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1.2.6e .pdf

Original filename: 1.2.6e.pdf

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Step-by-step solution

Step 1 of 3

Given a real number a, define

. By the definition of S, a is an upper

bound of S. We will show that a is the least upper bound of S by assuming that this is false and
deriving a contradiction.



Step 2 of 3

Let c be an upper bound of S such that
there is a rational number x in
Moreover, since

. Since the set of rational numbers is dense in

. Therefore, x is a rational number and


, so x is in S.

, c is not an upper bound of S as originally assumed.


Step 3 of 3
This contradiction implies that no number less than a is an upper bound of S. Thus,



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