Similar questions have been asked before.
This is problem of Foster's Analysis 1.
- Prove the Dedekind cut:
Let A, B non empty subsets of $\mathbb {R} $ and $A\cup B=\mathbb{R}$ such that $x \lt y$ for all $x \in A$ and $y \in B$, then there is exactly one $s\in \mathbb{R}$ with $ x\le s \le y$ for all $x \in A$ and $ y\in B$.
- Prove that in an ordered field the Dedekind cut implies the Archimedean Principle.
To 1):
A is bounded above, B is bounded below, $\sup A$, and $\inf B$ exist in $\mathbb {R}$, and they are equal, which is fairly easy to show.
To 2):
Archimedean Principle:
Let $x, y \in \mathbb{R}$, $x, y \gt 0$, then there is a $n\in \mathbb{N}$ such that $nx \gt y$.
Assume there is no $n\in \mathbb{N}$ with this property.
Then $A:= ${$nx| n \in \mathbb{N}$} is bounded above by y.
And let $B:=${$b| y \le b$}.
To use 1), i.e. to get a s, $s=\sup A $, we would need $A\cup B= \mathbb{R}$, which is not the case.
How can we infer the existence of a supremum to conclude a contradiction?
Stuck here. Any hints? Thank you.
Added:
In an ordered field the Dedekind cut implies completeness.
In an earlier chapter the nested interval principle is introduced and it is shown that it implies completeness.
Nested interval principle:
Let $I_{n+1}\subset I_n$ for $n \in \mathbb{N} $ be a descending sequence of closed intervals in $\mathbb{R}$ with \
$\lim_{n \rightarrow \infty}$ diam $(I_n) =0$, then there is exactly one real number $s$ with $s \in I_n$ for all $n \in \mathbb{N} $.
Let $I_n= [a_n, b_n]$ with $a_n \lt b_n$ ,$ n \in \mathbb{N}, $ be such a descending sequence of closed intervals with
\ $\lim_{n \rightarrow \infty} $diam $(I_n) =0$.
Consider
$B:= \cup_n [b_n, \infty]$ and
$A:=\mathbb{R}$ \ $B$.
We then have $a \lt b$ for
$a \in A$, and $b\in B$.
Dedekind cut:
There exists exactly one $s \in \mathbb{R} $ with $a \le s \le b$ for all $a, b$.
Hence $a_n \le s\le b_n$ for all n, i.e. $s \in I_n$ for all n.
Since $\lim_n$ diam $(I_n)=0$ there cannot be more than one such point, i.e. there is exactly one point, and we are done.
