequality |a-b| *leq* |a-c| + |c-b| is violated since the rhs is 2|a-b|/3. Hence a c in both neighborhoods with that epsilon cannot exist. 46 - 1 0 *leq* x for all x in S1, so 0 is a lower bound. by previous result a/2 < a for a>0, there is no lower bound >0 since multiplication by .5 leaves a number positive. Hence 0 is the infinum. Or noting 0 is a lower bound and also in S1 makes it the infinum There is the tacit assumption that any upper bound is in R, adding 1 to the bound produces a real number which is not less than the bound, so the existence of a bound is contradicted. (0<1 => u+0 < u+1 (where u is the upper bound). 6 - Let u* be an upper bound for S and u* in S. If v* is another upper bound, then v* is not less than u*, because u* is in S, hence less than or equal to every upper bound. Hence there is no upper bound less than u*, and u* is the lub by definition. 51 4a - Let u = inf S and v = inf {aS}. Let x in S; x>u => ax>au, so v *geq* au (i.e., au is a lower bound). Similarly, let y in {aS} y>v => y/a > v/a; so u *geq* v/a (v/a is a lower bound for S). the two inequalities taken together provide equality. (the sup can be done similarly) 8 - a) With x* fixed, the function becomes 2x*+y, for which the supremum is 2x*+1; the infinum of this function is not obtained, but is given when x=0 the infinum is 1 b) the infinum is 2 times 0 + y, the supremum of that is 1 The results are the same. 13 - I would prove by induction that 1/(2^n) < 1/n, and cite 2.5.3b

Sat Sep 6 1997