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Let $T$ be a tree. Let $T^0=T$. For an ordinal $\xi=\eta+1$ we define $T^\xi$ to be the complement of the set of all maximal elements of $T^\eta$. For a limit ordinal $\xi$ we set $T^\xi = \bigcap_{\beta<\xi}T^\beta$. Let me call the number $\min\{\xi\colon T^\xi = T^{\xi+1}\}$ the order of $T$.

Suppose that $T$ is an uncountable tree with countable order. Does $T$ contain a countable subtree with the same order?

It looks like some version of the Löwenheim–Skolem theorem, but I am not sure if this is applicable here.

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You're right, this is basically a Lowenheim-Skolem result.

The "pruning" process you've described comes equipped with a rank function from $T$ to the order of $T$ together with $\infty$:

  • If $\sigma\in T^\zeta$ but not $T^{\zeta+1}$, then $rk(\sigma)=\zeta$.

  • If $\sigma\in T^\xi$ for all $\xi$ (that is, if it lasts until the process stabilizes), then $rk(\sigma)=\infty$.

This rank function, in turn, gives us a family of unary predicates naming each "stage": consider the structure $$\mathcal{T}=(T; <, (U_\zeta)_{\zeta\in order(T)\cup\{\infty\}})$$ where $<$ is the tree order on $T$ and each $U_\zeta$ is a unary predicate naming $\{\sigma\in T: rk(\sigma)=\zeta\}$.

The structure $\mathcal{T}$ captures the tree structure and the pruning process; and if the order of $T$ is countable, then the language of $\mathcal{T}$ is also countable. So by Lowenheim-Skolem $\mathcal{T}$ has a countable elementary submodel, with underlying tree $S$. The key question now is:

Can you show that the order of $S$ is the same as that of $T$?

HINT: The order of a tree is the supremum of the non-$\infty$ ranks of its nodes. So by induction, can you show that the $S$-rank of a node $\sigma\in S$ is the same as the $T$-rank of $\sigma$? Certainly the $S$-rank can't exceed the $T$-rank, so it's enough to show that $S$ doesn't "kill nodes prematurely" . . .

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    That said, you don't need LS to prove this; there's a purely combinatorial construction of such a subtree. Just put in "enough" nodes of each rank . . . – 2017-02-12
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    Thanks Noah, this is clear now. The key thing was to introduce this extra structure. Doesn't it show more? That there exists a countable elementary submodel $M$ of some fragment of ZFC such that whenever $T$ is such a tree then ${\rm order}(T) = {\rm order}(T\cap M)$? – 2017-02-13
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    @TomekKania No - given any such $M$, you can find a tree which is disjoint from $M$. Then $T\cap M$ is empty. But if you swap the quantifiers around, the answer is yes: for each $T$ there is such an $M$. (Also, I'm a bit confused when you write "there exists a countable elementary submodel $M$ of some fragment of ZFC" - an elementary submodel is a submodel of a *model*, not a *theory*, and if the big model satisfies all of ZFC then so will the small one.) – 2017-02-13