A,bp and a<=b prove that a is unique greatest lower bound
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❤️❤️We assume that we know the Least Upper Bound Principle: Every non-empty set A which is bounded above has a least upper bound.
We want to show that every non-empty set B which is bounded below has a greatest lower bound.
Let A=−B. So A is the set of all numbers −b, where b ranges over B.
First we show that if B has a lower bound, then A has an upper bound. For let w be a lower bound for B. We show that −w is an upper bound for A.
Because w is a lower bound for B, we have w≤b for any b∈B. Multiply through by −1. This reverses the inequality, and we conclude that −w≥−b for any b∈B. The numbers −b are precisely the elements of A, so −w≥a for any a∈A.
Since A is bounded above, it has a least upper bound m. We show that −m is a greatest lower bound of B.
As usual, the proof consists of two parts (i) −m is a lower bound for B and (ii) nothing bigger than −m is a lower bound for B.
The proofs again use the fact that multiplying by −1 reverses inequalities. To prove (i), suppose that −m is not a lower bound for B. Then there is a b∈B such that b<−m. But then −b>m. Since −b∈A, this contradicts the fact that m is an upper bound of A.
To prove (ii), one uses the same strategy.❤️❤️
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