I was reading a paper by David Marker, whose main theorem was that if $T$ is a first-order theory which is not small, then $F_2\leq_B \cong_T$. That's not especially relevant to the question at hand, except to point out that it comes with no set-theoretic hypotheses and no "it is consistent that..." hedging. So the techniques involved shouldn't involve forcing without some sort of explanation and/or absoluteness argument, I would expect.

Anyway, along the way we have the following lemma:

For any countable set of Borel functions $\mathcal{F}$, there is a perfect $\mathcal{F}$-independent set.

Note here he's talking about functions on and subsets of Cantor space, so it should be applicable to any uncountable Polish space. The supplied proof is the following (verbatim):

If $P$ is a perfect set of suitably generic Cohen reals, then $P$ is $\mathcal{F}$-independent.

Now, I have no problem believing the lemma is true- there are a lot of reals, they tend to be transcendental over each other, and so on. But I have no idea what the proof means. When someone says "Cohen reals," my first thought is to say we can force this to be true, which feels believable, but that doesn't seem to be what he's saying.

So I have two questions.

- What is meant by "Cohen reals" in this context?
- How can we guarantee there is a perfect set of "suitably generic" ones?

I would be happy with a citation, too; no need to type out five pages of math if there's a published source out there that does a good job. But Google and my own math experience are giving me the same (unhelpful) answer.

*Note:* A set $I$ is said to be $\mathcal{F}$-indepedent if, for all $A\subseteq I$, $cl_{\mathcal{F}}(A)\cap I=A$.

That is, given some elements of $I$, and all the functions of $\mathcal{F}$, you can't get any *new* elements of $I$ just by using terms from $\mathcal{F}$. We are not concerned with any other notions of independence or closure (be they model-theoretic or topological).