Why is this true? A set is Dedekind infinite if and only if equipotent to the union with the set containing itself Put in notation: 
($A$ is Dedekind infinite)$\iff$($A\approx A\cup \{A\}$)
The book I'm reading has not yet covered the axiom of choice, it states that the above follows immediately from the definition, which it states as "A set is Dedekind infinite if and only if it is not Dedekind finite". 
The definition given for a set being Dedekind finite is "A set is Dedekind finite if and only if it is not equipotent to any of its proper subsets". 
I cannot see how this immediately follows, could anyone help? 
Book is "Axiomatic set theory" Suppes, 1970.
 A: If $h:A\to A\cup\{A\}$ is a bijection, the restriction of $h^{-1}$ to $A$ is a bijection between $A$ and a proper subset of $A$, so $A$ is Dedekind-infinite.
Now suppose that $A$ is Dedekind-infinite, and let $h:A\to A$ be injective but not surjective. Fix $a_0\in A\setminus h[A]$, and recursively define a sequence $\langle a_n:n\in\Bbb N\rangle$ in $A$ by setting let $a_{n+1}=h(a_n)$ for each $n\in\Bbb N$. Since $h$ is injective, so is the sequence: $a_m\ne a_n$ if $m,n\in\Bbb N$ and $m\ne n$. Now define
$$f:A\cup\{A\}\to A:x\mapsto\begin{cases}
a_0,&\text{if }x=A\\
a_{n+1},&\text{if }x=a_n\text{ for some }n\in\Bbb N\\
x,&\text{otherwise}\;;
\end{cases}$$
it’s easy to verify that $f$ is a bijection.
Added: By the way, this proves the harder direction of another useful characterization of Dedekind-infinite sets: $A$ is Dedekind-infinite if and only if there is an injection $f:\Bbb N\to A$.
A: $A$ is Dedekind finite if every injection $f:A \rightarrow A$ is a surjection. So a set is Dedekind infinite if there is an injection $f:A \rightarrow A$ which is not a surjection. 
Let $A$ be Dedekind infinite.
Now, there is an obvious injection from $A$ to $A \cup \{A\}$: The inclusion. So, to show that $A \cong A \cup \{A\}$ it suffices to find an injection from the right to the left (by Schroeder-Bernstein). 
Let $f$ be the injection that is not a surjection mentioned above. Now let $a \in A \setminus range(f)$. You can extend $f$ to $\hat{f}:A\cup \{A\}\rightarrow A$ by defining $\hat{f}(x) = f(x)$ if $x \in A$ and $\hat{f}(x) = a$ if $x = A$.
$\hat{f}$ is an injection, so we have $A \cong A \cup \{A\}$.
For the other direction, the restriction of any bijection $g: A\cup \{A\} \rightarrow A$ to $A$ is an injection from $A$ to $A$ which is not onto.
