Proof. (i) We are going to proceed by induction. Obviously $W_{s}\subseteq W_{t}$ for all $t\sqsubset s$ with $|s|=|t|$ . By definition of the tree scheme we have that $W_{s}\subseteq W_{t}$ for all $t\sqsubset s$ with $|s|=|t|+1$ . Suppose that for $n\in \unicode[STIX]{x1D714}$ we have $W_{s}\subseteq W_{t}$ for all $t\sqsubset s$ with $|s|=|t|+n$ . Let $t\sqsubset s$ with $|s|=|t|+n+1$ . There is $r\in T$ such that $t\sqsubset r\sqsubset s$ , $|r|=|t|+n$ and $|s|=|r|+1$ , and thus, $W_{r}\subseteq W_{t}$ and $W_{s}\subseteq W_{r}$ , and this completes the inductive step.

(ii) Let $x\in X$ . We set $t_{x}^{0}=\emptyset$ . Suppose that there is $t_{x}^{n}\in T^{n}$ with $x\in W_{t_{x}^{n}}$ . There is $s\in \operatorname{succ}(t_{x}^{n})$ such that $x\in W_{s}$ . We set $t_{x}^{n+1}=s$ . Thus, we can recursively define an increasing (with respect to $\sqsubset$ ) sequence $\{t_{x}^{n}\}_{n\in \unicode[STIX]{x1D714}}$ such that $|t_{x}^{n}|=n$ and $x\in W_{t_{x}^{n}}$ for all $n\in \unicode[STIX]{x1D714}$ . We conclude the result by defining $b_{x}=\bigcup _{n\in \unicode[STIX]{x1D714}}t_{x}^{n}$ .

Now let us examine the uniqueness of $b_{x}$ in the case where $(W_{t})_{t\in T}$ is disjoint. Suppose that there exist $b_{x}\neq b_{x}^{\prime }\in [T]$ such that $x\in \bigcap _{n\in \unicode[STIX]{x1D714}}W_{b_{x}|n}$ and $x\in \bigcap _{n\in \unicode[STIX]{x1D714}}W_{b_{x}^{\prime }|n}$ . There is a minimum $n\in \unicode[STIX]{x1D714}$ such that $b_{x}|n\neq b_{x}^{\prime }|n$ , so $W_{b_{x}|n}\cap W_{b_{x}^{\prime }|n}=\emptyset$ , which is a contradiction.

(iii) Follows easily from (ii).

(iv) Let $(W_{t})_{t\in T}$ be point-finite; the inverse direction is obvious. We are going to proceed by induction. Firstly, $\{W_{t}:|t|=0\}$ is point-finite in $X$ . Suppose that $\{W_{t}:|t|=n\}$ is point-finite in $X$ for some $n\in \unicode[STIX]{x1D714}$ . If $x$ belongs to infinitely many elements of $\{W_{t}:|t|=n+1\}$ , then there should be $t\in T^{n}$ such that $x\in W_{t}$ and $x\in W_{s}$ for infinitely many $s\in \operatorname{succ}(t)$ , which is a contradiction. Thus, $\{W_{t}:|t|=n+1\}$ is point-finite in $X$ .

(v) Let $(W_{t})_{t\in T}$ be a closed tree scheme on $X$ . Assume that $(W_{t})_{t\in T}$ is locally finite; the inverse direction is obvious. First of all, $\{W_{t}:|t|=0\}$ is locally finite in $X$ . We assume that $\{W_{t}:|t|=n\}$ is locally finite in $X$ and we will show that $\{W_{t}:|t|=n+1\}$ is locally finite in $X$ . Let $x\in X$ . By the inductive hypothesis there exists $U_{x}$ , an open neighborhood of $x$ , such that the set,

$$\begin{eqnarray}T_{U_{x}}^{n}=\{t\in T:U_{x}\cap W_{t}\neq \emptyset \text{ and }|t|=n\},\end{eqnarray}$$

is finite. Let also

$$\begin{eqnarray}T_{x}^{n}=\{t\in T:x\in W_{t}\text{ and }|t|=n\},\end{eqnarray}$$

