TITLE: How to derive the axiom no. 15 from the Cantor's and Archimedean axiom?
QUESTION [0 upvotes]: How could one substitute the (15th) axiom of completeness with Archimedean and Cantor's axiom?
We discussed Cantor's axiom as well as Archimedean in analysis lectures and were told this question might appear on the exam. The first 14 axioms which also work for $\Bbb Q$ are left, and I need to derive, as mentioned in comments, the axiom of completeness of the field $\Bbb R.$
I am looking for proof that the supremum of a non-empty $S\subset(0,+\infty)$ is unique, therefore I have to rewrite the statement that, no matter how small positive real number $\varepsilon$ I subtract from the supremum, there will be some $x\in S$ such that $$\sup S-\varepsilon<x.$$
But, since I need to use the Archimedean axiom, I tried to use it here, but I'm not sure how.
Then I wanted to replace the term supremum when describing the properties of the non-empty intersection segments in Cantor's theorem using the statement:
$\forall x,y\in \Bbb R^+, \exists n\in\Bbb N,$ such that $ny\ge x$
Am I at least at the right starting trail?

REPLY [0 votes]: We can use the Cantor-intersection theorem and the Archimedean axiom to prove the Dedekind's theorem (source:Enrico Gregorio's answer).

Suppose $A,B\subseteq\Bbb R$ are non-empty s. t. $(\forall a\in A)(\forall b\in B)\quad a\le B$ and $A\cup B=\Bbb R$. Then $\exists ! c\in\Bbb R$ s. t. $a\le c\le b, \forall a\in A,\forall b\in B$.

Let's abbreviate the above with $A\le B$ and $A\le c\le B$.
Since $A\ne\emptyset\space\land\space B\ne\emptyset$, let's choose $a_0\in A$ and $b_0\in B$.
Now, we choose $c_0=\frac{a_0+b_0}2$.
If $A\le c_0\le B$, we're done. Otherwise, it is either
$$c_0<a,\text{for some }a\in A\space\text {or}\space c_0>b\text{ for some }b\in B.$$
If $c_0<a$, then let $a_1=a, b_1=b_0$.
If $c_0>b$, then let $a_1=a_0, b_1=b$.
In either case, $a_0\le a_1\space\land\space b_0\ge b_1\implies [a_0,b_0]\supseteq[a_1,b_1]$.
Let $d=b_0-a_0$. We have $b_0-c_0=c_0-a_0=\frac{b_0-a_0}2=\frac{d}2$.
If $c_0<a$, then $\frac{d}2=b_0-c_0>b_0-a=b_1-a_1$.
If $c_0>b$, then $\frac{d}2=c_0-a_0>b-a_0=b_1-a_1$.
We can repeat the procedure to build a sequence of nested closed intervals:
$$[a_0,b_0]\supseteq[a_1,b_1]\supseteq\cdots\supseteq[a_n,b_n]$$
If at some point $A\le c_n\le B$, we're done. Otherwise, we get a sequence of closed nested intervals $[a_n,b_n]$ with $a_n\in A,b_n\in B$ and $b_n-a_n\le\frac{d}{2^n}.$
By the Cantor-intersection theorem (now assumed as an axiom), there is $c\in\bigcap_n[a_n,b_n]$.
Now we want to prove $A\le c\le B$. Let's assume the opposite: there is either
$$a\in A\text{ with } c<a\space\text{ or }\space b\in B\text{ with } c>b.$$
Suppose $a>c$.
Let's take $\varepsilon=\frac{a-c}2$.
We can take $n\in\Bbb N$ s. t. $\frac{d}{2^n}<\varepsilon$. $(*)$
Then we obtain:
$b_n<a+\varepsilon\le c+\varepsilon=c+\frac{a-c}2=\frac{a+c}2\le a$ which contradicts $A\le B$.
Analogously in case of $c>b$.

$(*)$ As $2^n>n, n\in\Bbb N$, we  have $\frac1{2^n}<\frac1n\implies\frac{d}{2^n}<\frac{d}n$. In order to make $\frac{d}{2^n}<\varepsilon$, we could take $n$ s. t. $\frac{d}n<\varepsilon$, which is guaranteed to hold true by the Archimedean axiom.

Now, we want to prove the Dedekind's theorem is equivalent to the axiom of completeness (like Henno Brandsma did here):
Every non-empty subset of $\Bbb R$ that is bounded above has a supremum in $\Bbb R$.
Direction $\boxed{\Rightarrow}$
Suppose Dedekind's theorem holds.
Let $S\subset\Bbb R$ with some upper bound $u$ and let's define:
$$A=\{x\in\Bbb R\mid\exists s\in S, x<s\}\\\text{and}\\ B=\{x\in\Bbb R\mid\forall s\in S, x\ge s\}$$
$B\ne\emptyset$ because $u\in B$.
Then our $c=\sup S$.
Direction $\boxed{\Leftarrow}$
Assume the axiom of completeness holds.
Then we take $c=\sup A$.
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