Prove the completeness axiom directly from the Nested Interval Theorem
Prove the completeness axiom directly from the Nested Interval Theorem
The Completeness Axiom, also known as the Least Upper Bound Property, states that every non-empty subset of real numbers that is bounded from above has a least upper bound (supremum) in the set of real numbers. To prove the Completeness Axiom directly from the Nested Interval Theorem, we need to show that any non-empty set of real numbers that is bounded from above has a least upper bound.
The Nested Interval Theorem, on the other hand, asserts that if we have a sequence of nested closed intervals in the real numbers, i.e., intervals [a₁, b₁] ⊇ [a₂, b₂] ⊇ [a₃, b₃] ⊇ ... such that the length of each interval approaches zero, lim (n → ∞) (bₙ - aₙ) = 0, then there exists a unique real number x that belongs to all these intervals.
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