I am trying to solve the following exercise:

Let $L=\{0,S\}$, where $0$ is a constant symbol and $S$ is an unary function symbol. Let $T$ be the set of following $L-$sentences. \begin{align*} & \forall x \forall y (S(x)= S(y) \rightarrow x = y) \\ & \forall y (y \neq 0 \rightarrow \exists x (S(x)=y)) \\ & \forall x (S(x) \neq 0) \\ & \forall x (S^n(x)\neq x) \quad , n \in \omega \end{align*} Show that $T$ has infinitely many pairwise non-isomorphic countable models.

We obtain a model $M$ for $T$ by setting $M := \{\mathbb{N};0_{\mathbb{N}},x+1\}$. We now have one of the infinitely many countable models. I shall use this as a starting point for all the others.

I want to define a model $M_n , n \in \omega$ over the universe $ \mathbb{N} \sqcup \mathbb{Z} \sqcup \dots \sqcup \mathbb{Z}$ with $n$ $\mathbb{Z}$'s. Then any $x \in M_n$ can be expressed as $x = (i_x, a_x), i \in \{0,...,n\}$ with the $a_x$ being a natural number for $i_x=0$ and an integer elsewise. The interpretation of the zero shall be $(0,0)$ and $S(i_x,a_x)=(i_x,a_x+1)$. Then $M_n$ models $T$.

It is left to show that the the $M_n, n\in \omega$ are pairwise non-isomorphic.

My approach is this: If I consider an ordering $<$ on $\mathbb{N}$ respectively $\mathbb{Z}$, I can show that the $M_n$ are pairwise non-order-isomorphic. I introduce a well-ordering on each $\mathbb{Z}$ by $0,1,-1,2,-2,...$. Then the order type of $M_n$ is the sum of $n+1$ $\omega$'s, hence all the $M_n$ are pairwise non-order-isomorphic.

I am afraid however that this is not actually a solution.

Problem 1: $L$ does not contain a symbol for an order. Is it therefore even legitimite to create new structure in the form of an order? And even if it is, is there a way of showing the non-isomorphy using what is given with $L$ alone?

Problem 2: This is probably connected to problem 1. If all of what I have done is actually ok, is there any way of identifying the isomorphism of models to the isomorphism of order types? Because what I actually want is the first (i.e. a bijective mapping between the two universes, s.t. all the structure is being preserved).

Problem 3: If we order $\mathbb{Z}$ as indicated, we need to make sure that all elements but the zero have a predecessor (resp. are successors), so the successur function would have to move along $ \mathbb{N} \sqcup \mathbb{Z} \sqcup \dots \sqcup \mathbb{Z}$ by

$0 \rightarrow 1 \rightarrow ...\rightarrow -m \rightarrow -m+1 \rightarrow ...0 \rightarrow 1 \rightarrow ...\rightarrow m \rightarrow m+1 \rightarrow ...\rightarrow-m, ...$

As opposed to the ordering $0,1,2,...,0,1,-1,2,...0,1,-1,...$. Can this lead to complications?

Remark: I know that there are similar questions on the successor function, however the answers don't go into detail about the part with the non-isomorphic $M_n$.

Any help is appreciated. Thank you.

  • $\begingroup$ Your statement of Problem 3 seems confused--there is nothing that requires $S$ to coincide with the usual order-theoretic "successor" operation for any order you might define on $M_n$. $\endgroup$ – Eric Wofsey Aug 3 '16 at 1:47
  • $\begingroup$ Can't you define x=[o,x) (generalizing the Von-Neumann ordinal)? In this universe of discourse you have equality iff isomorphic. $\endgroup$ – Jacob Wakem Aug 3 '16 at 2:02
  • $\begingroup$ @EricWofsey , maybe I should have written question instead of problem. I didn't see a specific problematic point (which makes sense as S and an ordering don't need to be related), but the more I thought about my approach the less convincing it seemed, so I guess, yes, I was confused. $\endgroup$ – Valerie Poulain Aug 4 '16 at 0:54

Let $M$ be a model of $T$, with $f$ the interpretation of the function symbol $S$.

If $a$ and $b$ are members of (the underlying set of) $M$, call $a$ and $b$ equivalent if there is a non-negative integer $k$ such that $b=f^{k}(a)$ or $a=f^{k}(b)$. This is an equivalence relation. In $M_n$, the number of equivalence classes is $1$ more than the number of copies of $\mathbb{Z}$.

Let $M$ and $M'$ be models of $T$, and let $\varphi$ be an isomorphism from $M$ to $M'$. Then $a$ is equivalent to $b$ if and only if $\varphi(a)$ is equivalent to $\varphi(b)$. Thus the cardinality of the number of equivalence classes is an isomorphism invariant, and therefore the $M_n$ are pairwise non-isomorphic.

Remark: It is legitimate to produce a partial order on any model of $T$, along the lines you described. And on your models, one can describe a total order, which though it has a somewhat arbitrary character, behaves reasonably well under isomorphism.

