Some days ago I have asked this question to which André Nicolas gave a link to this paper which contained a proof of the Least Upper Bound Axiom from Monotone Convergence Theorem via Archimedian Property and Dedekind Cut Property. But I was wondering, is there any proof of the Least Upper Bound Axiom using only Monotone Convergence Theorem?
I have already read this but the proof necessarily uses the Nested Interval Property without an explicit mention of it.
$$\color{blue}{\text{A proof using the same (or equivalent) ideas of the NIP isn't wanted here.}}$$
I tried to show this by considering the a non-empty set $A$ which is bounded above. I thought of the following three approaches but couldn't succed in any one of them.
I tried to construct a increasing sequence from $A$ which would converge to some limit and I tried to prove that this would be the $\sup A$. The construction was as follows. Since $A\ne \emptyset$ there exists an $a\in A$. We denote it by $x_1$. To construct $x_2$ we consider any uppebound of $A$ (such an upper bound exists since $A$ is bounded above), calling it by $y_1$. Then consider the intervals $\left[x_1,\dfrac{x_1+y_1}{2}\right]$ and $\left[\dfrac{x_1+y_1}{2},y_1\right]$. Now, $$x_2\begin{cases}=\text{Any}\ a\in \left[\dfrac{x_1+y_1}{2},y_1\right] & \text{if}\ A\cap \left[\dfrac{x_1+y_1}{2},y_1\right]\ne\emptyset\\=\text{Any}\ a\in \left[x_1,\dfrac{x_1+y_1}{2}\right] & \text{if}\ A\cap \left[\dfrac{x_1+y_1}{2},y_1\right]=\emptyset\end{cases}$$And we repeat this process. But this doesn't show that all members of $A$ is $\le$ of the limit, it only shows that all the members of the sequence $\{x_n\}$ are $\le$ the limit.
I tried to construct a decreasing sequence from $B$ (the set of all upperbounds of $A$). Using similar arguments arguments as above I have tried to show that the limit of the sequence is the $\sup A$ but here also similar problem arises.
The last approach was to modify the construction of (1). Instead of considering only one sequence I tried to consider multiple sequences. Let the first sequence be denoted as $\{x_{n,1}\}$. Now to construct the second sequence we first modify the our basis set. We let $$A_1:=A\setminus \{a\in A \land \exists m\in \mathbb{N} \mid a=x_{m,1}\}$$ and from this new set construct our sequence. Finally we consider a set of all the limits of the sequences (say $\mathfrak{L}$) obtained in this way. My guess is that the limit of all the members of $\mathfrak{L}$ will be the $\sup A$. I can't prove that also.
Any suggestion or hints as to how to proceed on any one of the three line of arguments or any new method?
Is there any way to prove the equivalence from (3)?
Added:-
One of my friend told me that the definition of the limit, say $\displaystyle\lim_{n\to\infty}a_n=a$ is logically equivalent to the statement, $$n\to \infty\implies a_n\to a$$ and so the unboundedness of natural numbers is implicitly assumed. More specifically $n\to\infty$ is a mathematical statement !!
But I argue that the definition of limit for the given example is, $$\forall\varepsilon>0, \ \exists n_0\in \mathbb{N}:\left\lvert a_n-a\right\rvert<\varepsilon, \ \forall n\in \mathbb{N}\land n\ge n_0$$ and so from this point of view their is no harm in seeing $n\to\infty$ as simply a symbol.
But I can't refute my friends statement in a rigorous way. Help on this will be great.
Best Answer
Combine the first two approaches. Start with $a_0\in A$ and an upper bound $b_0$. If $a_0=b_0$, you’re done; otherwise, let $x_0=\frac12(a_0+b_0)$. If $[x_0,b_0)\cap A\ne\varnothing$, pick any $a_1\in[x_0,b_0)\cap A$ and let $b_1=b_0$. If $[x_0,b_0)\cap A=\varnothing$, let $a_1=a_0$ and $b_1=x_0$. Continue in this fashion. At each stage you’ll have $a_n\in A$ and $b_n$ an upper bound for $A$, the sequence $\langle a_n:n\in\Bbb N\rangle$ is monotone non-decreasing, the sequence $\langle b_n:n\in\Bbb N\rangle$ is monotone non-increasing, and $b_{n+1}-a_{n+1}\le\frac12(b_n-a_n)$ for each $n\in\Bbb N$.