Could you help me with the following question?
Let TW denote the set of all transitive well-founded binary relations $R$ on the set of natural numbers. As any binary relation $R$ on natural numbers could be seen as a point of $\{0,1\}^{\mathbb{N}\times\mathbb{N}}$, we could think of TW as a subset of $\{0,1\}^{\mathbb{N}\times\mathbb{N}}$, endowed with its usual topology. For a countable ordinal $\alpha$ let TW$(\alpha)$ denote the set all those relations $R$ in TW that have a rank smaller than $\alpha$.
Is it true that TW$(\alpha)$ has Borel rank of at most $\alpha$?
Jech: Set theory, Lemma 25.10, says that the set of well-founded binary relations WF$(\alpha)$ with rank of at most $\alpha$ is a Borel set, but it is silent about its Borel rank.
Thanks a lot for your help!
This is a partial answer.
Consider $\mathrm{WO}(\alpha)$, the set of wellorderings of rank less than $\alpha$. It is clearly the intersection of $\mathrm{TW}(\alpha)$ with a closed set. Now, by Kechris (p.189, where he quotes Stern), $\mathrm{WO}(\omega^\alpha)$ is $\boldsymbol\Sigma^0_{2\cdot\alpha}$-complete, and each $\mathrm{WO}(\omega^\beta)$ for $\omega^\alpha<\beta<\omega^{\alpha+1}$ is not $\boldsymbol\Sigma^0_{2\cdot\alpha+1}$.
Thus, if $\alpha = \omega^\alpha$, then $\mathrm{TW}(\alpha)$ must be at least $\boldsymbol\Sigma^0_{\alpha}$, and we have a lower bound for such $\alpha$s.
I believe that the lower bound you are looking for can be obtained by reduction to $\mathrm{WF}(\alpha)$, but I have to double check this.