which is finite as a subset of $T_{U_{x}}^{n}$ . We observe that

$$\begin{eqnarray}U_{x}\setminus \bigcup \{W_{t}:t\in T_{U_{x}}^{n}\setminus T_{x}^{n}\}\end{eqnarray}$$

is an open (since $\{W_{t}:t\in T_{U_{x}}^{n}\setminus T_{x}^{n}\}$ is a finite family of closed subsets) neighborhood of $x$ which intersects $W_{t}$ with $|t|=n$ if and only if $t\in T_{x}^{n}$ . $\{W_{s}:s\in \operatorname{succ}(t)\}$ is locally finite in the subspace $W_{t}$ for all $t\in T$ , so for each $t\in T_{x}^{n}$ there exists an open (with respect to the relative topology of $W_{t}$ ) neighborhood $U_{t}(x)$ of $x$ , such that the set,

$$\begin{eqnarray}T_{U_{t}(x)}^{n+1}\cap \operatorname{succ}(t)=\{s\in \operatorname{succ}(t):U_{t}(x)\cap W_{s}\neq \emptyset \},\end{eqnarray}$$

is finite. Since $U_{t}(x)$ is open in $W_{t}$ , there exists an open (with respect to the topology of $X$ ) neighborhood $V_{t}(x)$ of $x$ such that $U_{t}(x)=V_{t}(x)\cap W_{t}$ . Observe that the set

$$\begin{eqnarray}T_{V_{t}(x)}^{n+1}\cap \operatorname{succ}(t)=\{s\in \operatorname{succ}(t):V_{t}(x)\cap W_{s}\neq \emptyset \}\end{eqnarray}$$

is equal to

$$\begin{eqnarray}T_{U_{t}(x)}^{n+1}\cap \operatorname{succ}(t)=\{s\in \operatorname{succ}(t):U_{t}(x)\cap W_{s}\neq \emptyset \}.\end{eqnarray}$$

Thus, the set

$$\begin{eqnarray}\biggl(U_{x}\setminus \bigcup \{W_{t}:t\in T_{U_{x}}^{n}\setminus T_{x}^{n}\}\biggr)\cap \biggl(\bigcap \{V_{t}(x):t\in T_{x}^{n}\}\biggr),\end{eqnarray}$$

is an open (with respect to the topology of $X$ ) neighborhood of $x$ that intersects finitely many elements of $\{W_{t}:|t|=n+1\}$ .◻

Proof. We are going to define the desirable tree schemes recursively. Set $F_{\emptyset }=Y$ and suppose that $F_{t}$ is closed in $Y$ and $F_{t}\subseteq U_{t}$ . Since $F_{t}$ is paracompact Hausdorff, as a subspace of a metrizable space, and $\{F_{t}\cap U_{s}:s\in \operatorname{succ}(t)\}$ is an open cover of $F_{t}$ in the relative topology, it has a locally finite open precise refinement $\{D_{s}:s\in \operatorname{succ}(t)\}$ such that $\operatorname{cl}_{F_{t}}(D_{s})\subseteq F_{t}\cap U_{s}$ for all $s\in \operatorname{succ}(t)$ . We define $F_{s}=\operatorname{cl}_{F_{t}}(D_{s})$ , which is closed in $Y$ , and we observe that

$$\begin{eqnarray}F_{t}=\bigcup \{\text{int}_{F_{t}}(F_{s}):s\in \operatorname{succ}(t)\}=\bigcup \{F_{s}:s\in \operatorname{succ}(t)\}.\end{eqnarray}$$

What is more, since $\{D_{s}:s\in \operatorname{succ}(t)\}$ is locally finite in $F_{t}$ , the same holds for $\{F_{s}:s\in \operatorname{succ}(t)\}$ and $\{\text{int}_{F_{t}}(F_{s}):s\in \operatorname{succ}(t)\}$ . The recursively constructed locally finite closed tree scheme $(F_{t})_{t\in T}$ is a precise refinement of $(U_{t})_{t\in T}$ ; by Lemma 3.3(v), it follows that $\{F_{t}:|t|=n\}$ is locally finite in $Y$ for all $n<\unicode[STIX]{x1D714}$ .

