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I have seen two "different" statements both of which are referred to as the Axiom of Completeness. They are equivalent, but I can not seem to see why they are equivalent. The statements are:

Axiom of Completeness 1: Suppose $L, H \subset \mathbb{R}$ where both $L$ and $H$ are non empty and satisfy $\forall l \in L, h \in H, l \le h$. Then, $\exists \alpha \in \mathbb{R}$ such that $\forall l \in L, h \in H, l \le \alpha \le h$.

Axiom of Completeness 2: Every non empty set of real numbers that is bounded above has a least upper bound.

From my introductory studies of real analysis, I have been mostly exposed to statement 2. Can someone explain why statement 1 is equivalent to 2?

user56031
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  • shouldn't version 1 not include that $L \cup H$ is all numbers ? – Henno Brandsma Jul 22 '17 at 09:37
  • It is from pg 31 de la Fuente "Mathematical Models" and does not make mention of $L \cup H$. – user56031 Jul 22 '17 at 09:41
  • I don't really use it anyway. But it's clearly inspired on Dedekind cuts, where this does hold. – Henno Brandsma Jul 22 '17 at 09:44
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    Statement 1 is Dedekind's theorem which is based on the Dedekind's construction of real numbers. And it is generally given with the additional constraint $L\cup H=\mathbb{R} $. This constraint ensures that the $\alpha$ in conclusion is unique. In your version there can be many numbers playing the role of $\alpha$. – Paramanand Singh Jul 22 '17 at 15:13

1 Answers1

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Suppose version 1 holds.

Let $A$ be a set of reals with an upperbound $u$. Define

$$L = \{x \in \mathbb{R}: \exists a \in A: x < a\}$$ and

$$H = \{x \in \mathbb{R}: \forall a \in A: a \le x\}$$

It is clear that $L$ and $H$ cover all reals and the condition on $L$ and $H$ is satisfied. Also $H$ is non-empty as $u \in H$. Then the promised $\alpha$ with $L \le \alpha \le H$ is equal to $\sup(A)$.

If version 2 is satified, we just take $\alpha = \sup(L)$.

Henno Brandsma
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