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Apparently, we can use compactness theorem to construct an infinitely large natural number such that it is not divisible by any (standard) natural number $n \in \mathbb{N}_{>1}$. And I must say that I have no idea how to do this.

I have seen the construction of a model of the theory of $\mathbb{N}_0$ containing an infinitely large natural number. The method is very similar to the one that is described in https://en.wikipedia.org/wiki/Non-standard_model_of_arithmetic. It states that a new constant $c$ is added in a set of axioms $P*$, which is defined in a language including the language of Peano arithmetic.

So I thought that I should use a similar method to construct an infinitely large natural number such that it is not divisible by any (standard) natural number $n \in \mathbb{N}_{>1}$. The problem is what new constant should I add?. What's the idea to find this new constant? Furthermore, let's say that this new constant is $x$. Should I just infinitely many new axioms $(n < x)$?

Vicky
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  • How about adding axioms $n<x$ and $n\nmid x$? – Angina Seng Oct 04 '20 at 20:11
  • Just multiply all prime numbers and add one to the resulting super-number. – markvs Oct 04 '20 at 20:18
  • @JCAA I had the same idea but that's not what they're looking for, they need a model theoretic explanation as to why the symbol $c$ with $c > n$ is not divisible by any prime. A divergent, monotone increasing product isn't useful in proving this. – CyclotomicField Oct 04 '20 at 20:20
  • If a space is compact then any sequence has a converging subsequence. – markvs Oct 04 '20 at 20:32
  • An easy way to do this is to take the theory $P*$ that you already have, add the one sentence "$c$ is prime", and then apply compactness as before. (Here $c$ is the new constant symbol that you're adding, of course.) – Mitchell Spector Oct 05 '20 at 09:19
  • @AnginaSeng I was thinking about this as well, but what would then be the interpretation of $x$ if you want to construct a model of a finite subset of axioms (including the new ones)? – Ruby Oct 05 '20 at 15:13
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    A hint that I got is to add a new constant $c$. Then, add new denumerably many new axioms $\mathcal{B}_n$, which states that $n< c$ and $c$ is not divisible by any integer $k < n$ and $k>1$. However, now I don't know how to interpret this $x$, just like what @Violet said. When we construct a model that contains an infinitely large natural number, we interpret $c$ as $\max{n_1, n_2, \dotsc, n_k} + 1$. But in this case, I don't think the same interpretation will work. – Vicky Oct 05 '20 at 21:31

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Fix $\mathcal{L} = \{+,\times,<,(c_i)_{i \in \mathbb{N}}\}$ and consider the theory of the $\mathcal{L}$-structure $M=(\mathbb{N},+,\times,<,(i)_{i \in \mathbb{N}})$. Now take the language $\mathcal{L}'=\mathcal{L} \cup \{c^*\}$ where $c^*$ is a constant symbol, and the theory given by $Th_{\mathcal{L}}(M) \cup \Psi$ where $\Psi$ consists of the following $\mathcal{L}'$-sentences:

  • for each $i \in \mathbb{N}$, the sentence $c_i < c^*$,
  • for each $i \in \mathbb{N}$, the sentence $\neg \exists x (c_i \times x = c^*)$.

Now show that every finite subset of $Th_{\mathcal{L}}(M) \cup \Psi$ has a model (hint: you can take $M$ and choose a suitable interpretation of $c^*$). By compactness you get that $Th_{\mathcal{L}}(M) \cup \Psi$ is consistent. So take a model $M^*$ of $Th_{\mathcal{L}}(M) \cup \Psi$. You can check that $M \prec M^*$ (after taking the $\mathcal{L}$-reduct of $M^*$) and therefore the interpretation of $c^*$ in $M^*$ can be thought of as a non-standard natural number with the desired properties.

When using compactness arguments you can only expect to show that something exists, you usually won't be able to hold it in your hands and 'interpret'.

Buchi Fan
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