Countable subsets of reals injecting into reals implies $\omega_1$ distinct reals I am working through this problem: work in ZF, assume there is an injection $f:\mathcal{P}_{\omega_1}(\mathbb{R})\rightarrow \mathbb{R}$, show that there are $\omega_1$ distinct reals. Here, $\mathcal{P}_{\omega_1}(\mathbb{R})$ denotes the set of all countable subsets of reals.
I think I have the right approach but I am stuck on finishing off the argument. Here's my attempt: without loss of generality we identify $\Bbb R$ with $\mathcal{P}(\omega)$; there is a surjection $\pi: \mathcal{P}(\omega)\rightarrow \omega_1$ by mapping codes of well-orderings to their ordertypes, and non-codes to $0$. Then, by taking pointwise preimage of $\pi$, there is an injection $g:\omega_1\to \mathcal{P(P}(\omega))$. I think the next thing to do is to somehow restrict the range of $g$ to $\mathcal{P}_{\omega_1}(\mathcal{P}(\omega))$ and then compose it with the injection we assume exists. But I don't know how to proceed.
 A: In order to build a well-ordered sequence of reals (which is how "$\omega_1$-many reals" should be thought of in the $\mathsf{ZF}$ setting), the most natural tool to look at is transfinite recursion. Without the axiom of choice, this is really the only good way we have of building long well-ordered sequences.
So suppose $f:\mathcal{P}_{\omega_1}(\mathbb{R})\rightarrow\mathbb{R}$ is injective. Do you see a way to "iteratively" use $f$ to build a long sequence of reals?
HINT: start with $f(\emptyset)$ ...

 We define a sequence of reals $(a_\eta)_{\eta<\omega_1}$ recursively as follows: $$a_\eta=f(\{a_\beta:\beta<\eta\}).$$

This $\omega_1$-sequence of reals is clearly well-defined, so all we have to do is show that it's injective: no real appears twice in the sequence. This is where we use the injectivity of $f$; think about how a real arises as an element of this sequence.

Note that here I've used the broad notion of countable, where finite sets are considered countable. What if we don't? That is what if $f$ only acts on countably infinite sets of reals?
Well, then we hit a bit of a problem: we have to worry about "stalling." Since our recursion can't start at $\emptyset$ anymore, we have to start by picking some countably infinite $X\subset\mathbb{R}$ to feed into $f$. The problem now is if $f$ spits out an element of $X$ - either now, or later on down the road - we'll stop generating new elements of our sequence. Put another way, we need to find an $X$ such that the recursively-defined sequence $$r_\alpha=f(\{r_\beta:\beta<\alpha\}\cup X)$$ never hits $X$.
If no such $X$ exists, then we can extract from $f$ a choice function for countably infinite sets of reals: just take $r_\alpha$ for the least $\alpha$ which yields an element of $X$. So the "strong" version of the question reduces to:

Does "There is a choice function for the set of countably infinite sets of reals" imply "There is an $\omega_1$-sequence of distinct reals?"

This actually seems nontrivial to me - by which I mean I don't know how to solve it. :P Presumably Asaf knows ...
