Prove that $|A|\ge |B|$, as $A$ and $B$ are two sets described as follows:
$A=\{A_i \vert \ i \in \mathbb{N}\}$, such that $\forall i,j\in\mathbb{N}:i\neq j: \ A_i\neq A_j\land A_i\cap A_j=\emptyset$,
$B$ is a set constructed of open intervals on $\mathbb{R}$,
such that for every $b_1, b_2 \in B: b_1 \cap b_2 = \emptyset$, $\forall b\in B: \ b = (x,y)\subseteq \mathbb{R}$ ($x\neq y$)
Prove: $|A| \ge |B|$, or in other words, Prove there exists a injective function $f: B \to A$.
My attempt:
As $B$ is a collection of open disjoint intervals on $\mathbb{R}$, then I'd like to define a partition over $\mathbb{Q}$ using the elements of $B$. Let this partition be called $\pi_{B}$.
By the definition of a partition, then $|\pi_B| = |\mathbb{Q}| =\aleph_0$
This allows me to know that there exists a bijection: $g: \pi_{B} \to A$, and as $\pi_B$ is defined using elements of $B$, then there exists an injective function $h: \pi_{B} \to B$.
I'd like to find a way to define a composition function of $h$ and $g$, such that this composition will be a one to one function: $f: B \to A$, and from that it will be possible to conclude that $|A| \ge |B|$.
But I don't know how to do such thing.
I will assume that $B$ contains no empty intervals.
Note that every interval $b\in B$ contains at least one rational number. This follows as $b$ is open and non-empty, hence we can find at least two elements in $b$, as a singleton set is closed. Since there is always a rational number between every two real numbers we can use the axiom of choice to find a map $f:B\rightarrow\mathbb{Q}$ such that $f(b)\in b$ for all $b\in B$. Note that this map is unique as the intervals in $B$ are pairwise disjoint. Let $g:\mathbb{Q}\rightarrow\mathbb{N}$ be an injection from $\mathbb{Q}$ to $\mathbb{N}$. Then $$h:B\rightarrow A,b\mapsto A_{g(f(b))}$$ is an injection.
To do this without the axiom of choice let $(a,b)$ be an open interval, Let $n$ be the smallest number such that $2^{-n}\leq b-a$. Then the set $$X_{(a,b)}=\{z2^{-n}\in (a,b):z\in\mathbb{Z}\}$$ is non-empty and we can define $f((a,b))=\min X_{(a,b)}$.