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:
- 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.
"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.