I wonder whether the following question have a positive answer within $ZFC$.



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> **Question** If $\{A\_n\}\_{n\in \omega}$ is a sequence of analytic sets so that $\bigcup\_n A\_n=2^{\omega}$, then there must be some $n$ so that $A\_n$ has a pointed subset.
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A pointed set is a perfect set $P$ of reals in which every member computes the representation of $P$.


Note that $PD$ implies a positive answer of the question.



Analytic sets and Turing determinacy
I believe this fails under $V=L$. If $P$ is a pointed perfect set and $X$ is a real, let $P(X)$ be the element of $P$ where at every split we choose according to the next bit of $X$. So $X\oplus P \equiv\_T P(X)$.


We'll build the $A\_n$ via an $\omega\_1$ length construction as follows. At stage $\alpha$, we consider $P\_\alpha$ the next pointed perfect set, and we choose $X\_0, X\_1, \dots$ which are self-constructible reals of $L$-rank at least $\alpha$ and computing $P\_\alpha$. We omit $P\_\alpha(X\_n)$ from $A\_n$.


We define $A\_n$ to consist of those reals which are never omitted. So $Y \not \in A\_n$ iff there is a well-founded $L\_\alpha \in \Delta^1\_1(Y)$ realizing $Y$'s omission, which is $\Pi^1\_1$. By appropriate choice of $X\_n$, no real is omitted from more than one $A\_n$, and so $\bigcup\_n A\_n = 2^\omega$. But clearly we have diagonalized against any $A\_n$ containing a perfect pointed set.