Now set $W_{\emptyset }=Y$ and suppose that $W_{t}$ is open in $Y$ and $W_{t}\subseteq F_{t}$ . Observe that

$$\begin{eqnarray}W_{t}\subseteq \,\text{int}_{X}(F_{t})\subseteq \bigcup \,\{\text{int}_{F_{t}}(F_{s}):s\in \operatorname{succ}(t)\}\end{eqnarray}$$

and

$$\begin{eqnarray}\text{int}_{X}(F_{t})\cap \,\text{int}_{F_{t}}(F_{s})\,\subseteq \,\text{int}_{X}(F_{s}).\end{eqnarray}$$

Hence, $\{W_{t}\cap \,\text{int}_{X}(F_{s}):s\in \operatorname{succ}(t)\}$ is an open cover of $W_{t}$ in the relative topology. Following the same argument as above, there is a locally finite open precise refinement $\{W_{s}:s\in \operatorname{succ}(t)\}$ . Thus, the recursively constructed open tree scheme $(W_{t})_{t\in T}$ is a precise refinement of $(F_{t})_{t\in T}$ , which implies that $\{W_{t}:|t|=n\}$ is locally finite in $Y$ for all $n<\unicode[STIX]{x1D714}$ .◻

Proof. (i) First we are going to construct a complete open tree scheme on $Y$ . Let $d$ be a complete compatible bounded metric on $Y$ and $\{U_{i}:i\in I\}$ a base for the topology of $Y$ consisted of non-empty sets. For each $i\in I$ we define

$$\begin{eqnarray}I(i)=\{j\in I:\overline{U_{j}}\subseteq U_{i}\text{ and }\operatorname{diam}_{d}(\overline{U_{j}})\leqslant {\textstyle \frac{1}{2}}\operatorname{diam}_{d}(U_{i})\}\end{eqnarray}$$

and we observe that $U_{i}=\bigcup _{j\in I(i)}U_{j}=\bigcup _{j\in I(i)}\overline{U_{j}}$ . Then we define the pruned tree $T\subseteq I^{{<}\unicode[STIX]{x1D714}}$ as follows:

$$\begin{eqnarray}t=(i_{1},\ldots ,i_{n})\in T\;\Leftrightarrow \;i_{k+1}\in I(i_{k})\quad \text{for all }k\in \{1,\ldots ,n-1\}\end{eqnarray}$$

and

$$\begin{eqnarray}\operatorname{diam}_{d}(\overline{U_{i_{1}}})\leqslant {\textstyle \frac{1}{2}}\operatorname{diam}_{d}(Y).\end{eqnarray}$$

We set $U_{\emptyset }=Y$ and $U_{t}=U_{i_{n}}$ for every $t=(i_{1},\ldots ,i_{n})\in T$ . We will show that $(U_{t})_{t\in T}$ is a complete open tree scheme on $Y$ . Let $t=(i_{1},\ldots ,i_{n})\in T$ . If $s\in \operatorname{succ}(t)$ , then, according to the definition of $T$ , there exists $i_{n+1}\in I(i_{n})$ so that $s=t^{\frown }i_{n+1}=(i_{1},\ldots ,i_{n},i_{n+1})$ , which implies that $\overline{U_{s}}=\overline{U_{i_{n+1}}}\subseteq U_{i_{n}}=U_{t}$ . Thus,

(1) $$\begin{eqnarray}\displaystyle \bigcup \,\{U_{s}:s\in \operatorname{succ}(t)\}\subseteq U_{t}. & & \displaystyle\end{eqnarray}$$

