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Prove that $(\mathbb{Q} \times \mathbb{Z}, \le) \equiv (\mathbb{R} \times \mathbb{Z}, \le)$ where $\equiv $ denotes elementary equivalence.

My approach:

$\equiv$ means elementary equivalence. It means that there exists always winning strategy for the duplicator. It also means that there exists partial $k$-isomporphism for $k \in \mathbb{N}$. Here, I would like ask:

  1. For every $k$ we have an isomorphism so we have an infinite isomorphism. It cannot be possible because their cardinality is not same. What I misunderstand?

Solution:

We know that $(\mathbb{R}, \le) \equiv (\mathbb{Q}, \le)$. So, we can take a partial $k-$isomorphism $f_k:(\mathbb{R}, \le) \to (\mathbb{Q}, \le)$.

Let's define for every $k \in \mathbb{N}$ partial isomorphism $g_k: (\mathbb{Q} \times \mathbb{Z}, \le) \to (\mathbb{R} \times \mathbb{Z}, \le)$ $$g_k((q,z)) = (f_k(q), z)$$

Now, we should show that $g_k$ is a partial isomorphism but it is obvious because $f_k$ was partial isomorphism.

Right & please ask 1.

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"For every $k$ we have an isomorphism so we have an infinite isomorphism."

This is false under any reasonable interpretation; and I think the confusion is stemming from what, exactly, a winning strategy for Duplicator gives you.

A $k$-partial isomorphism is a map $f$ from a subset of $A$ to a $subset$ of $B$, which preserves and reflects the truth of $k$-quantifier formulas. But such an $f$ need be nowhere near an isomorphism - for instance, the empty map is always a $k$-partial isomorphism.

Now, what a winning strategy for Duplicator in the $k$-round $EF$-game gives you is a family $I$ of partial isomorphisms with a bunch of properties:

  • $I$ is partitioned into "blocks" $I_0, I_1, . . . , I_k$; elements of $I_i$ are $i$-partial isomorphisms; and

  • $I$ satisfies the back-and-forth property.

But note that this isn't a single map! For example, if we take $k=0$, $A=(\mathbb{Q}, <$, and $B=(\mathbb{R}, <)$, then if we take $I_0$ to be the set of all maps $\{a\}\rightarrow\{b\}$ with $a\in\mathbb{Q},b\in\mathbb{R}$ - that is, the set of all $0$-partial isomorphism defined on a single element - then $\langle I_0\rangle$ is an object of the type above! And clearly it can't be extended, in any good way, to a single map from $\mathbb{Q}$ to $\mathbb{R}$ - since for each $a\in\mathbb{Q}$, there are many $b\in\mathbb{R}$ such that the map $\{(a, b)\}$ is in $I_0$. Similarly, even if Duplicator wins each EF-game, they don't have any way to "amalgamate" their back-and-forth systems to get an actual isomorphism.

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  • $\begingroup$ ". Similarly, even if Duplicator wins each EF-game, they don't have any way to "amalgamate" their back-and-forth systems to get an actual isomorphism.". I don't understand what do you mean here. especially, I don't know what is back-and-forth. $\endgroup$ – user376326 Feb 20 '17 at 19:34
  • $\begingroup$ @Logic A back-and-forth system (or $k$-back-and-forth system) is the object $I$; a family of partial isomorphisms with certain properties. The point is, there's no single function corresponding to Duplicator's strategy. For instance, in the EF game of length $k$ for the rationals vs. the reals, Duplicator could begin by sending $1$ to $5$ . . . or to $2$, or to $-{1\over 2}$, or anything else. There's no function here telling me where to send $1$; there's just a relation telling me where I'm allowed to send $1$. (cont'd) $\endgroup$ – Noah Schweber Feb 20 '17 at 19:52
  • $\begingroup$ I think the best way to see this is to actually try to do the thing you claim you can do: "For every $k$ we have an isomorphism so we have an infinite isomorphism." OK, convince me - build one. You'll quickly see that you can't. $\endgroup$ – Noah Schweber Feb 20 '17 at 19:53
  • $\begingroup$ you obviously right that EF-k-round-game gives a family of k-partial isomorphisms. But, what is the problem? We can just take a isomorphism from family. My reasoning is as follows: Based on isomorphism $f_k: (\mathbb{Q}, <) \to (\mathbb{R}, <)$ ( taken from family) we define k-partial family $g_k: (\mathbb{Q} \times \mathbb{Z} , <) \to (\mathbb{R} \times \mathbb{Z}, <)$ Super. We have a k-isomorphism. We cannot distuingish it with k-quantifier formula. We can do it for every $k$. Great! It is my reasoning. Why wrong? $\endgroup$ – user376326 Feb 20 '17 at 20:00
  • $\begingroup$ "OK, convince me - build one". I see now that it is impossbile. I don't claim ( now) that I can constuct infinite isomorphism. I claim that I can construct k-partial isomorphism for every $k$. Now, I see that I can. Indeed. But it is pointless. It doesn't mean that I cannot construct a k-quantifier formula which distiguish two structures. $\endgroup$ – user376326 Feb 20 '17 at 20:07

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