isomorphic and $Z$ is $p$–closed, then $Z$ is $q$–closed. If $Z$ is $p$–closed, $\alpha<{{\rm ht}}(p)$ and $p^*$ is an $\alpha$–component of $p$, then $Z\cap\alpha$ is $p^*$–closed.

Easy inductions on ${{\rm ht}}(p)$ (remember Lemma \[3.1\](2)).

\[UUpsilon\] Suppose that $p\in{{{\mathbb P}^\theta_\lambda}}$ and $Z\subseteq {{\rm ht}}(p)$ is a finite $p$–closed set. Let $Z=\{\alpha_0,\ldots,\alpha_{k-1}\}$ be the increasing enumeration.

1.  We define $$U[p,Z]\stackrel{\rm def}{=}\{j\in u^p: (\forall\beta<{{\rm ht}}(p))(h^p(j,\beta) 
    <\theta\ \Rightarrow\ \beta\in Z)\}.$$

2.  We let $$\Upsilon_p(Z)=\langle\zeta_\ell,\tau_\ell,n_\ell,\langle g_\ell,h^\ell_0,
    \ldots,h^\ell_{n_\ell-1}\rangle:\ell<k\rangle,$$ where, for $\ell<k$, $\zeta_\ell$ is an ordinal below $\theta$, $\tau_\ell$ is a Boolean term, $n_\ell<\omega$ and $g_\ell,h^\ell_0,\ldots,h^\ell_{
    n_\ell-1}:\ell\longrightarrow 2$, and they all are such that for every (equivalently: some) $\alpha_\ell+1$–component $q=\langle\zeta^*,\tau^*,
    n^*,u^*,\langle q_\xi,v_\xi:\xi<\theta\rangle\rangle$ of $p$ we have:$\zeta_\ell=\zeta^*$, $\tau_\ell=\tau^*$, $n_\ell=n^*$ and if $v_\xi=\{j_0,
    \ldots,j_{n_\ell-1}\}$ (the increasing enumeration) then $$(\forall m<n_\ell)(\forall\ell'<\ell)(h^\ell_m(\ell')=h^q(j_m,
    \alpha_{\ell'})),$$ and if $i_0=\min(u^{q_{\zeta^*}}\setminus u^*)$ then $(\forall\ell'<\ell)(
    g_\ell(\ell')=h^q(i_0,\alpha_{\ell'}))$. (Note that $\zeta_\ell,\tau_\ell,
    n_\ell$, $g_\ell,h^\ell_0,\ldots,h^\ell_{n_\ell-1}$ are well-defined by Lemma \[3.4x\]. Necessarily, for all $m<n_\ell$ and $\beta\in
    \alpha_\ell\setminus Z$ we have $h^q(i_0,\beta),h^q(j_m,\beta)\geq\theta$; remember that $Z$ is $p$–closed.)

Note that if $Z\subseteq{{\rm ht}}(p)$ is a finite $p$–closed set, $\alpha=\max(Z)$ and $p^*$ is the $\alpha+1$–component of $p$ satisfying $p^*\leq_{\rm pr} p$ (see \[3.4x\](3)), then $U[p,Z]\subseteq u^{p^*}$.

\[3.6.x\] Suppose that $p\in{{{\mathbb P}^\theta_\lambda}}$ and $Z_0,Z_1\subseteq{{\rm ht}}(p)$ are finite $p$–closed sets such that $\Upsilon_p(Z_0)=\Upsilon_p(Z_1)$. Then ${{\rm otp}}(U[p,Z_0])=
