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Let A ={1-1/n | n is a natural number} Prove that 0 is a lower bound...

Let A ={1-1/n | n is a natural number}

Prove that 0 is a lower bound and 1 is an upper bound:  Start by taking x in A.  Then x = 1-1/n for some natural number n.  Starting from the fact that 0 < 1/n < 1 do some algebra and arithmetic to get to 0 < 1-1/n <1.


Prove that lub(A) = 1:  Suppose that r is another upper bound.  Then wts that r<= 1.  Suppose not.  Then r<1.  So 1-r>0. Use Archimedes to show that there is a natural number n such that 1/n< 1-r.  Use this to see that r<1-1/n.  Thus...


Prove that glb(A) = 0: Suppose that s is another lower bound.  Then wts that 0 <=s.  This is easy to show because 0 is an element of A.


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