Fact. Assume \(\mathbb{R}\neq \mathbb{R}^L\), then any uncountable \(\Pi^1_1\) set has a perfect subset if and only if it is not a subset of \(L\).

This is exercise-left-to-reader from

  • Mansfield, Richard. “Perfect subsets of definable sets of real numbers.” Pacific Journal of Mathematics 35, no. 2 (1970): 451-457.

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Proof. The Mansfield-Solovay Theorem states that if a \(\Sigma^1_2\) set has an unconstructible element, then it has a perfect subset coming from a perfect tree coded in \(L\) (and this relativizes). So this takes care of the right-to-left direction. On the other hand, if \(X\) is \(\Pi^1_1\) and has a perfect subset, then by Shoenfield absoluteness there is a perfect tree in \(L\) whose branches are (a subset) of the set defined by the same \(\Pi^1_1\) formula. Using any \(r\in 2^\omega \smallsetminus L\) as a guide for going left or right going down that tree, we can obtain a real that cannot be in \(L\). This is because if it were in \(L\), then we’d be able to recover \(r\) and that’s a contradiction.