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Let $M$ be the model $\langle \mathbb{Z}, s^M\rangle$ where $s$ is a unary function symbol interpreted as the standard successor function ($s^M(x)=x+1$). Let $F$ be a nonprincipal ultrafilter on $\omega$ and let $N$ be the ultrapower $N:=M^{\omega}/F$. I'm struggling to answer the following question:

Prove that for every $i <\omega$ there is an injective homomorphism $f_i: M \to N$ such that $f_i(M) \cap f_j(M) = \emptyset$ for every $i \neq j$.

I really don't how to define the image of $n$ under $f_i$. The sequence should somehow codes $i$ and $n$. All the 'obvious' guesses ($ n \mapsto [\langle 0, \ldots, 0, n, n, \ldots\rangle]_F$ or $n \mapsto [\langle 0, \ldots 0, n, n+1, n+2, \ldots \rangle]_F$) are clearly wrong.

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2 Answers 2

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Let $d_i : n\mapsto i\cdot n$. Then $[d_i]_F$ is an element of $N$.

For $m\in\mathbb N$, let $\hat m : n \mapsto m$ be the constant map.

Note that $f_i : m\mapsto [d_i+\hat m]_F$ is an embedding of $M$ in $N$.

Now, let $i<j$. I claim that $f_i[M]$ is disjoint of $f_j[M]$.

Pick two arbitrary elements $[d_i+\hat m_1]_F\in f_i(M)$ and $[d_j+\hat m_2]_F\in f_i(M)$.

Note that $(d_i+\hat m_1)(n)= i\cdot n+m_1$ and $(d_j+\hat m_j)(n)= j\cdot n+m_2$.

Hence $(d_i+\hat m_1)(n)<(d_j+\hat m_j)(n)$ for almost all $n$. Therefore $[d_i+\hat m_1]_F<[d_j+\hat m_2]$.

As $m_1$ and $m_2$ are arbitrary, the claim follows.

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  • $\begingroup$ @Mockingbird Alex Kruckman has been quicker than me. Note that the answer is exactly the same. $\endgroup$ Jan 17 at 15:05
  • $\begingroup$ Yes @PrimoPetri he was quicker you provided a little more detail. I take some time to choose the answer to accept $\endgroup$ Jan 17 at 15:46
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How about $f_i(n)=[\langle n, i+n, 2i+n, 3i+n,\dots\rangle]_F$?

It's worth verifying for yourself that this works, and also spending some time thinking about how I could have come up with this solution.

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