The Proofs I have seen so far are using the fact (Completeness Axiom) that every non-empty set in $\mathbb{R}$ with an upperbound also has a Supremum. However there are Theorems/Propositions (I don't know the difference) which are equivalent to the Completeness Axiom.
For example that $\mathbb{R}$ has no gaps
[[section$:= \mathbb{Q}= A\coprod B ,A:=\{q\in \mathbb{Q|q\leq\mathbb{r}}\},B:=\{q\in\mathbb{Q}|q>r\},r\in\mathbb{R}$]gap:=section if $r$ is irrational]
Or that nested Intervalls $(I_i)_{i\in \mathbb {N}}$ always converge to a unique number $r$ in $\mathbb{R}$ such that $r \in I_i,\forall i\in\mathbb{N}$
I want to ask you if my proof using nested Intervalls is correct.
Let $k\in\mathbb{N},c\in\mathbb{R}$
Define sequence recurisively : $(I_i)_{i\in\mathbb{N}}=([a_i,b_i])_{i\in\mathbb{N}}$
$I_1:= [0,|c|+1]$
$I_{n+1}:=[\frac{a_n + b_n}{2},b_n]$, if $(\frac{a_n + b_n}{2})^k\leq |c|$
$I_{n+1}:=[a_n,\frac{a_n + b_n}{2}]$, if $(\frac{a_n + b_n}{2})^k\geq |c|$
To Show: $a) a_1\leq...a_n\leq b_n ... \leq b_1$ and $b)|b_n - a_n| \rightarrow 0$
$a)$
Inductionbase: $0<|c|+1$ True
Inductionstep:
If $(\frac{a_n + b_n}{2})^k\leq |c|$ then $a_{n+1} = \frac{a_n + b_n}{2},b_{n+1}= b_n$, IH $\Rightarrow a_n\leq b_n\Rightarrow a_n\leq a_{n+1},a_{n+1}\leq b_{n+1}$
If $(\frac{a_n + b_n}{2})^k\geq |c|$ then $a_{n+1}=a_n,b_{n+1} = \frac{a_n + b_n}{2}$, IH $\Rightarrow a_n\leq b_n\Rightarrow a_n\leq a_{n+1},a_{n+1}\leq b_{n+1}$
$\Rightarrow a_1\leq...a_n\leq b_n ... \leq b_1$
$b)$
$a)\Rightarrow |b_n-a_n|\geq |b_{n+1}-a_{n+1}|=: |b_n - \frac{a_n + b_n}{2}|$ or $|\frac{a_n + b_n}{2}-a_n|$
In both cases $|b_{n+1}-a_{n+1}|= |\frac{b_n-a_n}{2}|$
The successor is always half the value of the predecessor therefore $|b_n-a_n|$ must coverge to $0$
$|b_n - a_n| \rightarrow 0$
Because nested intervals converge to a unique number = $x$, this number must be the root since
$a_n\leq x \leq b_n, \forall n\in \mathbb{N},(a_n)^k\leq |c| \leq (b_n)^k, \forall n\in \mathbb{N}\Rightarrow x^k=|c|$
q.e.d ?