For more general models of $T$, there are many non-isomorphic extensions of the natural partial order to a total order, so imposing such a total order is not useful in discussing isomorphism and non-isomorphism.

I do not understand the third problem. In any model of $T$, any equivalence class that does not contain the interpretation of $0$ has natural order isomorphic to the usual order on $\mathbb{Z}$. There is no reason to introduce a well-ordering, and no natural way to do so.

  • $\begingroup$ I'm not sure in what sense there is a total order that "behaves reasonably well under isomorphism". There is in fact no total order on $M_n$ for $n>0$ that is preserved by all automorphisms of $M_n$, since $M_n$ has automorphisms that swap pairs of elements. $\endgroup$ – Eric Wofsey Aug 3 '16 at 1:22
  • $\begingroup$ @EricWofsey: With a finite number of equivalence classes, all orderings of equivalence classes are order isomorphic (though there is the little matter of detail about where we put the copy of $\mathbb{N}$). So there is a small family of orderings, which can be used for the discussion of isomorphisms. This breaks down badly if the number of equivalence classes is infinite. $\endgroup$ – André Nicolas Aug 3 '16 at 1:29
  • $\begingroup$ Of course! Using an equivalence relation is much easier. I actually worked with one for a different part of the exercise but never thought to apply it here. I shall use this instead, thank you. $\endgroup$ – Valerie Poulain Aug 4 '16 at 0:58
  • $\begingroup$ @ValeriePoulain: You are welcome. $\endgroup$ – André Nicolas Aug 4 '16 at 2:08

As André Nicolas's answer indicates, a simpler way to solve this problem is to use an equivalence relation, rather than an ordering. Let me elaborate on what happens when you approach this problem using an ordering.

As alluded to in your Problem 1, if you introduce an ordering, you need to be sure that it is respected by isomorphisms. That is, you can define an order $<$ on $M_n$, but to be able to use these orders to tell the $M_n$ apart, you would need to know that any isomorphism $f:(M_m,S)\to (M_n,S)$ is also an isomorphism of ordered sets $f:(M_m,<)\to (M_n,<)$. Typically, the way you check a condition like this is by observing that $<$ is defined using the operation $S$ and no other structure of the sets $M_n$, and so it is routine to verify that a bijection preserving the operation $S$ must also preserve the relation $<$. However, this is not obviously true in your case: your definition of $<$ uses your explicit description of the set $M_n$ as consisting of copies of $\mathbb{Z}$ (and $\mathbb{N}$), and is not stated in terms of the operation $S$.

In fact, it turns out that your order $<$ does not satisfy the condition above. That is, there are isomorphisms between the structures $(M_n,S)$ that do not preserve the order $<$. For instance, consider the map $f:M_1\to M_1$ given by $f(0,a)=(0,a)$ and $f(1,a)=(1,a+1)$. This is a bijection that preserves $S$, but it does not preserve your ordering $<$, (for instance, $(1,0)<(1,-1)$ but $f(1,0)=(1,1)>(1,0)=f(1,-1)$).

So unfortunately, there is no apparent way to use your ordering to show the $M_n$ are not isomorphic. However, you can use a different ordering. Let me define a partial order $\prec$ on any model $M$ of $T$ as follows. We say that $x\prec y$ if there exists a positive integer $n$ such that $y=S^n(x)$. On $M_n$, this corresponds to the usual ordering on each $\mathbb{Z}$ (however, $\prec$ is not a total order if $n>0$, since elements in different copies of $\mathbb{Z}$ or $\mathbb{N}$ are incomparable).

Since $\prec$ is defined for any model of $T$ using only the operation $S$, it is easy to see that it is preserved by isomorphisms. Explicitly, if $f:(M,S)\to (N,S)$ is an isomorphism of models of $T$ and $x,y\in M$, then if $x\prec y$ in $M$ then $x=S^n(y)$ so $f(x)=f(S^n(y))=S^n(f(y))$ so $f(x)\prec f(y)$ (and similarly conversely).

So if $M_n$ and $M_m$ are isomorphic, then the ordered sets $(M_n,\prec)$ and $(M_m,\prec)$ would be isomorphic. But you can verify that $(M_n,\prec)$ is isomorphic to a disjoint union of one copy of $\omega$ and $n$ copies of $\mathbb{Z}$ (with their usual orders). Since these partially ordered sets are not isomorphic for different values of $n$, the $M_n$ are not isomorphic for different values of $n$. (But this raises the question of how you would actually prove that these partially ordered sets are not isomorphic, and to do so you would probably end up using the equivalence relation of André Nicolas's answer anyways.)

  • $\begingroup$ Thanks for the detailed answer. I couldn't quite put my finger on what would go wrong with my approach but you've stated it very clearly. I shall accept André Nicolas answer, as using the eqivalence relation seems to be the easiest way to go and because the non-isomorphy of $M_n$ and $M_m$ with the partial order is yet to be shown. $\endgroup$ – Valerie Poulain Aug 4 '16 at 1:05

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