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# 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Multi-Lipschitz Transformations

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## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Multi-Lipschitz Transformations

Still arguing in $\mathbf{L}$, we let $\mathbf{L I P}^{\mathcal{I}}$ be the $\mathcal{I}$-product of the group LIP (see Section 2.3), this will be our second family of transformations, called multi-Lipschitz. Thus, a typical element $\lambda \in \mathbf{L I P}^{\mathcal{I}}$ is $\lambda=\left{\lambda_v\right}_{v \in|\lambda|}$, where $|\lambda|=\operatorname{dom} \lambda \subseteq \mathcal{I}^{+}$has $\omega_1$-size, $\lambda_v \in \operatorname{LIP}, \forall v$. Define the action of any $\lambda \in \operatorname{LIP}^{\mathcal{I}}$ on:

• systems $U:|\lambda \cdot U|:=|U|$, and $(\lambda \cdot U)(v):=\lambda_v \cdot U(v)$ for all elements $v \in|\lambda| \cap|U|$, but $(\lambda \cdot U)(v):=U(v)$ for all $v \in|U|>|\lambda|$;
• conditions $p \in \mathbf{Q}^:|\lambda \cdot p|^{+}=|p|^{+}$, if $-1 \in|p|^{+}$then $\boldsymbol{b}_{\lambda \cdot p}=\boldsymbol{b}_p$, if $v \in|p| \cap|\lambda|$ then $(\lambda \cdot p)(v)=\lambda_v \cdot p(v)$, but if $v \in|p| \backslash|\lambda|$, then $(\lambda \cdot p)(v)=p(v)$; $-\operatorname{sets} G \subseteq \mathrm{Q}^: \lambda \cdot G:={\lambda \cdot p: p \in G}$;
• names $\tau \in \mathbf{S N}_\omega^\omega\left(\mathbf{Q}^*\right): \lambda \cdot \tau:={\langle\lambda \cdot p,\langle n, k\rangle\rangle:\langle p,\langle n, k\rangle\rangle \in \tau} ;$
In the first two items, we refer to the action of $\lambda_V \in$ LIP on sets $u \subseteq$ FUN and on forcing conditions, as defined in Section 2.3.

Lemma 15 (routine). If $\lambda \in \mathbf{L I} \mathbf{P}^{\mathcal{I}}$ then $p \longmapsto \pi \cdot p$ is an order-preserving bijection of $\mathbf{Q}^$ onto $\mathbf{Q}^$, and if $U$ is a system then $p \in \mathbf{Q}[U] \Longleftrightarrow \lambda \cdot p \in \mathbf{Q}[\lambda \cdot U]$.

Lemma 16. Suppose that $U, V$ are systems, $|U|=|V|, p \in \mathbf{Q}[U], q \in \mathbf{Q}[V],|p|=|q|$, and sets $F_p^{\vee}(v)$, $F_q^{\vee}(v)$ are $i$-similar for all $v \in|p|=|q|$. Then there is $\lambda \in \operatorname{LIP}^{\mathcal{I}}$ such that $|\lambda|=|U|=|V|, \lambda \cdot U=V$, and $F_q^{\vee}(v)=F_{\lambda \cdot p}^{\vee}(v)$ for all $v \in|p|=|q|$.

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Jensen—Solovay Sequences

Arguing in L, let $U, V$ be systems. Suppose that $M$ is any transitive model of $\mathbf{Z F C}2^{-}$. Define $U \preccurlyeq_M U^{\prime}$ iff $U \preccurlyeq U^{\prime}$ and the following holds: (a) the set $\Delta\left(U, U^{\prime}\right)=\bigcup{v \in|U|}\left(U^{\prime}(v) \backslash U(v)\right)$ is multiply SEQ-generic over $M$, in the sense that every sequence $\left\langle f_1, \ldots f_m\right\rangle$ of pairwise different functions $f_{\ell} \in \Delta\left(U, U^{\prime}\right)$ is generic over $M$ in the sense of $\mathrm{SEQ}=\omega_1^{<\omega_1}$ as the forcing notion in $\mathbf{L}$, and
(b) if $v \in|U|$ then $U^{\prime}(v) \backslash U(v)$ is dense in FUN, therefore uncountable.
Let JS, Jensen-Solovay pairs, be the set of all pairs $\langle M, U\rangle$ of:

a transitive model $M \models \mathbf{Z F C}_2^{-}$, and a system $U$,such that the sets $\omega_1$ and $U$ belong to $M$-then sets $\mathrm{SEQ}, \mathbf{Q}[U]$ also belong to $M$.
Let sJS, small Jensen-Solovay pairs, be the set of all pairs $\langle M, U\rangle \in \mathbf{J S}$ such that both $U$ and $M$ have cardinality $\leq \omega_1$. We define:
$\langle M, U\rangle \preccurlyeq\left\langle M^{\prime}, U^{\prime}\right\rangle \quad\left(\left\langle M^{\prime}, U^{\prime}\right\rangle\right.$ extends $\left.\langle M, U\rangle\right)$ iff $M \subseteq M^{\prime}$ and $U \preccurlyeq_M U^{\prime}$;
$\langle M, U\rangle \prec\left\langle M^{\prime}, U^{\prime}\right\rangle$ (strict extension) iff $\langle M, U\rangle \preccurlyeq\left\langle M^{\prime}, U^{\prime}\right\rangle$ and $\forall v \in \mathcal{I}\left(U(v) \varsubsetneqq U^{\prime}(v)\right)$.

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Jensen—Solovay Sequences

(b)如果 $v \in|U|$ 然后 $U^{\prime}(v) \backslash U(v)$ 在FUN中是密集的，因此不可数。

$\langle M, U\rangle \preccurlyeq\left\langle M^{\prime}, U^{\prime}\right\rangle \quad\left(\left\langle M^{\prime}, U^{\prime}\right\rangle\right.$扩展$\left.\langle M, U\rangle\right)$、$M \subseteq M^{\prime}$和$U \preccurlyeq_M U^{\prime}$;
$\langle M, U\rangle \prec\left\langle M^{\prime}, U^{\prime}\right\rangle$(严格扩展)、$\langle M, U\rangle \preccurlyeq\left\langle M^{\prime}, U^{\prime}\right\rangle$和$\forall v \in \mathcal{I}\left(U(v) \varsubsetneqq U^{\prime}(v)\right)$。

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