Proving the Archimedean Property of reals and NIP (Nested Interval Property) using MCT I'm doing exercise 2.4.4 in the book Understanding Analysis by Stephen Abbott. I'd like to ask, if my proof (especially part (a) of the question) is rigorous and technically correct.

(a) In a previous section, we used the Axiom of Completeness(AoC) to prove the Archimedean property of real numbers $\mathbf{R}$. Show that the Monotone Convergence Theorem can also be used to prove the Archimedean property without making any use of AoC.
(b) Use the Monotone Convergence Theorem to supply a proof for the Nested Interval Property that doesn't make use of the AoC.
These two results would suggest that we could have used the Monotone Convergence Theorem in place of the AoC as our starting axiom for building a proper theory of the real numbers.

Proof.
(a) The Archimedean property states that, $\mathbf{N}$ is an unbounded set that sits in $\mathbf{R}$. Given any real number $x$, there exists $n \in \mathbf{N}$ such that $x < n$.
By contradiction, let us assume that $\mathbf{N}$ is bounded subset of $\mathbf{R}$. Then, there exists an upper bound $x \in \mathbf{R}$, such that $n < x$ for all natural numbers $n \in \mathbf{N}$. The natural numbers are recursively defined using the sequence $x_{n+1} = 1 + x_n$. This is a monotonically increasing sequence. Hence, $\mathbf{N}$ is monotonic increasing and bounded, by monotone convergence theorem, the sequence $(x_n)$ is convergent. Let $\lim x_{n+1} = \lim x_{n} = x$. Taking limits on both sides, we have:
\begin{align*}
    x &= 1 + x \\
    0 &= 1
\end{align*}
This is a false statement and violates Peano's axioms. Moreover, we cannot find an interval $V_\epsilon(x)$ such that the values of the sequence $(x_n)$ will eventually lie in this interval. So, $(x_n)$ is not a convergent sequence $\implies$ $(x_n)$ is unbounded $\implies$ $\exists n \in \mathbf{N}$, such that $x < n$ for all reals $x$.
(b) The nested interval property states that the real number line $\mathbf{R}$ contains no gaps. I reproduce the statement of NIP for completeness.

For each $n \in \mathbf{N}$, assume that we are given a closed interval $I_n = [a_n, b_n] = \{x \in \mathbf{R}: a_n \le x \le b_n\}$. Assume also that each $I_n$ contains $I_{n+1}$. Then, the resulting nested sequence of closed intervals
\begin{align*}
    I_1 \supseteq I_2 \supseteq I_3 \supseteq I_4 \ldots
\end{align*}


have a non-empty intersection; that is
\begin{align*}
    \bigcap_{n=1}^{\infty} I_n \ne \phi
\end{align*}

Since, $I_n \supseteq I_{n+1}$, that is $[a_n,b_n] \supseteq [a_{n+1},b_{n+1}]$, the sequence consisting of the left hand end-points $a_1,a_2,a_3,\ldots, a_n$; $(a_n)$ is monotonically increasing. Moreover, the sequence $(a_n)$ is bounded above by $b_n$. By the Monotone Convergence Theorem, the sequence is convergent and $\lim_n (a_n) = s = \sup \{a_n : n \in \mathbf{N}\}$.
Since, $s$ is an upper bound for the sequence $(a_n)$, $a_n \le s$ for all $n \in \mathbf{N}$. Moreover, as $s$ is the least upper bound $s \le b_n$ for all $n \in \mathbf{N}$. Consequently, $a_n \le s \le b_n$ for all $n \in \mathbf{N}$. Thus, $s \in I_n$ for all $n \in \mathbf{N}$, and the countable intersection of these intervals is non-empty.
 A: What you’ve done is mostly correct, but there are a few fairly minor problems. First, your derivation of the contradiction in (a) could be made clearer. Once you have $x_{n+1}=1+x_n$ and $\lim_{n\to\infty}x_n=x$, you can write
$$x=\lim_{n\to\infty}x_n=\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}(1+x_n)=1+\lim_{n\to\infty}x_n=1+x\,;$$
this makes it very clear how you deduce that $x=1+x$ and hence that $0=1$. You should stop there: the remainder of that last paragraph is unnecessary. Moreover, part of it is false: it is simply not true that there is an $n\in\Bbb Z^+$ such that $x<n$ for all reals $x$. It isn’t even true that there is an $n\in\Bbb Z^+$ such that $x\le n$ for all reals $x$: just take $x=n+1$ to get a counterexample.
The idea in (b) is fine, but some of what you’ve written is rather sloppy. For instance, you write that

… the sequence $(a_n)$ is bounded above by $b_n$.

Here you are using the letter $n$ to mean two different things in the same statement: in $(a_n)$ it’s an index that runs over all positive integers, and in $b_n$ it refers to some specific index. What you mean is that for each $n\in\Bbb Z^+$, the sequence $\langle a_k:k\in\Bbb Z^+\rangle$ is bounded above by $b_n$. In particular, it is bounded above, so it converges to some $s=\lim_{n\to\infty}a_n=\sup\{a_n:n\in\Bbb Z^+\}\in\Bbb R$. The remainder of that last paragraph is fine.
