See [[Jech - Set theory_ The third millennium edition]] pg. 557. **Theorem (Jensen).** There is a generic extension $L[a]$ of $L$ such that a is a $\Delta_3^1$ real. >[!note] Remark >By [[Shoenfield's Absoluteness Theorem]], every $\Pi_2^1$ or $\Sigma_2^1$ real is constructible; on the other hand $0^{\sharp}$ is a $\Delta_3^1$ real, but it cannot be obtained by forcing over $L$. # Definition Recall the definition of [[Sacks forcing|forcing with perfect trees]]. $\mathbb{P}$ is a set of perfect trees, ordered by $\subseteq$, which is carefully constructed using the $\diamondsuit$-[[Diamond|principle]] in $L$. A generic $G\subseteq \mathbb{P}$ corresponds to a branch $a\in 2^{\omega}$ such that in $L[a]$: - $a$ is the unique $\mathbb{P}$-generic - $\{ n<\omega \mid a(n)=1\}$ is a $\Delta_{3}^{1}$ definable set. %% , and ![[Sacks forcing#^stem|Sacks real]] >[!info] Definition > Let $T=$ $\{T(s): s \in \operatorname{Seq}(\{0,1\})\}$ be a collection of perfect trees. We say that $T$ is *fusionable* if that for every $s$, > 1. $T(s)$ is a perfect tree whose stem has length $\geq \operatorname{length}(s)$. > 2. $T\left(s^{\frown} 0\right) \subset T(s)$ and $T\left(s^{\frown} 1\right) \subset T(s)$. > 3. $T(s^\frown 0)$ and $T\left(s^{\frown} 1\right)$ have incompatible stems. > and we let the *fusion of $T$* be > $ > \mathcal{F}(T)=\bigcap_{n=0}^{\infty} \bigcup_{s \in\{0,1\}^n} T(s) . > $ For each fusionable $T, p=\mathcal{F}(T)$ is a perfect tree; and for each $s$, if $t$ is the stem of $p_s=T(s)$, then $p \upharpoonright t$ is stronger than both $p$ and $p_s$. %% # Properties - $\mathbb{P}$ and $\mathbb{P}\times\mathbb{P}$ satisfy [[Chain conditions#^ccc|ccc]]