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