Size of infinite sets equality #$A$=#($A\cup B$) and counterexample for $A\cap B\ne\emptyset$

Question

Let $A$ be an infinite set and $B$ a countable set and let them be disjoint.

Then there exists an injection $j:\Bbb N\hookrightarrow{} A$.

We can call its image $C:=j(\Bbb N)\subset A$. Then we have that $\Bbb N\cong C$ ($C$ is equipotent $i.e.$ there is a bijection between the two sets).

Since $B$ is countable and $C$ is countable and infinite, $B\cup C$ is countable and infinite.

So $B\cup C\cong \Bbb N\cong C\implies\exists \varphi:B\cup C\xrightarrow{\text{bij}} C$

Now I have two questions:

1. Why is the function defined as $$\psi:A\cup B\rightarrow A\ , x\rightarrow\begin{cases}\varphi(x)\ \text{if}\ x\in C\cup B\\ x\ \text{if}\ x\in A\end{cases}$$ bijective?

2. Is there an example of $A,B$ that are not disjoint and $A\cup B \ncong A$?

Answer 1

To answer question 1, note that $\varphi \colon A \cup B \to A$ is bijective when we view $\varphi$ as the restriction onto $B \cup C$, which is bijective by construction. Now, since $A \cap B = \emptyset$ then $x \in A \iff x \notin B$. Therefore $\psi$ is just the identity map on $A$, so it is bijective on $A$ as well.

Answer 2

Perhaps this will clarify the ideas : Let $j:\Bbb N\to C\subseteq A$ be injective. Let $B$ be countable.

Since $B\setminus A$ is countable there exists $M\subset \Bbb N$ such that (i) $B\setminus A$ is bijective to $M$ and (ii) $\Bbb N \setminus M$ is infinite. (E.g. if $B\setminus A$ is infinite let $M=\{2n: n\in \Bbb N\}$).

$(I).$ Let $f(x)=x$ for $x\in A\setminus C.$

$(II).$ Let $f$ map $C$ bijectively to $\{j(n):n\in \Bbb N\setminus M\}$... This is possible because $C$ and $\{j(n):n\in \Bbb N\setminus M\} $are each countably infinite.

$(III).\;$ $B\setminus A$ is bijective to $M$ and $M$ is bijective to $\{j(n):n\in M\},$ so we can let $f$ map $B\setminus A$ bijectively to $\{j(n):n\in M\}.$

Then $f$ maps $A\cup (B\setminus A)=A\cup B$ bijectively to $A.$