On the other hand, let $y\in U_{t}$ , where $t=(i_{1},\ldots ,i_{n})\in T$ . There is $\unicode[STIX]{x1D700}>0$ such that $2\unicode[STIX]{x1D700}<\frac{1}{2}\operatorname{diam}_{d}(U_{i_{n}})$ and $\overline{B_{d}(y,\unicode[STIX]{x1D700})}\subseteq U_{t}=U_{i_{n}}$ . Since $\{U_{i}:i\in I\}$ is a base for the metric topology, there exists $j\in I$ such that $y\in U_{j}\subseteq B_{d}(y,\unicode[STIX]{x1D700})$ . Hence, $\overline{U_{j}}\subseteq U_{i_{n}}$ and $\operatorname{diam}_{d}(U_{j})\leqslant 2\unicode[STIX]{x1D700}<\frac{1}{2}\operatorname{diam}_{d}(U_{i_{n}})$ , which implies $j\in I(i_{n})$ . So $y\in U_{s}$ , where $s=(i_{1},\ldots ,i_{n},j)\in \operatorname{succ}(t)$ , and thus,

(2) $$\begin{eqnarray}\displaystyle U_{t}\subseteq \bigcup \,\{U_{s}:s\in \operatorname{succ}(t)\}. & & \displaystyle\end{eqnarray}$$

From (1) and (2) we have

$$\begin{eqnarray}U_{t}=\bigcup \,\{U_{s}:s\in \operatorname{succ}(t)\}\quad \text{for all }t\in T.\end{eqnarray}$$

So $(U_{t})_{t\in T}$ is an open tree scheme. It remains to show that $(U_{t})_{t\in T}$ is complete. Let $b\in [T]$ . We observe that $\bigcap _{n\in \unicode[STIX]{x1D714}}U_{b|n}=\bigcap _{n\in \unicode[STIX]{x1D714}}\overline{U_{b|n}}$ and that $\{\overline{U_{b|n}}:n\in \unicode[STIX]{x1D714}\}$ is a decreasing sequence of non-empty closed subsets with vanishing diameter; therefore, since $d$ is complete, it follows that there is $y_{b}\in Y$ so that $\bigcap _{n\in \unicode[STIX]{x1D714}}U_{b|n}=\{y_{b}\}$ . Moreover, $\{U_{b|n}:n\in \unicode[STIX]{x1D714}\}$ is a neighborhood base at $y_{b}$ , since, as we observed above, $\operatorname{diam}(U_{b|n})\rightarrow 0$ .

Now, by applying Lemma 3.6, $(U_{t})_{t\in T}$ has an open precise refinement $(W_{t})_{t\in T}$ such that $\{W_{t}:|t|=n\}$ is locally finite in $Y$ for all $n<\unicode[STIX]{x1D714}$ . It is easy to see that $(W_{t})_{t\in T}$ is also complete.

(ii) If $Y$ is separable, then the initial base $\{U_{i}:i\in I\}$ in the proof of (i) can be chosen to be countable, and thus, $T$ in (i) would be countably branching.

(iii) From (i) we get a complete open tree scheme $(W_{t})_{t\in T}$ on $Y$ such that $\{W_{t}:|t|=n\}$ is locally finite in $Y$ for all $n<\unicode[STIX]{x1D714}$ . We are going to define recursively a clopen disjoint precise refinement of $(W_{t})_{t\in T}$ . We set $G_{\emptyset }=X$ . Let $G_{t}\subseteq W_{t}$ be clopen and observe that $G_{t}=\bigcup _{s\in \operatorname{succ}(t)}(G_{t}\cap W_{s})$ . Since $G_{t}$ is normal with $\operatorname{dim}(G_{t})=0$ and $\{G_{t}\cap W_{s}:s\in \operatorname{succ}(t)\}$ is locally finite in $G_{t}$ , there exists a clopen precise refinement $\{U_{s}:s\in \operatorname{succ}(t)\}$ . We consider a strict well-order ${<}$ on $\operatorname{succ}(t)$ and we define $G_{s}=U_{s}\setminus \bigcup _{s^{\prime }<s}U_{s^{\prime }}$ , which is clopen since $\{U_{s}:s\in \operatorname{succ}(t)\}$ is locally finite in $G_{t}$ . So the recursively defined $(G_{t})_{t\in T}$ is a clopen disjoint tree scheme, and it is easy to see that, as a precise refinement of $(W_{t})_{t\in T}$ , is also complete.◻

