Exercise 3.  Suppose that  .   Show that  .   

Solution 3.

See text and/or instructor's solution manual.

Solution.  Method I.    

Let    be given.   Since  ,  there exists    such that if    then  ,    i.e.,   .   

But since  ,  this implies that if     , then  .   Therefore  .   

Solution.  Method II.    

Let   ,   with   ,    and   .  

By Theorem 4.1, both      and      exist, and  

                    ,   and

                    .  

Thus,      exists,   and   .   

Therefore,   

                    

This solution is complements of the authors.

(c) 2008 John H. Mathews, Russell W. Howell