The following fact, which I've heard being called "soft version of Moschovakis's lemma" (see top answer here) is the following:

Under AD, if there is a surjection $\Bbb R\rightarrow\alpha$, then there is a surjection $\Bbb R\rightarrow\mathcal{P}(\alpha)$.

I remember few days ago I have seen a really nice proof of this result, which directly used AD (and possibly DC, can't remember) and not some complex workaround using scales and things like that. However, when yesterday I wanted to show the proof to a friend of mine, I couldn't have find it anywhere, and I would be very thankful if anyone pointed out where I could have seen this proof.

The rough idea of the proof is the following:

- Fix a surjection $f:\Bbb R\rightarrow\alpha$.
- For each $X\subseteq\alpha$, using $f$, define a game $G(X)$ on $\omega$.
- Show that, for any $X\neq Y$, no winning strategy for $G(X)$ is a winning strategy for $G(Y)$.
- Thus, the function $F:\Bbb R\rightarrow\mathcal{P}(\alpha)$, defined as $F(r)=X$ if $r$ codes a winning strategy for $G(X)$, and $F(r)=\varnothing$ if $r$ codes no winning strategy, is well-defined.
- By AD, every $G(X)$ has a winning strategy for some player, so $F$ is surjective.

Hope this helps to find the proof I meant. Thanks in advance.

specificproof of the fact. $\endgroup$ – Wojowu Jun 15 '15 at 5:58