Proof. For (i) it suffices to observe that $\unicode[STIX]{x1D6F7}^{l}(Y)=X$ and that the lower inverse image under $\unicode[STIX]{x1D6F7}$ preserves arbitrary unions; (ii) and (iii) follow immediately from the definition of $\unicode[STIX]{x1D6F4}$ -measurable or continuous multifunction respectively.◻

Proof. (i) Let $(W_{t})_{t\in T}$ be a complete open tree scheme on $Y$ and $(G_{t})_{t\in T}$ a disjoint precise refinement of $(\unicode[STIX]{x1D6F7}^{l}(W_{t}))_{t\in T}$ . By Lemma 3.3(iii), for each $x\in X$ there exists a unique $b_{x}\in [T]$ such that $x\in \bigcap _{n\in \unicode[STIX]{x1D714}}G_{b_{x}|n}$ . Notice that $W_{b_{x}|n}\neq \emptyset$ for all $n\in \unicode[STIX]{x1D714}$ and $(W_{t})_{t\in T}$ is complete, therefore we can define $\unicode[STIX]{x1D719}:X\rightarrow Y$ by $\{\unicode[STIX]{x1D719}(x)\}=\bigcap _{n\in \unicode[STIX]{x1D714}}W_{b_{x}|n}$ . We will show that $\unicode[STIX]{x1D719}$ is a selection of $\unicode[STIX]{x1D6F7}$ . Let $x\in X$ . Since $\unicode[STIX]{x1D6F7}$ is closed-valued, it suffices to show that $\unicode[STIX]{x1D719}(x)\in \overline{\unicode[STIX]{x1D6F7}(x)}$ . But this follows immediately from the fact that $\{W_{b_{x}|n}:n\in \unicode[STIX]{x1D714}\}$ is a neighborhood base at $\unicode[STIX]{x1D719}(x)$ and $\unicode[STIX]{x1D6F7}(x)\cap W_{b_{x}|n}\neq \emptyset$ for all $n\in \unicode[STIX]{x1D714}$ .

(ii) Let $W$ be an open subset of $Y$ . For each $x\in \unicode[STIX]{x1D719}^{-1}(W)$ there exists $n_{x}\in \unicode[STIX]{x1D714}$ such that $\unicode[STIX]{x1D719}(x)\in W_{b_{x}|n_{x}}\subseteq W$ . Then for each $x\in \unicode[STIX]{x1D719}^{-1}(W)$ we have that $x\in G_{b_{x}|n_{x}}\subseteq \unicode[STIX]{x1D719}^{-1}(W_{b_{x}|n_{x}})\subseteq \unicode[STIX]{x1D719}^{-1}(W)$ , and thus, $\unicode[STIX]{x1D719}^{-1}(W)=\bigcup _{x\in \unicode[STIX]{x1D719}^{-1}(W)}G_{b_{x}|n_{x}}$ , which is an open subset of $Y$ since each $G_{b_{x}|n_{x}}$ is open.

(iii) As we proved in (ii) for each $W\subseteq Y$ open we have $\unicode[STIX]{x1D719}^{-1}(W)=\bigcup _{x\in \unicode[STIX]{x1D719}^{-1}(W)}G_{b_{x}|n_{x}}$ . Moreover, $\unicode[STIX]{x1D6F4}$ is closed under countable unions, $\{b_{x}|n_{x}:x\in \unicode[STIX]{x1D719}^{-1}(W)\}$ is countable ( $T$ is countably branching) and each $G_{b_{x}|n_{x}}$ belongs to $\unicode[STIX]{x1D6F4}$ , and thus, $\unicode[STIX]{x1D719}^{-1}(W)\in \unicode[STIX]{x1D6F4}$ .◻

Proof. (i) It suffices to observe that $\{V_{t}:t\in T\}$ is a base for the topology of $[T]$ , the lower inverse image under $\unicode[STIX]{x1D6F7}$ preserves arbitrary unions and finally $\unicode[STIX]{x1D6F7}^{l}(V_{t})=U_{t}\in \unicode[STIX]{x1D6F4}$ for all $t\in T$ .

