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# 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Application 2: Nonconstructible Self-Definable $\Delta_n^1$ Reals

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## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Application 2: Nonconstructible Self-Definable $\Delta_n^1$ Reals

Note that the set $a$ as in Theorem 2(i) is definable in the generic extension of $\mathbf{L}$, considered in Section 7.2, by means of other reals in that extension, including those which do not necessarily belong to $\mathbf{L}[a]$. Claim (ii) of Theorem 2 achieves the same effect with the advantage that $a$ is definable inside $\mathbf{L}[a]$.

The key idea (originally from [9] Section 4) can be explained as follows. Recall that a set of the form $a_0=a_G(0)$ was made definable in a generic extension of the form $L\left[G \mid z_G\right]$ by means of the presence/absense of other sets of the form $S_G(v), v<\omega$, in $\mathbf{L}[G \mid z]$, see Sections 7.2 and 7.3. Our plan will now be to make each of the according sets $a_G(v) \in \mathbf{L}[G \mid z]$ (note that $a_G(v) \subseteq \omega \backslash{0}$, see Definition 9), as well as the whole sequence of them, $\Delta_{n+1}^1$-definable in $\mathbf{L}[G \mid z]$. In order to do this, we need to develop a suitable coding construction.

Assumption 4. We continue to assume $\mathbf{V}=\mathbf{L}$ in the ground universe. We fix an integer $\mathrm{n} \geq 2$, for which Theorem 1(ii) will be proved, and make use of a system $\mathbb{U}$ and the forcing notion $\mathbb{P}=\mathbf{P}[\mathbb{U}]$ as in Definition 16; both $\mathbb{U}$ and $\mathbb{P}$ belong to $\mathbf{L}$.

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Nonconstructible Self-Definable $\Delta_{n+1}^1$ Reals: The Model

Here we begin the proof of Theorem 2(ii). Recall that $\omega^\omega=\left{s_k: k<\omega\right}$ is a fixed recursive enumeration of strings of natural numbers, such that $s_0=\Lambda$, the empty string, and $s_k \subseteq s_{k^{\prime}} \Longrightarrow k \leq k^{\prime}$. Let $\ell_i^k=\operatorname{num}\left(s_k \frown i\right)$, thus $s_{\ell_i^k}=s_k \frown i$. Then we have:

Each set $L(k)=\left{\ell_i^k: i<\omega\right} \subseteq \omega$ is countably infinite, $k<\min i \ell_i^k$, $k \neq k^{\prime} \Longrightarrow L(k) \cap L\left(k^{\prime}\right)=\varnothing$ and $i \neq j \Longrightarrow \ell_i^k \neq \ell_j^k$, and finally each $m \geq 1$ is equal to $\ell_i^k$ for exactly one pair of indices of $i, k<\omega$. Define a partial order $\ll$ on $\omega$ so that $i \ll k$ iff $s_i \subset s_k$. Obviously $k \ll \ell_i^k$ for all $i, k \in \omega$, and 0 is the $\ll$-least element. For any sequence $\vec{a}=\left{a_k\right}{k<\omega}$ of sets $a_k \subseteq \omega$, we define a set $\zeta_{\vec{a}} \subseteq \omega$ so that:
1) $0 \in \zeta_{\vec{a}} ;$
2) if $k \in \zeta_{\vec{a}}$ then, for every $i$ : if $i \in a_k$ then $\ell_{2 i}^k \in \zeta_{\vec{a}}$ and $\ell_{2 i+1}^k \notin \zeta_{\vec{a}}$, but if $i \notin a_k$ then $\ell_{2 i}^k \notin \zeta_{\vec{a}}$ and $\ell_{2 i+1}^k \in \zeta_{\vec{a}}$;
3) if $k \notin \zeta_{\vec{a}}$ then $\ell_i^k \notin \zeta_{\vec{a}}$ for all $i$.
The next theorem obviously implies Theorem 2(ii).

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Nonconstructible Self-Definable $\Delta_{n+1}^1$ Reals: The Model

1) $0 \in \zeta_{\vec{a}} ;$
2)如果$k \in \zeta_{\vec{a}}$则对每一个$i$:如果$i \in a_k$则$\ell_{2 i}^k \in \zeta_{\vec{a}}$和$\ell_{2 i+1}^k \notin \zeta_{\vec{a}}$，如果$i \notin a_k$则$\ell_{2 i}^k \notin \zeta_{\vec{a}}$和$\ell_{2 i+1}^k \in \zeta_{\vec{a}}$;
3)如果$k \notin \zeta_{\vec{a}}$，那么$\ell_i^k \notin \zeta_{\vec{a}}$为所有$i$。

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