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I try to prove Dedekind gapless property with nested interval theorem and Archimedean property, and here is my idea:

$(A,B)$ is a partition of $\mathcal{R}$ with the property that $a<b$ for all $a\in A$ and $b\in B$.

Since $\mathcal{R}$ is ordered, I pick a increasing sequence $\{a_n\}$ in $A$ and decreasing sequence $\{b_n\}$ in $B$. We observe the intervals $[a_n,b_n]$, and the intersection of the set of the intervals is nonempty by means of Nested interval theorem.

I assume $r\in\cap[a_n,b_n]$, and r is a upper bound of $A$ and a lower bound of $B$. By definition of partition, either $r\in A$ or $r\in B$ holds. r is one of the extreme values of $A$ and $B$. If extreme values of $A$ and $B$ exist together, then the sum of them divided by 2 doesn't exist.

I hesitate whether I can pick sequence in this way and say that max$A$ or min$B$ exist.

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  • $\begingroup$ Please specify a definition of "Dedekind gapless property". Does this mean the "least upper bound property", i.e., completeness? $\endgroup$
    – Gary
    Commented Mar 11, 2022 at 17:55
  • $\begingroup$ Gapless property means for ($A$,$B$) a Dedekind cut of $\mathcal{R}$, either max$A$ or min $B$ exist. (All of them are equivalent to completeness of real number, but I try to avoid the others of completeness.) $\endgroup$
    – xfireskyx
    Commented Mar 12, 2022 at 6:55
  • $\begingroup$ So $A,B\subset{}\mathbb{R}$? $\endgroup$
    – Gary
    Commented Mar 13, 2022 at 19:23
  • $\begingroup$ Yeah, since the union of Dedekind cut equals to the universal set, so $A$, $B$ are subsets. $\endgroup$
    – xfireskyx
    Commented Mar 14, 2022 at 5:20
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    $\begingroup$ Also, are you saying you hope to prove the claim without reference to the completeness property and without reference to representation of real numbers as Dedekind cuts? $\endgroup$
    – Gary
    Commented Mar 14, 2022 at 18:51

1 Answer 1

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I know the question has been posted a while ago, but here's my solution. Please correct me if I'm wrong. I will use the Archimedean property and the nested intervals propery to prove that if $A,B\subseteq \mathbb{R}$ are nonempty sets such that $a < b\; \forall a\in A, b\in B$, then $\exists c\in \mathbb{R}: a\leq c\leq b\; \forall a\in A, b\in B$.

Take any $b\in B$, then the Archimedean property implies $$ \exists n\in \mathbb{N}: b<n \Rightarrow a < n\; \forall a \in A. $$ Therefore, the set $$ M := \{ k \in \mathbb{Z}: a < k\; \forall a \in A\} $$ is nonempty. Moreover, it is bounded from below, and we can define $m:= \min M$. Thus, $a < m\; \forall a \in A$ and there exists a number $a'\in A$ such that $m-1 \leq a'$. Define $a_0:=m-1$ and $$ a_n := \left\{\begin{array}{ll} a_{n-1} + \frac{1}{2^n} & \text{if } \exists a\in A: a_{n-1} + \frac{1}{2^n} \leq a \\ a_{n-1} & \text{otherwise} \end{array} \right.,\quad n \in \mathbb{N}. $$ One can show by induction that $$a < a_n + \frac{1}{2^n}\quad \forall a\in A, n\in \mathbb{N}. \quad (*) $$

Similarly, there exists $m'\in \mathbb{Z}$ such that $m' < b\; \forall b\in B$ and $b' \leq m'+1$ for some $b'\in B$. We define $b_0:=m'+1$ and $$ b_n := \left\{\begin{array}{ll} b_{n-1} - \frac{1}{2^n} & \text{if } \exists b\in B: b_{n-1} - \frac{1}{2^n} \geq b \\ b_{n-1} & \text{otherwise} \end{array} \right.,\quad n \in \mathbb{N}. $$

By construction, for any $n,m\in \mathbb{N}$ there exist some $a\in A, b\in B$ such that $a_n \leq a,\, b_m \geq b$. Note that $a\in A, b\in B$ implies $a<b$, and therefore, $a_n < b_m$. Thus, $\{[a_n,b_n]\}_1^\infty$ is a valid sequence of nested intervals, hence they have a common point $c\in \mathbb{R}$.

We will now show that $a \leq c\; \forall a \in A$. If there was $\hat{a}\in A$ such that $c<\hat{a}$ then Archemedian property would imply the existence of $n\in \mathbb{N}$ such that $\frac{1}{2^n}\leq \hat{a}-c$ which in turn leads to $$ \hat{a}\geq c+\frac{1}{2^n}\geq a_n+\frac{1}{2^n} \stackrel{(*)}{>} \hat{a}, $$ which is a contradiction. Similarly, one can show $c\leq b\; \forall b\in B$, which finishes the proof.

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