Extensions of branches to nonstandard trees

Question

This is sort of inspired by the "cute" nonstandard proof of the fact that an infinite but finitely branched rooted tree has an infinite branch. An infinite branch is found by taking nonstandard element in a nonstandard extension of the tree and looking at all the elements which lie below it. I am wondering when that initial element is in the nonstandard extension of that branch.

To be more precise:

Suppose we have a rooted tree $T$ with vertices arbitrarily far from the base point. Also note that rooted trees have a natural partial order with $v\leq w$ when $v$ lies on the shortest path from $w$ to the base.

Let $^*T$ denote the nonstandard enlargement of $T$. Since $T$ has vertices of arbitrary distance from the root, there is a vertex $v$ in the enlargement which is some nonstandard distance (i.e. infinite distance) from the root. Viewing $T$ as a subtree of $^*T$, let $B$ be the branch of $T$ given by those $t\in T$ such that $t {\leq^*} v$.

The set $B$ has a nonstandard extension $^*B$ which we can view as a subset of $^*T$. When is it the case that $v$ is in $^*B$?

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I don't think $v$ has to be in $^*B$, and it seems like there is a chance it never is.

Since being a branch is first-order, $^*B$ is also a branch. Let's say we have $v\rightarrow v'$ and $v'' \rightarrow v'$ in $^*T$. Then branch $B$ described above which lies below $v$ is the same as the one lying below $v''$. But because $^*B$ is a branch, only one of $v,v''$ could be in $^*B$. I don't see how either possibility could be distinguished! Is it possible that, with the construction above $v$ is never in $^*B$?

Answer 1

Fix an infinite, finitely-branching tree $T$. Let $X\mapsto {}^*X$ be the usual "nonstandard analogue" map. For $v\in {}^*T\setminus T$, let $\mu(v)=\{w\in T: w<_{{}^*T}v\}$ be the standard part of the original tree $T$ lying below $v$. Finally, let $P(v)={}^*\mu(v)$. You're asking about when we have $v\in P(v)$.

The key point is this:

$P(v)$ is always a maximal path in ${}^*T$. Moreover, every maximal path $B$ in $T$ has exactly one extension to a maximal in the range of $P(v)$, namely ${}^*B$ itself.

The first sentence is just an application of transfer to the property "is a maximal path:" since $\mu(v)$ is a maximal path in $T$, its nonstandard analogue will be a maximal path in $T$'s nonstandard analogue. The second point is basically just a rephrasing of the definition of $P$ itself, the point being that (per the argument this question was motivated by) such a $B$ has some nonstandard extension $v$ which satisfies $P(v)={}^*B$.

This gives us the following:

Suppose $v,w\in {}^*T\setminus T$. If $w\in P(v)$ then $P(w)=P(v)$ and so $w\in P(w)$.

This tells us that while a given nonstandard node may not lie in the nonstandard branch it creates, all nonstandard nodes on any nonstandard branch so created will in fact generate that branch.

In the other direction, via overspill the "$v\not\in P(v)$"-phenomenon is essentially describing the non-isolated paths:

The following are equivalent, for $v\in {}^*T\setminus T$:

• There are infinitely many splitting nodes in $\mu(T)$.
• There are infinitely many maximal paths through ${}^*T$ extending $\mu(T)$.
• There is some $u$ with $\mu(u)=\mu(v)$ (and so $P(u)=P(v)$) but $u\not\in P(u)$.