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About LinkBacksFind and Prove the inf and sup:
{x is a rational number: x^2 < 2 }
If you were taking x in the real numbers, then your supremum would beand your infimum would be
.
However, these are both irrational numbers. There does not exist a supremum (least upper bound) in the rational numbers. Say it did exist, and call it a. Then. Define
. Then b is also an upper bound of your set, and
, which contradicts our assumption that a was our LEAST upper bound.
A similar argument works for the infimum.
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