I need to prove that for countably infinite sets $A_1,A_2,…,A_m\subset\mathbb{R}$, $\bigcup A_i$ is also countable.
My Attempt: First, consider two countable sets $A_1$ and $A_2$, and define $B_2=A_2\backslash A_1=\lbrace x\in A_2 | x\notin A_1\rbrace$. We then have two cases, one where $|B_2|$ is infinite, and one where $|B_2|$ is finite. If $|B_2|$ is inifite, then $B_2$ is countable, because there exists natural numbers $n_i$ and a function $f:\mathbb{N}\rightarrow A_2$ such that $f(n_i)=b_i$ for all $b_i\in B_2$. Then, a function $g:\mathbb{N}\rightarrow B_2$ can be defined such that $g(k)=f(n_k)$, proving that $B_2$ is countable. We now prove that the union $A_1\cup B_2=A_1\cup A_2$ is countable. Because $A_1$ and $B_2$ are countable, there exists bijective functions $f_1:\mathbb{N}\rightarrow A_2$ and $f_2:\mathbb{N}\rightarrow B_2$. Let $O_k$ be the $k$th odd natural number, and $E_k$ be the $k$th even natural number. We can simply define a new funtion $g:\mathbb{N}\rightarrow A_1\cup A_2$ where $g(O_k)=f_1(k)$ and $g(E_k)=f_2(k)$. This proves that $A_1\cup A_2$ is countable for the case $|B_2|=\infty$. Next, let $|B_2|=k$ and label each element $b_1,b_2,…,b_k$. Because $A_1$ is countable, we have the same bijective function $f_1:\mathbb{N}\rightarrow A_1$. To prove that $A_1\cup B_2$ is countable. we define a new function $g:\mathbb{N}\rightarrow A_1\cup A_2$ where $g(n)=b_n$ and $g(k+n)=a_n$ for all $a_n\in A_1$. This completes the proof.
I would like to know first if this proof is correct, and second if there is a more efficient way to write all of this.
Best Answer
With a cursory examination, it looks like the OP's proof is in the ballpark. They asked for a proof that would be more efficient, so here we will highlight the main ideas found in the OP's proof, presented as a logical progression.
In what follows we do not assume that our sets are contained in $\mathbb R$.
Proposition 1: If $A_1$ and $A_2$ are two disjoint countably infinite sets, then the union $A_1 \cup A_2$ is countably infinite.
Proof:
Let $F_1: \mathbb N \to A_1$ and $F_2: \mathbb N \to A_2$ be bijective mappings, where $\mathbb N = \{0, 1, 2, \dots, n, \dots \}$. Define $F: \mathbb N \to A_1 \cup A_2$ as follows:
$$ F(m) = \left\{\begin{array}{lr} F_1(\frac{m}{2}), & \text{for even } m \in \mathbb N\\ F_2(\frac{m+1}{2}-1), & \text{for odd } m \in \mathbb N \end{array}\right\} $$ It is easy to check that $F$ describes a bijective correspondence. $\quad \blacksquare$
The following sequence of 'add-on' propositions are not difficult to prove:
Proposition 2: Let $A_1$ and $A_2$ be any two sets. Then there exist a set $B$ satisfying the following:
$\tag 1 B \subset A_1 \text{ and } B \subset A_2$
$\tag 2 A_1 \cup A_2 \text{ is the disjoint union } (A_1 \backslash B) \cup B \cup (A_2 \backslash B)$
Proposition 3: If $A_1$ is a countably infinite set and $A_2$ is any finite set, then the union $A_1 \cup A_2$ is countably infinite..
Proposition 4: Any subset of a countably infinite set is either finite or countably infinite.
Proof (sketch)
Use the lemma found here.
Proposition 5: The union of a finite number of finite sets is finite.
We now prove the main proposition.
Proposition 6: The union of any two countably infinite sets $A_1$ and $A_2$ is countably infinite.
Proof:
By proposition 2, 4 & 5, $A_1 \cup A_2$ can be written as a disjoint union
$\tag 3 C_1 \cup B \cup C_2$
where each of the sets is either countably infinite or finite, and at least one of the sets is countably infinite.
By using the commutativity and associativity laws of the union operation and proposition 1 and/or proposition 3, we can simplify (3) in two steps so that a single countably infinite set remains that is equal to $A_1 \cup A_2$. $\quad \blacksquare$
The OP might want answer this question again at some point using the following:
Theorem: If $f$ is a surjective function from $\mathbb N$ onto a set $A$, then $A$ is either finite or countably infinite.