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