Prove that a set is denumerable if and only if the set is equipotent to any of its infinite subsets I have shown the 'only if' part. Can anyone give a nice proof for 'if' part?
 A: For the "if" part: Let $S$ be equipotent to each of its infinite subsets. Either $S$ is finite in which case we're done or $S$ is infinite. Let's now consider the case where $S$ is infinite: Then $|S| \geq |N|$. Therefore there is an injective function $f: N \rightarrow S$ and $f(N) \subset S$ and $f(N)$ is infinite and $|f(N)| = |N|$. But then because $S$ is equipotent to $f(N)$ it follows that $|S| = |f(N)| = |N|$.
A: $\Longrightarrow$: Let $A$ be a denumerable set. Then $|A|=\aleph_0$; let $B$ be an infinite subset of $A$.
Since $B\subseteq A$, $|B|\leq|A|=\aleph_0$. Since $B$ is infinite, $|A|=\aleph_0\leq|B|$
By the Cantor-Bernstein theorem, $B$ has the cardinality of $A$
$\Longleftarrow$: There are several cases:


*

*$A$ is finite


Then obviously none of its proper subsets will be equipotent with $A$


*

*$A$ is infinite


Then $A$ is not empty, say $a_0\in A$. Then the proper subset $B=A\setminus\{a_0\}$ is equipotent with $A$ by hypothesis; let $f:A\longrightarrow B$ be an bijective function from $A$ onto $B$. Let $g:\mathbb{N}\longrightarrow A$ be the sequence defined by recursion:
$$\begin{cases}
  g(0)=a_0\\
  g(n+1)=f(g(n))
\end{cases}$$
We will prove that $g$ is injective. so $\text{Im}(g)$ will be a countably infinite subset of $A$.
Suppose that $g$ is not injective. Then there exists a couple of natural numbers $m,n$ with $m<n$ such that $g(m)=g(n)$. We have that the set
$$\mathcal{N}=\{n\in\mathbb{N}\;|\;\text{there exists }m>n\text{ such that }g(m)=g(n)\}$$
is a nonempty subset of $\mathbb{N}$. Let $n_0$ be its first element.


*

*In first place, $n_0\not=0$, because $g(0)=a_0,\;a_0\not\in B$, and if $n>0,\;g(n)\in\text{Im}(f)=B$

*Since $n_0>0$, we have that if $m\in\mathcal{N}$ is such that $m>n_0$ and $g(n_0)=g(m)$, then


$$g(n_0)=f(g(n_0-1))=f(g(m-1))=g(m)$$
And, since $f$ is injective
$$g(n_0-1)=g(m-1)$$
Which contradicts the choice of $n_0$ as the first element of $\mathcal{N}$
