Cardinality of the set of all finite subsets of an infinite set. Let $I$ be an infinite set. And let $\mathcal{I}$ be a set of finite subsets of $I$, this is $\mathcal{I}=\{J\subset I: |J|<\infty\}$. It is a known fact that  $|I|=|\mathcal{I}|$. This is equivalent to saying that there exists injective functions $f:I\longrightarrow \mathcal{I}$ and $g:\mathcal{I}\longrightarrow I$. One is very easy to find:
\begin{align*}
f:I&\longrightarrow \mathcal{I}\\
i&\longmapsto\{i\}
\end{align*}
But how can I define an injective function:
\begin{align*}
g:\mathcal{I}&\longrightarrow I\\
J&\longmapsto g(J)
\end{align*}
Any hint or idea would be appreciated.
 A: This fact, in the generality presented here, is equivalent to the axiom of choice. So a simple injection from $\cal I$ into $I$ cannot be written out explicitly when you're taking the whole generality.

Let's limit ourselves to sets that can be well-ordered, or rather to ordinals, or rather to initial ordinals (since we care about cardinality). Assuming the axiom of choice every set is equipotent with an initial ordinal, so that's fine.
If $\omega_\alpha$ is an initial ordinal, then every finite set can be thought of as a finite sequence. The knee-jerk reaction is to look at the ordinal $(\omega_\alpha)^\omega$, here the exponent is an ordinal exponentiation. That's absolutely fine, and we can prove that if $\alpha$ is an infinite ordinal then $\alpha$ and $\alpha^\omega$ have the same cardinality (again, ordinal exponentiation!)
But I want to go to a different route. Let's look at $\omega^{\omega_\alpha}$. We can prove, quite easily depending on the definitions and theorems at hand, that this ordinal is $\omega_\alpha$ itself. So it is enough to find an injection from the finite subsets of $\omega_\alpha$ into $\omega^{\omega_\alpha}$.
A direct definition of $\alpha^\beta$ is the order type of finite partial functions from $\beta$ to $\alpha$ which are decreasing. But this gives us a dead giveaway for an injection: given a finite set $J\subseteq\omega_\alpha$, write it in increasing order as $\{\xi_1,\dots,\xi_n\}$, and then simply map $\xi_i$ to $i$.
Therefore we found an injection from $\cal I$ to $\omega^{\omega_\alpha}$ as wanted.
