Prove that if two strucutres are not distinguishable then also modifed structures are distinguishable

Question

Let $\mathbb{A}= \langle A, \le_A \rangle$ ,$\mathbb{B}= \langle B, \le_B \rangle$ will be partial orders. Prove that if $\forall_n $ and $\forall_{\mathbb{A},\mathbb{B}}$ $\mathbb{A}\equiv_n \mathbb{B}$ then also $\mathbb{A'}\equiv_n \mathbb{B'}$, where $\mathbb{X'}$ denotes $\mathbb{X}$ with added the largest element (so to both structure we add per one element such that it is the biggest.

I am trying to prove it using EF-games. From graphical point of view structure is tree - undirected acyclic connected graph. However, the idea is to use strategy of duplicator for $\mathbb{A}\equiv_n \mathbb{B}$ - let call it as old strategy.

If spoiler doesnt pick up the largest element (root of tree) then duplicator uses old strategy. In case of picking up root, duplicator also pick up root.

Why does it work ?
Following old strategy duplicator prevent relation between old elements. In both structure root is in the same relation between old elements it is largest than each old element. So if duplicator pick up root iff spoiler do it he manage to win.

What do you think ? Maybe someone can solve it in other way ?

Answer 1

Looks good to me!

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There's also a very silly solution via ultrapowers - this is definitely more work than the EF-games solution, so there's no earthly reason why anyone would do this, but I think it's neat so I'll put it here.

There truly is no good reason to do this at all - I'm honestly just putting it down since I think some readers will find it neat, as I did. It uses nontrivial facts about ultrapowers, so it definitely takes work to follow. Understanding it is completely unnecessary, so don't worry if it doesn't make sense yet.

By the Keisler-Shelah theorem, since $A\equiv B$ there are ultrafilters $\mathcal{U}, \mathcal{V}$ on index sets $I, J$ such that $$\prod_IA/\mathcal{U}\cong \prod_JB/\mathcal{V}.$$ Call these ultrapowers $M$ and $N$ respectively.

Now consider $M'$ and $N'$, built in the same way as $A'$ and $B'$ (just add a new top element to $M$ and $N$, respectively). I claim that $$(*)\quad \quad M'\cong\prod_IA'/\mathcal{U}\quad\mbox{and}\quad N'\cong\prod_JB'/\mathcal{V}.$$ To see why this is enough, let $f: M\rightarrow N$ be an isomorphism; then $f$ extends to an isomorphism $\hat{f}: M'\rightarrow N'$ (this is true for general structures - just map the new top element to the new top element, and otherwise do what $f$ does). By Los' Theorem, if two ultrapowers are elementarily equivalent, then so are the ultraroots (= the structures we're taking the ultrapowers of); since $M'\cong N'$ via $\hat{f}$, we have a fortiori that $M'\equiv N'$, so this will show $A'\equiv B'$.

So how do we prove $(*)$? I'll show $M'\cong \prod_{I}A'/\mathcal{U}$; the $N'$-version is identical. An element of $M'$ is an equivalence class of a sequence $\alpha=(a_i)_{i\in I}$ with $a_i\in A'$. Now, for such a sequence, let $New=\{i: a_i\mbox{ is the new top element}\}$. Either $New\in\mathcal{U}$ or $I\setminus New\in\mathcal{U}$; if the former holds, then $\alpha$ corresponds to the new top element of $M'$, and if the latter holds, then $\alpha$ corresponds to one of the elements of $M$ (since $\alpha$ is $\mathcal{U}$-equivalent to some $\beta=(b_i)_{i\in I}$ whose terms all come from $A$: just get $\beta$ from $\alpha$ by changing each of the "new" terms to some random thing from $A$, and however you do this you'll have $\beta\approx_\mathcal{U}\alpha$ since they'll agree on $I\setminus New$, which is in $\mathcal{U}$). This shows that the underlying set of $M'$ is exactly the underlying set of $\prod_IA'/\mathcal{U}$, and it's not hard to show via similar reasoning that the natural bijection here is in fact an isomorphism.