Does there exist this type of sequence of subsets of $\mathbb{R}$? Let $A_n$ be a sequence of subsets of $\mathbb{R}$ with the following properties.


*

*$A_n$ is unbounded for all $n$

*The union of all $A_n$ is $\mathbb{R}$

*No two $A_n$ share elements

*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.

Here are the definitions for the properties. Please let me know if you can spot mistakes.
$$ \tag{1}{(\forall ~ n ~ \nexists ~ M \in \mathbb{R} ~ \forall ~ x \in A_n : M \geq x) \land (\forall ~ n ~ \nexists ~ m \in \mathbb{R} ~ \forall ~ x \in A_n : m \leq x)} $$
$$ \tag{2}{\bigcup_n A_n = \mathbb{R}}$$
$$ \tag{3}\forall ~ m \neq n : A_m \cap A_n = \emptyset $$
$$ \tag{4}\forall ~ n ~ \forall ~ x_1, x_2 \in A_n ~ \exists ~ y \in A_n : x_1 < x_2 \implies x_1 < y < x_2 $$
 A: 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 $\mathbb{Q}(x_1,...,x_m)\subseteq \mathbb{Q}(x_1,...,x_{n-1})$.
A: Partition the strictly positive integers $\mathbb N$ into a sequence of sets $(S_n)_{n \ge 0}$, with the properties:


*

*$S_n$ is infinite  

*The union of all $S_n$ is $\mathbb N$

*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.
A: 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 $m\in \mathbb{N}$.
