Question about countable and uncountable map correspodence So I was solving the following question:

If $B$ is uncountable with countable subset $A\subset B$, prove that there exists a one-to-one correspondence between $B$ and $B-A$.

So here is how I proved it:

Since $B$ is uncountable, $B-A$ is uncountable, so there exists a countable subset $G \subset B-A$, and we can write $G=\{g_1,g_2,\cdots\}$ and $A=\{a_1,a_2,\cdots\}$. Define $\phi:B\to B - A$ by
  $$\begin{cases}\phi(b) = b & b \in B - (G \cup A) \\
\phi(a_i) = g_{2i} & a_i \in A \\
\phi(g_i) = g_{2i + 1} & g_i \in G \end{cases}$$
  By definition it is $1$-$1$, and it is clear that is onto as $\big(B-(G\cup A)\big)\cup A\cup G = B$.

Now I want to discuss a generalization. Suppose we have some countable family $F$ of subsets of $B$ and then consider $B-\cup F$. Can we prove there is $1$-$1$ correspondence between $B-\cup F$ and $B$ by expanding along $G$ along coprime powers? Also, how far can we stretch it until it break down? That is, if we consider some arbitrary union of countable sets that is inside $B$ can we still prove $B-\cup F$ is one-to-one correspondence to $B$? 

Let $B$ be an uncountable set and for each $n \in \mathbb N$ let $A_n = \{a_1^n, a_2^n, \ldots \} \subseteq B$ be pairwise disjoint. Let $G = \{g_1,g_2, \ldots\}$ be a countable subset of $B$ disjoint from all $A_n$. Let $\mathbb P = \{p_1,p_2, \ldots\} \subseteq \mathbb N$ be the set of prime numbers. Let $D = \{d_1, d_2, \ldots \} =\mathbb N \setminus \{p_i^k \mid i,k \in \mathbb N \}$. (Note that $D$ is in fact countably infinite.)
Then
$$
\phi: B \rightarrow B \setminus \bigcup_{n=0}^\infty A_n , x \mapsto  \begin{cases} g_{p_k^i} \text{, if } x = a_i^k \\
g_{d_i} \text{, if } x = g_i \\
x \text{, otherwise} \end{cases}
$$
defines a bijection.
(Note that in order to list all the $A_n$, I implicitly used countable choice.)
