Does there exist this type of sequence of subsets of $\mathbb{R}$?

Question

Let $A_n$ be a sequence of subsets of $\mathbb{R}$ with the following properties.

1. $A_n$ is unbounded for all $n$
2. The union of all $A_n$ is $\mathbb{R}$
3. No two $A_n$ share elements
4. For all $n$, given any two distinct elements of $A_n$, there exists an element of $A_n$ in between them

Question: Does such a sequence of subsets of $\mathbb{R}$ exist?

I can only think of finitely many sets collectively exhibiting these properties, such as $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$. I would appreciate hearing if any of these properties have better or more widely used names.

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Here are the definitions for the properties. Please let me know if you can spot mistakes.

$$ \tag{1}{(\forall ~ n ~ \nexists ~ M \in \mathbb{R} ~ \forall ~ x \in A_n : M \geq x) \land (\forall ~ n ~ \nexists ~ m \in \mathbb{R} ~ \forall ~ x \in A_n : m \leq x)} $$ $$ \tag{2}{\bigcup_n A_n = \mathbb{R}}$$ $$ \tag{3}\forall ~ m \neq n : A_m \cap A_n = \emptyset $$ $$ \tag{4}\forall ~ n ~ \forall ~ x_1, x_2 \in A_n ~ \exists ~ y \in A_n : x_1 < x_2 \implies x_1 < y < x_2 $$

Answer 1

Let $x_n$ denote a sequence of elements for which $x_{n+1}$ is transcendental over the field $\mathbb{Q}(x_1,....,x_n)$. Such a sequence exists because any field of the form $\mathbb{Q}(y_1,...,y_k)$ is countable.

Now, let $A_n = \mathbb{Q} + x_n = \{q+x_n: q\in \mathbb{Q}\}$ and let $A_0 = \mathbb{R}\setminus \bigcup A_n$. Then, I claim that $\{A_0, A_1,....\}$ satisfies your requirements.

First note that $A_0$ contains the rationals, because if $q\in\mathbb{Q}\cap A_n$, then $q = p + x_n$ for some rational number $p$, so $x_n = q-p\in\mathbb{Q}$, contradicting the choice of $x_n$.

This implies that $A_0$ satisfies 1 and 4. Similarly, because the $A_i$ for $i > 0 $ are just translates of $\mathbb{Q}$, they satisfy 1 and 4.

Condition 2 is satisfies for free, because by definition $A_0$ contains everything in $\mathbb{R}$ which hasn't been used up by an $A_i$ with $i > 0$ already.

So, it remains to check $3$. But $A_0$ is disjoint from each $A_n$ by construction. So, assume for a contradiction that we can find an $x\in A_n \cap A_m$ with $n > m > 0$. Then $x = q + x_n = p + x_m$ for some rationals $q$ and $p$.

This implies that $x_n = p-q + x_m$, so $x_n$ is an element of $\mathbb{Q}(x_1,...,x_m)$, contradicting the fact that $x_n$ is transcendental over $\mathbb{Q}(x_1,...,x_m)\subseteq \mathbb{Q}(x_1,...,x_{n-1})$.

Answer 2

Partition the strictly positive integers $\mathbb N$ into a sequence of sets $(S_n)_{n \ge 0}$, with the properties:

1. $S_n$ is infinite
2. The union of all $S_n$ is $\mathbb N$
3. No two $S_n$ share elements

For instance, we can set $S_n = $ all odd multiples of $2^n$.

Then the sets $A_n = \bigcup_{r\in S_n}[r-1,r) \cup (-r, -r+1])$ satisfy your four conditions.

Answer 3

Take your favorite countably-infinite partition of $[0,1)$ into dense non-empty subsets without maximum elements, say $(B_n)_{n\in \mathbb{N}}$ and define $A_n$ by $A_n=\{k+b \mid k\in \mathbb{Z} \wedge b\in B_n\}$. For example, given any strictly-decreasing sequence $(c_n)_{n\in \mathbb{N}}$ of elements of $[0,1)$ that converges to $0$ the sequence $([c_n,c_{n-1}))_{n\in \mathbb{N}}$ satisfies these conditions.

Since each $B_n$ is non-empty, it follows that for every real number $r>0$ there exists a natural number $n$ such that $n>r$ and thus given $b\in B_n$ we have $n+b>r$ and $-(n+1)+b<-r$, showing that each $A_n$ is unbounded.

Given any $r$ in $\mathbb{R}$ let $m=\lfloor r\rfloor$ and $\{r\}=r-m$ be the fractional part of $r$. Then $\{r\}\in [0,1)$ and there exists $n$ such that $\{r\}\in B_n$, and thus $r\in A_n$ since $r=m+\{r\}$. This shows that $(A_n)_{n\in \mathbb{N}}$ covers $\mathbb{R}$.

To see that $A_n\cap A_m=\emptyset$ for $n\neq m$, suppose for sake of a contradiction that there is $a\in A_n\cap A_m$. It follows that $\{a\}$ is an element of $B_n$ and $B_m$, where $\{a\}$ is the fractional part of $a$. This contradicts the assumption that $(B_n)_{n\in \mathbb{N}}$ is a partition of $[0,1)$ (which requires that $B_n\cap B_m=\emptyset$ for every $n\neq m$).

Finally, given $r,s\in A_n$ with $r<s$, let $r=k_1+b_1$ and $s=k_2+b_2$ where $k_1,k_2\in \mathbb{Z}$ and $b_1,b_2\in B_n$. Because $r<s$, it follows that either $k_1<k_2$ or $k_1=k_2$ and $b_1<b_2$. If the latter condition holds, then because $B_n$ is dense we find there is $b\in B_n$ such that $b_1<b<b_2$ and thus $t=k_1+b\in A_n$ satisfies $r<t<s$. If the former condition holds, then either $b_2=0$ or $b_2>0$. Let $b$ be any element of $B_n$ such that $b_1<b$, which exists by the assumption that $B_m$ has no maximum for any $m\in \mathbb{N}$. Then $t=k_1+b\in A_n$ and $r<t<s$. Thus, it follows that $A_m$ is densely ordered for every $m\in \mathbb{N}$.