In the paper ''**A Nonstandard Model of Arithmetic Constructed by means of Forcing Method**'', Zhang Jinwen states the following in his abstract:



> 
> The first nonstandard model of arithmetic was given by Skolem. A. Robinson has introduced the concepts of standard, internal and external objects (sets, relations, functions, etc.) on the compactness theorem and concurrent relations, and has proved that if a set S is infinite, then S contains nonstandard internal objects. It is interesting to ask whether this is a common property of all non-standard modes of arithmetic. The author's answer to this question is in the negative. We have proved the theorem that there exists a nonstandard model of formal arithmetic in which there are infinitely many infinite internal subsets containing no nonstandard elements. This means that these infinite internal subsets are composed exclusively of finite natural numbers. In order to obtain this theorem we have made use of Cohen's forcing method.
> 
> 
> 


The paper is in Chinese, and I could not understand it (I have a copy of it). 


**Question.** I am wondering if someone can explain the main idea of his proof. 




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**Remark.** Based on Mathscinet, the paper is translated in English, but I could not find a copy of it.


The file can be seen here: [A Nonstandard Model of Arithmetic Constructed by means of Forcing Method](https://www.dropbox.com/s/wl5q1k8xltbmtth/A%20Nonstandard%20Model%20of%20Arithmetic%20Constructed%20by%20means%20of%20Forcing%20Method.pdf?dl=0)



On a theorem of Zhang Jinwen about models of arithmetic
(The final two sections are translated modulo tweaks...)




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I do not know the answer to your question, but I can try to translate (part of) this paper. There are some terms that are unfamiliar to me (e.g., "infinite internal sets" as remarked in a comment) so I will indicate where I am quite confused by using brackets.


I am making this **community wiki** as I am sure that a logician who speaks native Chinese could complete this much more quickly, and so I hope others (including non-Chinese speakers who are well versed in mathematical logic!) will not hesitate to edit. I will translate the first two sentences immediately after the abstract, and then attempt the sections in reverse order. The paper is pretty short, but remember that Chinese characters can represent entire words; so, it would be a lot to translate the entire manuscript. I think a more reasonable aim is the main ideas of sections 3-6.


No guarantee on a timeline, although it'll obviously be faster if others assist. I don't recall MO being used in this way before, so if it's un-welcome I can delete.




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**p. 7**: In order to keep the manuscript self-contained, the article at hand briefly delves into the Peano Axioms and arithmetical definability in Section 1, and provides an overview of a few results around non-standard models of arithmetic in Section 2. Sections 3 through 6 concern our main result. 




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**p. 11**:


$$5, \text{Main Lemma and the structure of the model}^{\*}M(G)$$


With regard to forcing notions, following the work of Cohen $[2]$ it is not difficult to prove the lemmata below.


**Lemma 1.** For any $Q$ and $A$, $Q$ cannot simultaneously force $A$ and $\neg A$.


**Lemma 2.** If $Q \Vdash A$ and $Q' \supseteq Q$, then $Q' \Vdash A$.


**Lemma 3.** For any $Q$ and $A$, there exists $Q' \supseteq Q$ for which $Q' \Vdash A$ or $Q' \Vdash \neg A$.


**Lemma 4.** $Q \Vdash A$ if and only if for all $Q' \supseteq Q$ it is the case that $Q' \not\Vdash \neg A$.


**Definition.** A sequence of forcing conditions $Q\_0, Q\_1, Q\_2, \ldots, Q\_n, \ldots$ is **generic** if for any statement $A$ in $^{\*}L(G)$ there is a standard natural number $m$ for which $Q\_m \Vdash A$ or $Q\_m \Vdash \neg A$.


**Lemma 5.** There exists a sequence of generic forcing conditions.


**Proof.** Observe that there are countably many formulae in $^{\*}L(G)$. Therefore, one can enumerate the formulae of $^{\*}L(G)$ as $A\_1, A\_2, A\_3, \ldots, A\_n, \ldots$. Pick any $Q\_0$ and suppose one already has $Q\_{n-1}$; for any $A\_n$, there must exist $Q \supseteq Q\_{n-1}$ for which $Q \Vdash A\_n$ or $Q \Vdash \neg A\_n$ holds. Defining $Q\_n := Q$ and continuing in this fashion, one obtains a generic forcing sequence $Q\_0, Q\_1, Q\_2, \ldots$.


Now, pick a generic forcing sequence $Q\_0, Q\_1, \ldots$, from which we define generic sets $S\_1, S\_2, \ldots, S\_i, \ldots$ as below, for $i < \delta$:


$$S\_i = \{k|\text{there exists }m \text{ for which }O\_m \Vdash \bf{k} \in F\_i\}$$ 


In this way, we have defined a generic nonstandard arithmetical model $^{\*}M(G) =$ $^{\*}M(S\_0, S\_1, \ldots, S\_i, \ldots)$, with underlying structure [?] $(^{\*}N, \textbf{,}, +, \cdot, S\_0, S\_1, \ldots, S\_i, \ldots)$ where $i < \delta \leq \omega$. For any formula $A$ in $^{\*}L(G)$, the truth value of $A$ in the model $^{\*}M(G)$ is defined as follows:


$$^{\*}M \vDash A \iff (\exists m) Q\_m \Vdash A$$


By lemmas 1 through 5, the model $^{\*}M(G)$ is well-defined.




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**pp. 11-12**: 


$$6, \text{Characteristics of the model }^{\*}M(G)$$


The lemma below follows clearly from the definition of forcing.


**Lemma 6.** For any statement $A$ in $^{\*}L$, we have


$$^{\*}M\_1 \vDash A \iff ^{\*}M(G) \vDash A$$


Given that $M$, $^{\*}M$, and $^{\*}M\_1$ are all models of $P$ (Peano Arithmetic), it follows from Lemma 6 that $^{\*}M(G)$ is also a model of $P$, and it is a non-standard model; however, $^{\*}M(G)$ is distinct from $^{\*}M$, and $^{\*}M\_1$; $^{\*}M(G)$ contains as a subset $S\_i$ $(0 \leq i < \delta)$, and the $S\_i$ are generic infinite sets, and their elements are all standard natural numbers, which, in turn, proves our main theorem.


By the result of Feferman $[3]$, $S\_i$ is hyperarithmetic, thus, the following propositions hold:


1. If $S \subset N$ is any arithmetic set, then a necessary and sufficient criterion for $S^{\*}$ to contain a nonstandard number is that $S$ be an infinite set.
2. $N$ is an external subset of $^{\*}M(G)$.
3. $^{\*}N - N$ is an external subset of $^{\*}M(G)$.


