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Let's have this axiom of completeness for the reals:

Axiom: Let $(I_n)$ be a sequence of closed intervals in $\mathbb{R}$ such that $\forall n\in\mathbb{N}:I_{n+1}\subset I_n$. Also let $\lim_{n\to\infty} |I_n|=0$. Then $$\bigcap_{n=1}^\infty I_n\neq \emptyset$$

Now, I wish to prove the least upper bound property from this. That is, each set $A\subset \mathbb{R}$ bounded above has a least upper bound (supremum). By definition, $c$ is a supremum if $c$ is upper bound and for any other upper bound $c'$ it holds, that $c\leq c'$. We assume that $(\mathbb{R},+,\cdot,\leq)$ is given axiomaticaly together with the completeness axiom given. My go:

Let $A\subset \mathbb{R}$ be non-empty and bounded above by $M$. Choose arbitrary $a_0\in A$ and $b_0:=M$. Set $m:=\frac{a_0+b_0}{2}$. If $m$ is an upper bound, then set $b_1:=m$ and $a_1:=a_0$, else $b_1:=b_0$ and then pick $a_1\in A$ such that $a_1>m$. Now, inductively construct a sequence of self-contained closed intervals $I_n=\langle a_n,b_n\rangle$ whose lenghts converge to $0$. By Axiom, $\bigcap I_n$ is non-empty, so has some element $c$. Now, I claim that $c=\sup{A}$.

Now, from here I am a bit stuck. I wish to show that this found $c$ really is a supremum. By construction, all of $b_n$'s are upper bounds for $A$ now, how do I proceed? Now, the step "pick '$a_1$' is also weird for me, since we do not know anything about the set (it doesnt have to be interval), so I would like a constructive proof too.

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    $\begingroup$ When you say '...pick $a_1 \in A$' it starts sounding a bit spooky (no precise algorithm to sink your teeth into). If you change your question, asking for an alternate construction-proof, I will be able to supply an answer. $\endgroup$ – CopyPasteIt Oct 16 '18 at 15:23
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    $\begingroup$ I was actually unsure about that step, looked weird for me aswell. I have the supremum part now, but edited the original post. $\endgroup$ – Michal Dvořák Oct 16 '18 at 15:26
  • $\begingroup$ You don't need the hypotheses $\lim _{n\to \infty} |I_n|=0$ as the conclusion $\bigcap\limits _{n=1}^{\infty} I_n\neq\emptyset $ is true without that assumption. $\endgroup$ – Paramanand Singh Oct 17 '18 at 0:58
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Throughout this post we are analyzing an ordered field $R$ satisfying a slightly modified form of the OP's axiom:

$\text{Axiom (P1) =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=}$

Let $(I_n=[a_n,b_n])$ be any sequence of closed intervals in $R$ such that $\forall n\in\mathbb{N}:I_{n+1}\subset I_n$. Also assume that for every integer $n\gt0$ an interval $I_k$ can be found such that

$\quad b_k−a_k \lt \frac{1}{n}$

Then the intersection of all the intervals is a singleton.

$\text{=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=}$

Proposition 1: The ordered field $R$ satisfies the following property:

$\tag A (\forall a \in R, a \gt 0) \; (\exists n \in \mathbb N, n\gt0) \text{ such that } \frac{1}{n} \lt a$
Proof
By $\text{(P1)}$ the interesection of the intervals $[-\frac{1}{n},+\frac{1}{n}]$ can only be equal to $\{0\}$.
But this is equivalent to $\text{(A)}$. $\quad \blacksquare$

So $R$ satisfies the axiom of Archimedes and the following is also true.

Proposition 2: Every $r \in R$ belongs to exactly one interval of the form $[n, n+1)$ with $n \in \mathbb Z$.

We are now ready to show that $R$ also satisfies the least-upper-bound property.

