If $A\sim B$(both dedekind infinite), is it then that $A\sim B\cup \{x\}$

Question

If the symbol $A\sim B $signifies that there is a bijection between A and B, and We take our sets to be dedekind infinite, then is the following correct? If not, what is the counter example?:$$A \sim B \Rightarrow A \sim B\cup\{x\}$$

Answer 1

Let $f : A \to B$ be a bijection, and fix an injection $g : \mathbb{N} \to A$, which exists since $A$ is Dedekind-infinite, and let $a_n=g(n)$ for each $n \in \mathbb{N}$.

Define $f' : A \to B \cup \{ x \}$ by letting $$f'(a) = \begin{cases} f(a) & \text{if } a \not\in \mathrm{im}(g) \\ x & \text{if } a=a_0 \\ f(a_{n-1}) & \text{if } a=a_n \text{ for some } n>0 \end{cases}$$ You need to prove that $f'$ is a bijection.

Answer 2

Assuming the axiom of choice, cardinality is a total order; equivalently, every set is well ordered. Now think of ordinals and each element associated with exactly one ordinal.