Members of (lightface) Borel sets I'm fairly certain this question has a very simple answer, and that I've learned it before; I just can't seem to remember it.
Suppose I have a nonempty lightface Borel set $X\subseteq 2^\omega$. What is something we can say about how simple some member of $X$ must be?
For example, if $X$ is lightface $\Pi^0_1$, then by the low basis theorem $X$ has a low element, and if $X$ is lightface $\Sigma^0_1$, $X$ has a computable element by obviousness.
My recollection is that every nonempty lightface Borel subset of $2^\omega$ has a hyperarithmetic member, but I can't seem to prove this myself or find this result in Moschovakis' book, so I'm beginning to suspect I'm wrong.
(I'd also like to know - if my recollection is correct - when was this proved, and by whom?)


Does every nonempty lightface Borel subset of $2^\omega$ have a hyperarithmetic member?

There is an arithmetical subset of $2^\omega$ with no hyperarithmetical member.
It is known that there is a $\Pi^0_1$ class $P$ in $\omega^\omega$ which has no hyperarithmetical member (Theorem III.1.1, Sacks, Higher Recursion Theory). 
Let $Q$ be subclass of $2^\omega$ such that $A \in Q$ if and only if $A$ is the graph of a total function $f_A\colon\omega\to\omega$, under some canonical effective pairing operation, and $f_A \in P$. It is straightforward to check that $Q$ is arithmetical, using only the fact that $P$ is arithmetical. If there was a hyperarithmetical set $B$ in $Q$ then the function $f_B$ would also be hyperarithmetical, which is impossible. 
The key point here is that $\omega^\omega$ is recursively isomorphic (and homeomorphic) to a $\Pi^0_2$ subclass of $2^\omega$. We can use either the class of graphs of total functions, as above, or the class of infinite subsets of $\omega$. Thus most descriptive set theory results for $\omega^\omega$ transfer to $2^\omega$ as soon as we are willing to accept $\Pi^0_2$ quantification, for example if we are looking at arithmetical or Borel classes. If we limit ourselves to $\Pi^0_1$ classes in $2^\omega$ we get stronger results than can be obtained for $\omega^\omega$, because nonempty $\Pi^0_1$ subsets of $2^\omega$ are compact. 
