# 8.5: The Least Upper Bound Property

• • Bob Dumas and John E. McCarthy
• University of Washington and Washington University in St. Louis
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DEFINITION. Upper bound Let $$X \subset \mathcal{D}$$. We say that $$X$$ is bounded above if there is $$M \in \mathcal{D}$$ such that $(\forall x \in X) x \leq M .$ In this event we say that $$M$$ is an upper bound for $$X$$.

DEFINITION. Least upper bound Let $$X \subset \mathcal{D}$$ be bounded above. Suppose $$M$$ is an upper bound for $$X$$ such that for any upper bound $$N$$ for $$X, M \leq N$$. Then the number $$M$$ is called the least upper bound for $$X$$.

Lower bound and greatest lower bound are defined analogously.

THEOREM 8.3. Least Upper Bound Property If $$X$$ is a non-empty subset of $$\mathcal{D}$$ and is bounded above, then $$X$$ has a least upper bound. If it is bounded below, then it has a greatest lower bound.

PROOF. Let $$X \subset \mathcal{D}$$ be bounded above. Let $M=\bigcup_{L \in X} L \subseteq \mathbb{Q} .$ The set $$M$$ is bounded above (why?), and hence $$M \neq \mathbb{Q}$$. Any element of $$M$$ is an element of some $$L \in X$$, and consequently cannot be a maximal element of $$L$$. Therefore $$M$$ has no largest element. If $$a \in M$$, $$c \in \mathbb{Q}$$ and $$c<a$$ then $$c \in M$$. Therefore $$M$$ is a Dedekind cut. For any $$L \in X, L \subseteq M$$ and hence $L \leq M .$ That is, $$M$$ is an upper bound for $$X$$.

Let $$K<M$$. Then there is $$a \in M \backslash K$$. So $$a$$ is in some $$L_{0}$$ in $$X$$. Therefore $$L_{0}$$ is not contained in $$K$$ and $$K$$ is not an upper bound for $$X$$. It follows that $$M$$ is the least upper bound for $$X$$.

We leave the argument for the existence of a greatest lower bound to the reader.

The least upper bound property is the essential property of real numbers that permits the main theorems of calculus. It is the reason we use this large set, rather than, say, the algebraic numbers. It uniquely characterizes the real numbers as an extension of the rational numbers - see Theorem $$8.23$$ for a precise statement.

Now that we have proved this key property, we shall use $$\mathbb{R}$$ to denote the set of real numbers, identifying a real number $$\alpha$$ with the Dedekind cut $$(-\infty, \alpha) \cap \mathbb{Q}$$. We shall no longer need to concern ourselves with Dedekind cuts per se.

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