Analytic boundedness for well-founded trees

If A is empty use $\gamma =1$. Otherwise F1 makes the ambient tree space Polish, so F3 with A1 gives continuous $f:\mathcal{N}\to \mathrm{T}\mathrm{r}$ with image A. Form the synchronous tree S of pairs $(s,t)$ of equal-length natural words such that some a extending s has $t\in f(a)$. Taking prefixes of a witness proves prefix closure. Pair the two natural letters into one natural number at each coordinate; this codes S as a tree on $\mathbb{N}$.

Suppose $(a,b)$ were a branch of S. Fix m. The map assigning membership of $b\upharpoonright m$ in f(a) is continuous with values in the discrete two-point space, by F1 and continuity of f. Choose n at least m so that this membership is constant on $N_{a\upharpoonright n}$. Because $(a\upharpoonright n,b\upharpoonright n)\in S$, one witness a' in that cylinder has $b\upharpoonright n\in f(a^{\mathrm{\prime }})$, hence also $b\upharpoonright m\in f(a^{\mathrm{\prime }})$. Constancy gives $b\upharpoonright m\in f(a)$. This holds for every m, so b is a branch of f(a), contradicting F2 since f(a) is well-founded. Each m used one existential witness; no family of witness choices is needed. Thus S has no branch and is well-founded by F2.

By F2 and A1 let $\delta =r(S)<{\omega }_1$. For every a with nonempty f(a), the map $t\mapsto (a\upharpoonright \mathrm{\mid }t\mathrm{\mid },t)$ takes f(a) into S and preserves proper extensions, sending root to root. F2 implies $r(f(a))\le \delta$. Empty f(a) has rank zero by F1, also at most $\delta$, even if S is empty. Hence $\gamma =\delta +1<{\omega }_1$ strictly bounds all ranks in A, since every member is f(a). QED.