Nested interval theorem and Archimedean property implies Dedekind gapless property

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 $$a for all $$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.

• Please specify a definition of "Dedekind gapless property". Does this mean the "least upper bound property", i.e., completeness?
– Gary
Commented Mar 11, 2022 at 17:55
• 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.) Commented Mar 12, 2022 at 6:55
• So $A,B\subset{}\mathbb{R}$?
– Gary
Commented Mar 13, 2022 at 19:23
• Yeah, since the union of Dedekind cut equals to the universal set, so $A$, $B$ are subsets. Commented Mar 14, 2022 at 5:20
• 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?
– Gary
Commented Mar 14, 2022 at 18:51

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 $$\exists n\in \mathbb{N}: b Therefore, the set $$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, 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.