TITLE: Is it possible to prove without choice axiom that the set of finite subsets of any infinite ordinal is injectable in itself?
QUESTION [7 upvotes]: To be more precise, let $\alpha$ be any infinite ordinal. I want to show that there exists an injection from the set of all finite subsets of $\alpha$ into $\alpha$. With the choice axiom this is clear, because the union of $\aleph_0 \leq | \alpha|$ many sets of cardinality $|\alpha^n| =|\alpha |$, is again of cardinality $\leq | \alpha |$ (by a well-known theorem that uses the choice axiom), and clearly the set of all finite subsets of $\alpha$ can be injected into $\bigcup_{1 \leq n \in \omega} \alpha^n$. But is this result still true without the choice axiom? I've been trying to define a bijection explicitly ("coding") but it doesn't seem to work out. I know such proofs for the case $\alpha = \omega$. But unless I am mistaken these do not seem to generalize straightforwardly to bigger ordinals.

REPLY [3 votes]: The axiom of choice is not needed to prove that the set of all finite subsets of an infinite ordinal $\alpha$ admits a one-to-one map into $\alpha$. Here's an overkill proof: The ordinal $\alpha$ and the set of its finite subsets are members of Gödel's constructible universe $L$. The axiom of choice holds in $L$, so $L$ contains an injection $f$ of the desired sort. But then $f$ is also an injection of the desired sort in the full universe $V$. 
I think I've also seen completely explicit constructions of injections (or even bijections) of the desired sort, but I can't find a reference just now. (By "completely explicit", I mean something better than just "the constructibly first such bijection".)