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 $w\le b$ for any $b\in B$. Multiply through by $-1$. This reverses the inequality, and we conclude that $-w\ge -b$ for any $b\in B$. The numbers $-b$ are precisely the elements of $A$, so $-w\ge a$ for any $a\in A$.
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 $b\in B$ such that $b\lt -m$. But then $-b\gt m$. Since $-b\in A$, this contradicts the fact that $m$ is an upper bound of $A$.
To prove (ii), one uses the same strategy.
Let's review some definitions:
A set S of real numbers is called bounded from above if there is a real number k such that k ≥ s for all s in S. The number k is called an upper bound of S. The terms bounded from below and lower bound are similarly defined.
The supremum of a subset S of a totally or partially ordered set T is the least element of T that is greater than or equal to all elements of S. Consequently, the supremum is also referred to as the least upper bound. The infimum or greatest lower bound is similarly defined.
completeness of the real numbers asserts every nonempty subset of the set of real numbers that is bounded above has a supremum that is also a real number.
In our given problems, the set T in the supremum definition is ℝ. To be bounded above, each set S, do we have a number k ≥ for all s in S? Sure. It's 10. So it is bounded above and by completeness, it has a supremum that is also a real number. So your intuition about having the greatest upper bound property is correct. Now what is the least element of ℝ that is greater than or equal to all elements of S?
Similarly, what is the greatest element of ℝ that is less than or equal to all the elements of S?
Best Answer
$X \subset \mathbb{R}$. It is a property of $\mathbb{R}$ that the Least Upper Bound Axiom holds. The least upper bound axiom states that any nonempty, bounded subset of $\mathbb{R}$ has a least upper bound.
So $X$ has a least upper bound. Now we'll do the greatest lower bound: Consider the set $-X = \{-x | x \in X\}$. Again, by the LUB property, it has a least upper bound $L$. Then $-L$ is the greatest lower bound for $X$.
Another way to prove the existence of the greatest lower bound of $X$ is to let $B$ be the set of all lower bounds for $X$. So $B \subset \mathbb{R}$. $B$ is bounded above, so its supremum will be the greatest lower bound for $X$.