I have a question regarding some boundedness theorems in descriptive set theory.
First, consider the following result:
Theorem. (Boundedness) Let $A\in\Pi_1^1\setminus\Delta_1^1$ and let $\varphi:A\to\operatorname{ORD}$ be a $\Pi_1^1$-rank. Let $B\subseteq A$, $B\in\Sigma_1^1$. Then $B$ is bounded, i.e. there is $x_0\in A$ such that for all $x\in B$ we have $\varphi(x)\le\varphi(x_0)$.
The proof of this is relatively simple (can someone confirm I understand it correctly?): Assume for contradiction $B$ is not bounded. Then we can write $$x\in A\iff \exists y (y\in B\wedge x\le_{\varphi}^{\Sigma_1^1} y),$$ where $\le_{\varphi}^{\Sigma_1^1}$ is a $\Sigma_1^1$ relation such that $\varphi(x)\le\varphi(y)\iff x\le_{\varphi}^{\Sigma_1^1} y$ (this just arises from the definition of a rank). This means $A$ is a $\Sigma_1^1$ set, contradiction.
As a corollary we get the following: If $A\subseteq WO$ is $\Sigma_1^1$, then $|A|=\sup\{|\alpha|:\alpha\in A\}<\omega_1$. Here $WO$ represents the set of all wellorders on $\mathbb{N}$ (we construct this by associating with each element $w$ of the Baire space an order $n <_w m\iff w(\langle n,m\rangle)=0$, and then taking $WO=\{w\in\mathcal{N}:<_w\text{ is a wellorder on } \mathbb{N}\}$).
Now, the effective version of the boundedness theorem was presented as such.
Theorem. (Effective boundedness) If $B\subseteq WO$, $B\in\Sigma_1^1$, then $\sup\{|\alpha|:\alpha\in B\}<\omega_1^{CK}$, where $\omega_1^{CK}:=\sup\{|\alpha|:\alpha\in WO, \alpha \text{ recursive}\}$ is the Church-Kleene ordinal.
The proof of this is given as follows: Let $A\subseteq\mathbb{N}, A\in\Pi_1^1\setminus\Sigma_1^1$. $WO$ is $\Pi_1^1$-complete, so let $f:\mathbb{N}\to\mathcal{N}$ be recursive such that $n\in A\iff f(n)\in WO$. If, towards a contradiction, $\sup\{|\alpha|:\alpha\in B\}\ge\omega_1^{CK}$, then $n\in A\iff\exists\alpha(\alpha\in B\wedge f(n)\le_{|\cdot|}^{\Sigma_1^1}\alpha)$. Then $A$ is $\Sigma_1^1$, contradiction.
Doesn't this say exactly the same thing as the corollary to the non-effective boundedness theorem? My suspicion is that $A$ should be a boldface $\mathbf{\Sigma}_1^1$ set in the statement of the corollary, but then how do we apply the regular boundedness theorem? I'm guessing there is something I'm not understanding in the two proofs presented.
Also, I'm wondering if someone can explain nicely what the difference/relation is between these results and this result due to Kunen, which is also a boundedness result: Let $(X,\prec)$ be a well-founded relation. Put $\rho_{\prec}:X\to\operatorname{ORD}=\sup\{\rho_{\prec}(y) + 1: y\prec x\}$, where $\rho_{\prec}(x) = 0$ if $x$ is minimal, and put $\rho(\prec):=\sup\{\rho_{\prec}(x)+1:x\in X\}$. Then, if $(X,\prec)$ is a $\Sigma_1^1$ well-founded relation, then $\rho(\prec)<\omega_1^{CK}$.
Any help is appreciated, and I hope I am not just completely confused.