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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.

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    $\begingroup$ Every non empty subset of $\mathbb{R}$ which is bounded above has a supremum in $\mathbb{R}$.If $\mathbb{N} $ is bounded above, then it also has a supremum, you denote it by $\alpha$ . Here key thing to notice, $\alpha$ is a real number. If you consider any subset of $\mathbb{Q}$ it may be bounded above but it may not have supremum in $\mathbb{Q}$ but it may have supremum in $\mathbb{R}$ $\endgroup$
    – Sourav Ghosh
    Dec 1, 2021 at 18:29
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    $\begingroup$ We are allowed to use AoC here because we are looking at $\mathbb{N}$ as a subset of $\mathbb{R}$ here. $\endgroup$
    – Юрій Ярош
    Dec 1, 2021 at 18:31
  • $\begingroup$ Isn't it the case that both the AoC and unboundedness of $\mathbb{N}$ can be proven if the integers are our entire universe? For AoC, just use the fact that every non-empty subset of $\mathbb{N}$ has a least element. For unboundedness, assume $n$ is an upper bound, then $n<n+1$; contradiction. I think the point here is that $\mathbb{N}$ is also unbounded in $\mathbb{R}$, which isn't obvious, because maybe there's a real number that's larger than all the integers. $\endgroup$
    – march
    Dec 1, 2021 at 19:07
  • $\begingroup$ @march In case it helps, the same book says: "In fact, as disorienting as it may sound, there are ordered field extensions of $\mathbb{Q}$ that include "numbers" bigger than every natural number. [Archimedean Property] asserts that the real numbers do not contain such exotic creatures." $\endgroup$
    – independentvariable
    Dec 2, 2021 at 10:59

2 Answers 2

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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.

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  • $\begingroup$ Thanks so much for your reply! Exactly, that was my point of view. The typical counter-example for the rational numbers, that is, $A = \{ a \in \mathbb{Q} : a^2 < 2\} \subset \mathbb{Q}$ does not have a supremum when we restrict ourselves to rational numbers. So we need to define $\mathbb{R}$ first by using $\mathbb{Q}$ and $\mathbb{N}$ sequentially. However, if the definition of $\mathbb{R}$ depends on these two, how can we see $\mathbb{N}$ as a subset of $\mathbb{R}$ in this proof (as in your first sentence), where $\mathbb{R}$ is not well-defined without fixing $\mathbb{N}$? $\endgroup$
    – independentvariable
    Dec 1, 2021 at 18:36
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    $\begingroup$ Can't versions of AoC and unboundedness be proven inside Peano? AoC (in the sense of every bounded subset of $\mathbb{N}$ has a maximal element) would be provable because every non-empty subset of t$\mathbb{N}$ has a least element (so, choose this set to be all the integers larger than every integer in our target set $A$, and the least element is then the least upper bound). Unboundedness is simple: let $n$ be the upper bound; then $n<n+1$; contradiction. The point then is that we have to show that no real number is larger than every natural number. $\endgroup$
    – march
    Dec 1, 2021 at 19:21
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    $\begingroup$ Regarding the theory of $\Bbb R$: We can extend $\Bbb R$ to a larger ordered field $\Bbb R^*$ that has members that are $>0$ but smaller than every member of $\Bbb R^+$. Their reciprocals are larger than every member of $\Bbb R$... In $\Bbb R^*,$ the set $\Bbb N$ has an upper bound but no $ lub....$ To define $\Bbb R$ we produce a construction of an ordered field $F$ that satisfies AoC, and we show that any ordered field that has AoC is isomorphic to $F$, and so we call $F$ "the" reals. $\endgroup$
    – DanielWainfleet
    Dec 2, 2021 at 4:16
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Here I think is the crux of the issue: once you've embedded the natural numbers into the real numbers, there could be a real number that's larger than all of the natural numbers. We need to prove that that's not true.

The reason this feels strange (apart from the fact that it's "obvious" that there's no real number larger than all the integers) is that both the Axiom of Completeness and the unboundedness property of $\mathbb{N}$ are provable if our entire universe is just $\mathbb{N}$.$^1$ However, once we've embedded the natural numbers into the real numbers, it's no longer obvious that any set of natural numbers has a least upper bound (because we allow real numbers to bound the sets as well), and it's also no longer guaranteed that that the natural numbers are unbounded, because we have embedded them in a larger set.

$^1$For AoC, just use the fact that every non-empty subset of integers (i.e. the set of all integers greater than or equal to every integer in our target set $A$) has a least element: then that element is the smallest upper bound of $A$. For unboundedness, assume that $n$ is the largest integer; then $n<n+1$, which is a contradiction.

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