Proposition 3: Let a non-empty subset $A$ of $R$ be bounded above by $M \in R$. Then there exists a least upper bound for $A$.
Proof
The set $C = \{n \in \mathbb Z \, | \, n \text{ is an upper bound for the set } A \}$ must have a least element $c_0$. Set $b_0 = c_0 -1$, so that $A$ has points in the closed interval $[b_0,c_0]$.

We continue using recursion/induction.

Let the interval $[b_k,c_k]$ satisfy the following:

$\tag 1 c_k \text{ is an upper bound for the set } A$
$\tag 2 \text{The set } A \text{ has points in } [b_k,c_k]$
$\tag 3 c_k - b_k = 2^{-k}$

Let $\mu = \frac{b_k+c_k}{2}$. If $A$ has points in $[\mu,c_k]$, define $b_{k+1} = \mu$ and $c_{k+1} = c_k$; else, define $b_{k+1} = b_k$ and $c_{k+1} = \mu$.

So we have a constructed a $\text{decreasing}_{↓0}$ chain $[b_n, c_n]$ of closed intervals satisfying (1), (2) and (3) with the intersection being a singleton set $\{\gamma\}$.

It is immediate that if $\lambda$ and $\kappa$ are any two distinct numbers, they can't both belong to all the intervals $[b_n, c_n]$. So if $a \in A$, it must be less than or equal to $\gamma$. Otherwise, $b_n \le \gamma \lt a \le c_n$ is true for all $k$, but then $a = \gamma$, a contradiction.

So $\gamma$ is an upper bound for the set $A$.

Suppose $\rho$ is another upper bound for $A$ that is stricly less than $\gamma$. Then for any $n$, $b_n \le \rho \lt \gamma \le c_n$. Again, this implies that $\rho = \gamma$, a contradiction.$\quad \blacksquare$

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  • $\begingroup$ This looks good, just to clarify. We assume this $c_0$ to be the integer given by the archimedean property, right? $\endgroup$ – Michal Dvořák Oct 16 '18 at 15:36
  • $\begingroup$ @MichalDvořák Using the fact that $\mathbb N$ is well-ordered. $\endgroup$ – CopyPasteIt Oct 16 '18 at 16:15
  • $\begingroup$ @MichalDvořák and axiom (P1). $\endgroup$ – CopyPasteIt Oct 17 '18 at 0:38
  • $\begingroup$ You can essentially say, that for any $\epsilon>0$ there is $n\in\mathbb{N}$ such that $|I_n|<\epsilon$. There is no need for that $1/n$. $\endgroup$ – Michal Dvořák Oct 17 '18 at 21:49
  • $\begingroup$ Yes. But I want to get rid of the limit stuff and anyway, looks like we do need singleton. See math.stackexchange.com/questions/1855286/… and other links $\endgroup$ – CopyPasteIt Oct 17 '18 at 22:16
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You forgot to say that you pick $a_1\in A$.

In order to prove that $c=\sup A$ you have to rule out (i) the existence of an $a\in A$ with $a>c$ and (ii) the existence of a $b<c$ which is an upper bound of $A$. Note that both $a$ and $b$ here would have a positive distance from $c$, whereas $|I_n|\to0$. This will allow you to produce a contradiction in both cases (i) and (ii).

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  • $\begingroup$ It is clear that $(a_n)$ is non-decreasing and $(b_n)$ is non-increasing. Now, by construction, $b_n$ is bounded below by all of $a_n$'s and $a_n$'s are all bounded above by all of $b_n$, (by monotone convergence theorem, which I am able to prove from the given Axiom) so this implies $\lim a_n=\lim b_n$. So, if we denote this limit $l$, then is bounded by $l$ from above and $b_n$ is bounded by $l$ from below. But all of $a_n$'s are, by construction, in $A$ so are not suprema for $A$... Well, I think I am running in circles :/ $\endgroup$ – Michal Dvořák Oct 16 '18 at 13:06
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    $\begingroup$ It's much simpler: You get the contradictions by looking at an $[a_n,b_n]$ with sufficiently large $n$. $\endgroup$ – Christian Blatter Oct 16 '18 at 14:59

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