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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$:= mathbb{Q}= Acoprod B ,A:={qin mathbb{Q|qleqmathbb{r}}},B:={qinmathbb{Q}|q>r},rinmathbb{R}$]gap:=section if $r$ is irrational]

Or that nested Intervalls $(I_i)_{iin mathbb {N}}$ always converge to a unique number $r$ in $mathbb{R}$ such that $r in I_i,forall iinmathbb{N}$

I want to ask you if my proof using nested Intervalls is correct.

Let $kinmathbb{N},cinmathbb{R}$

Define sequence recurisively : $(I_i)_{iinmathbb{N}}=([a_i,b_i])_{iinmathbb{N}}$

$I_1:= [0,|c|+1]$

$I_{n+1}:=[frac{a_n + b_n}{2},b_n]$, if $(frac{a_n + b_n}{2})^kleq |c|$

$I_{n+1}:=[a_n,frac{a_n + b_n}{2}]$, if $(frac{a_n + b_n}{2})^kgeq |c|$

To Show: $a) a_1leq...a_nleq b_n ... leq b_1$ and $b)|b_n - a_n| rightarrow 0$

$a)$

Inductionbase: $0<|c|+1$ True

Inductionstep:

If $(frac{a_n + b_n}{2})^kleq |c|$ then $a_{n+1} = frac{a_n + b_n}{2},b_{n+1}= b_n$, IH $Rightarrow a_nleq b_nRightarrow a_nleq a_{n+1},a_{n+1}leq b_{n+1}$

If $(frac{a_n + b_n}{2})^kgeq |c|$ then $a_{n+1}=a_n,b_{n+1} = frac{a_n + b_n}{2}$, IH $Rightarrow a_nleq b_nRightarrow a_nleq a_{n+1},a_{n+1}leq b_{n+1}$

$Rightarrow a_1leq...a_nleq b_n ... leq b_1$

$b)$

$a)Rightarrow |b_n-a_n|geq |b_{n+1}-a_{n+1}|=: |b_n - frac{a_n + b_n}{2}|$ 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

$a_nleq x leq b_n, forall nin mathbb{N},(a_n)^kleq |c| leq (b_n)^k, forall nin mathbb{N}Rightarrow x^k=|c|$

q.e.d ?