In the book "Understanding Analysis, second edition" by Stephen Abbot, the unboundedness of the set of natural number $\mathbb{N}$ is proven as the following proof:
Assume, for contradiction, that $\mathbb{N}$ is bounded above. By the Axiom of Completeness (AoC), $\mathbb{N}$ should then have a least upper bound, and we can set $\alpha = \sup \mathbb{N}$. If we consider $\alpha – 1$, then we no longer have an upper bound [referring to the definition of supremum], and therefore there exists an $n \in \mathbb{N}$ satisfying $\alpha – 1 <n$. But this is equivalent to $\alpha < n+1$. Because $n+1 \in \mathbb{N}$, we have a contradiction to the fact that $\alpha$ is supposed to be an upper bound for $\mathbb{N}$. (Notice that the contradiction here depends only on AoC and the fact that $\mathbb{N}$ is closed under addition.)
What I do not understand is the following: How can we assume AoC holds for natural numbers? Especially in this book, we start by defining $\mathbb{N}$, and then say that $\mathbb{Q}$ (the set of rational numbers) is an extension of $\mathbb{N}$. Then, it is shown that $\sup A$ may not exist for bounded $A \subset \mathbb{Q}$. Then, we finally define AoC to "close the holes" of $\mathbb{Q}$, which is an axiomatic way of defining $\mathbb{R}$. So, how can we now go back to $\mathbb{N}$, and assume AoC holds to prove $\mathbb{N}$ is unbounded. In summary, I believe (probably I am mistaken) the way we prove this is some sort of paradox since AoC comes after defining $\mathbb{N}$ as a property of real sets. What if $\mathbb{N}$ was indeed bounded, but AoC does not hold for $\mathbb{N}$, hence this proof is wrong? In my view, we should define an AoC argument for natural numbers (that any bounded subset of $\mathbb{N}$ admits a $\sup$, and indeed this is a member of the set, hence we have a $\max$), then use this in our proof.
Best Answer
AoC is one of the axioms of $\Bbb R$. So this argument applies to a set $\Bbb N$ defined as a subset of $\Bbb R$ (e.g., as the intersection of all sets that contain $1$ and are closed under "$+1$").
When we axiomatize $\Bbb N$ per se without the context of $\Bbb R$, i.e., with the Peano axioms, then there is no AoC and no upper bound, indeed.