(ii) Let $V$ be an open subset of $[T]$ and $S\subseteq T$ such that $V=\bigcup _{t\in S}V_{t}$ . Obviously $T$ is countable and $\unicode[STIX]{x1D6F7}^{l}(V)=\bigcup _{t\in S}\unicode[STIX]{x1D6F7}^{l}(V_{t})$ , and thus, $\unicode[STIX]{x1D6F7}^{l}(V)\in \unicode[STIX]{x1D6F4}$ .

(iii) Let $x\in X$ . It is easy to see that $\unicode[STIX]{x1D6F7}(x)=[T_{x}]$ , where $T_{x}=\{t\in T:x\in U_{t}\}$ is a pruned subtree of $T$ . In other words, $\unicode[STIX]{x1D6F7}(x)$ is a body of a pruned subtree of $T$ , and thus, is a closed subset of $[T]$ .

(iv) If, moreover, $(U_{t})_{t\in T}$ is point-finite, then $T_{x}=\{t\in T:x\in U_{t}\}$ is a pruned finitely branching subtree of $T$ , and thus, $\unicode[STIX]{x1D6F7}(x)=[T_{x}]$ is compact.

(v) If $(U_{t})_{t\in T}$ is locally finite and clopen, then, due to (i), it remains to show that $\unicode[STIX]{x1D6F7}$ is upper semicontinuous. Let $V$ be an open subset of $[T]$ and $x\in \unicode[STIX]{x1D6F7}^{u}(V)$ . For every $b\in \unicode[STIX]{x1D6F7}(x)$ there exists $n_{b}\in \unicode[STIX]{x1D714}$ such that $V_{b|n_{b}}\subseteq V$ , so $\{V_{b|n_{b}}\}_{b\in \unicode[STIX]{x1D6F7}(x)}$ is an open cover of $\unicode[STIX]{x1D6F7}(x)$ . From (iv) we have that $\unicode[STIX]{x1D6F7}(x)$ is compact, and hence, there is a subcover $\{V_{t_{i}}\}_{i=1}^{k}$ . We can assume that $|t_{i}|=n_{0}$ for some $n_{0}\in \unicode[STIX]{x1D714}$ and for all $i=1,\ldots ,k$ . Since $(U_{t})_{t\in T}$ is locally finite and clopen, by virtue of Lemma 3.3(v) there exists an open neighborhood $U$ of $x$ such that the set

$$\begin{eqnarray}T_{U}^{n_{0}}=\{t\in T:|t|=n_{0}\text{ and }U\cap U_{t}\neq \emptyset \}\end{eqnarray}$$

is finite. Let

$$\begin{eqnarray}T_{x}^{n_{0}}=\{t\in T:|t|=n_{0}\text{ and }x\in U_{t}\}\end{eqnarray}$$

and

$$\begin{eqnarray}G=U\setminus \bigcup \,\{U_{t}:t\in T_{U}^{n_{0}}\setminus T_{x}^{n_{0}}\}.\end{eqnarray}$$

First we observe that $G$ is an open neighborhood of $x$ . Finally, by the definition of $\unicode[STIX]{x1D6F7}$ , we observe that $G\subseteq \unicode[STIX]{x1D6F7}^{u}(V)$ , since we have that $T_{x}^{n_{0}}\subseteq \{t_{1},\ldots ,t_{k}\}$ and $\unicode[STIX]{x1D6F7}(x)\subseteq \bigcup _{i=1}^{k}V_{t_{i}}$ .◻

Proof. (i) $\Rightarrow$ (ii) Let $(U_{t})_{t\in T}$ be a $\unicode[STIX]{x1D6E4}(X)$ tree scheme on $X$ . Set $G_{\emptyset }=X$ and suppose that $G_{t}\subseteq U_{t}$ and $G_{t}\in \unicode[STIX]{x1D6E5}(X)$ . Since $\{U_{s}:s\in \operatorname{succ}(t)\}\subseteq \unicode[STIX]{x1D6E4}(X)$ , we observe that $\{G_{t}\cap U_{s}:s\in \operatorname{succ}(t)\}\subseteq \unicode[STIX]{x1D6E4}(G_{t})$ covers $G_{t}$ , so it has a disjoint precise refinement $\{G_{s}:s\in \operatorname{succ}(t)\}\subseteq \unicode[STIX]{x1D6E4}(G_{t})$ . From (1) and (2) we conclude that the recursively defined disjoint precise refinement $(G_{t})_{t\in T}$ belongs to $\unicode[STIX]{x1D6E4}(X)$ .

