I have had no formal training in analysis before, and am using Understanding Analysis by Abbott (2015) for self-study.
My questions concern his presentation of the Archimedean property, and I would appreciate if members of the community could supply me with the tacit context I am clearly missing here.
Context.
The statement of the theorem he uses is as follows:
Theorem 1.4.2 (Archimidean Property).
(i) Given any number $x \in \mathbb{R}$ there exists an $n \in \mathbb{N}$ satsifying $n > x$.
(ii) Given any real number $y > 0$ there exists and $n \in \mathbb{N}$ satisfying $1 /n < y$.
The author then goes on to state that
Part $(i)$ of the proposition [in Theorem 1.4.2] states that $\mathbb{N}$ is not bounded above.
His proof by contradiction of part $(i)$ of Theorem 1.4.2 states that
Assume for contradiction that $\mathbb{N}$ is bounded above,. By the axiom of completeness (AoC), $\mathbb{N}$ should 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, (see Lemma 1.3.8) and therefore there exists an $n \in \mathbb{N}$ satisfying $\alpha – 1 < n.$
Where his statement of the axiom of completeness is
Axiom of Completeness. Every nonempty set of real numbers that is bounded above has a least upper bound.
And the lemma he references is
Lemma 1.3.8.. Assume $s \in \mathbb{R}$ is an upper bound for the set $A \subseteq \mathbb{R}$. Then $s = \sup A$ if and only if for every choice of $\epsilon > 0$, there exists an element $a \in A$ satisfying $s – \epsilon < a$.
Queries.
1. Why does part $(i)$ of Theorem 1.4.2 mean that $\mathbb{N}$ is not bounded above?
I unable to see this intutively or formally. Intuitively, I parse part $(i)$ of Theorem 1.4.2 as "for any given real number we can always find a natural number that is bigger than said real number." Formally, Abbott's definition of an upper bound is stated in terms of $\mathbb{R}$. His definition is
Definition 1.3.1. A set $A \subseteq \mathbb{R}$ is bounded above if there exists a $b \in \mathbb{R}$ such that $a \leq b$ for all $a \in A$. The number $b$ is an upper bound for $A$.
Using this definition I cannot see how $\mathbb{N}$ is not bounded above, nor do I understand how his definition can be applied to $\mathbb{N}$, given that it is stated in terms of $\mathbb{R}$.
2. Given that Abbott's statement of the AoC and Lemma 1.3.8 are couched in terms of the real numbers $\mathbb{R}$ rather than $\mathbb{N}$, what step is he relying on to justify the applications of both to the extract of the proof of contradiction I have included?
Whilst I understand that he has assumed that $\mathbb{N}$ is bounded above to show that yields a contradiction, I fail to see how he can invoke his statement of AoC, given that it is an axiom about real numbers $\mathbb{R}$, not natural numbers $\mathbb{N}$, and similarly with Lemma 1.3.8, which is also a statement couched in terms of $\mathbb{R}$, not $\mathbb{N}$.
Best Answer
If $\mathbb N$ were bounded above it would mean that we could find a real number that is larger than (or equal to) each natural. Then, to say that $\mathbb N$ is not bounded above is to say that we cannot find such a real number -- or in other words, that no matter which real number we pick, the number we picked is not one that is larger than (or equal to) each of the naturals.
But 1.4.2(i) says exactly that no matter which element of $\mathbb R$ you pick, there is a larger natural. This means that the real was not larger than that particular natural, and therefore especially it fails to be larger than all the naturals.