Problem in understanding the proof of "every nonempty set of natural numbers has a smallest member". I was reading Real Analysis by Royden and Fitzpatrick; there the authors introduced the theorem

Theorem 1:
Every non-empty set of natural numbers has a smallest member.

Then

Proof:
Let $E$ be a non-empty set of natural numbers. Since, the set $\{x\in \mathbb R| ~ x\geq 1\}$ is inductive, the natural numbers are bounded below by $1\;.$ Therefore, $E$ is bounded below by $1\;.$ As a consequence of the Completeness Axiom, $E$ has an infimum; define $c= \textrm{inf}~ E\;.$ Since, $c+1$ is not a lower bound for $E,$ there is an $m\in E$ for which $m\lt c+1\;.$ We claim $m$ is the smallest member of $E\;.$ Otherwise, there is an $n\in E$ for which $n\lt m\;.$ Since, $n\in E, c\leq n\;.$ Thus, $c\leq n\leq m \leq c+1$ and therfore, $m-n\lt 1\;.$ Therefore, the natural number$m$ belongs to the interval $(n,n+1)\;.$ an induction argument shows that for every natural number $n,$ $(n, n+1)\cap \mathbb N= \emptyset\;.$ This contradiction confirms that $m$ is the smallest element.

I'm having some problems in conceiving the proof:
$\bullet$  How is the fact $E$ being bounded below by $1$ related to the fact that the set $\{x\in\mathbb R|~x\geq 1\}$ being inductive? I'm not getting that.
$\bullet$ What is the necessity in the proof to show that $E$ is bounded below by $1\;?$ I'm not seeing how it helped the later part of the proof.
$\bullet$ The authors didn't show how $(n, n+1)\cap \mathbb N= \emptyset\;.$ How could I show that?
 A: I suppose that this book defines an "inductive set" to be a set $S$ of real numbers such that $1 \in S$ and if $k \in S$ then $k + 1$ in $S$.
(There is more than one definition of an "inductive set"; see the article in MathWorld.
But this definition seems to be the one that makes sense in this context.)
The fact that the set $\{x\in \mathbb R \mid x\geq 1\}$
is inductive then implies that the natural numbers are a subset of
$\{x\in \mathbb R \mid x\geq 1\}$. Presumably, the book has already shown
that $\{x\in \mathbb R \mid x\geq 1\}$ is bounded below by $1$,
so the natural numbers also are bounded below by $1$.
Regarding why it is important that $E$ is bounded below by $1$,
consider the integers, which are a lot like the natural numbers except
that you can count down indefinitely (not bounded below by $1$) as well as count up. It is easy to construct a subset of the integers with no least element.
Regarding the specific steps of the book's proof, if we ignore this fact about $E$, then we lose the reason by which the book invokes the Completeness Axiom, so we lose the definition of $c$ and we lose every fact that uses $c$.
The authors presumably mean for you to prove that $(n,n+1)\cap\mathbb N = \emptyset$ by induction on $n$ with a base case of $n=1$
(for which you must prove that $(1,2)\cap\mathbb N = \emptyset$)
and an inductive case in which you prove that if
$(n,n+1)\cap\mathbb N = \emptyset$
then $(n+1,n+2)\cap\mathbb N = \emptyset$.
A: Logicians have the notion of begging the question, from the archaic meaning of beg, to assume something you have no right to use or possess. If you say maple syrup is good for you, because it comes from the sap of a tree, you are begging the question: Socrates drank hemlock and it wasn't good for him. I think the proof is correct, but begs the question. The authors assume that somehow you have been able to come upon the real numbers, without ever using that every non-empty set of natural numbers has a least member. And amazingly you are asked to accept that every nonempty set of real numbers with a lower bound has an infimum, while at the same time holding in abeyance until proven, that every nonempty set of natural numbers has a least member. In short, I believe the axiom of completeness is really a theorem, but the theorem you are considering is not really a theorem; it's an axiom.