(ii) $\Rightarrow$ (i) Let $Y\in \unicode[STIX]{x1D6E5}(X)$ . We have to show that $\unicode[STIX]{x1D6E4}(Y)$ satisfies the weak reduction principle. Let $\{A_{i}\}_{i\in I}\subseteq \unicode[STIX]{x1D6E4}(Y)$ such that $Y=\bigcup _{i\in I}A_{i}$ . From (2) we have that $\{A_{i}\}_{i\in I}\subseteq \unicode[STIX]{x1D6E4}(X)$ . Set $U_{\emptyset }=X$ . Consider that the immediate successors of $\emptyset$ are exactly $t_{1},t_{2}$ and set $U_{t_{1}}=Y$ and $U_{t_{2}}=X\setminus Y$ . Then consider that the immediate successors of $t_{1}$ is the set $I$ and for every $i\in \operatorname{succ}(t_{1})$ set $U_{i}=A_{i}$ . Finally, consider that $U_{t}=U_{t_{2}}$ for any $t\in \operatorname{succ}(t_{2})$ , and $U_{t}=U_{i}$ for any $t\in \operatorname{succ}(i)$ and $i\in I$ . It is easy to see that in this way we can define a pruned tree $T$ (on some set) and a $\unicode[STIX]{x1D6E4}(X)$ tree scheme $(U_{t})_{t\in T}$ , which, by hypothesis, has a $\unicode[STIX]{x1D6E4}(X)$ disjoint precise refinement $(G_{t})_{t\in T}$ . To complete the proof it suffices to observe that $\{G_{i}\}_{i\in I}\subseteq \unicode[STIX]{x1D6E4}(X)$ is a disjoint precise refinement of $\{A_{i}\}_{i\in I}$ .◻

Proof. (i) $\Rightarrow$ (ii) Let $(U_{t})_{t\in T}$ be a ${\mathcal{A}}(\unicode[STIX]{x1D6E4}(X))$ tree scheme on $X$ . Set $G_{\emptyset }=X$ and suppose that $G_{t}\subseteq U_{t}$ and $G_{t}\in \unicode[STIX]{x1D6E5}(X)$ . Since $\{U_{s}:s\in \operatorname{succ}(t)\}\in {\mathcal{A}}(\unicode[STIX]{x1D6E4}(X))$ , by applying (3) we observe that $\{G_{t}\cap U_{s}:s\in \operatorname{succ}(t)\}$ belongs to ${\mathcal{A}}(\unicode[STIX]{x1D6E4}(G_{t}))$ and, moreover, covers $G_{t}$ , so it has a disjoint precise refinement $\{G_{s}:s\in \operatorname{succ}(t)\}\in \unicode[STIX]{x1D6E4}(G_{t})$ . From (2) we have that the recursively defined disjoint precise refinement $(G_{t})_{t\in T}$ is a $\unicode[STIX]{x1D6E4}(X)$ tree scheme.

(ii) $\Rightarrow$ (i) Let $Y\in \unicode[STIX]{x1D6E5}(X)$ . We have to show that ${\mathcal{A}}(\unicode[STIX]{x1D6E4}(Y))$ satisfies the weak reduction principle. Let $\{A_{i}\}_{i\in I}\in {\mathcal{A}}(\unicode[STIX]{x1D6E4}(Y))$ such that $Y=\bigcup _{i\in I}A_{i}$ . From (4) we have that $\{A_{i}\}_{i\in I}\in {\mathcal{A}}(\unicode[STIX]{x1D6E4}(X))$ . Set $U_{\emptyset }=X$ . Consider that the immediate successors of $\emptyset$ are exactly $t_{1},t_{2}$ and set $U_{t_{1}}=Y$ and $U_{t_{2}}=X\setminus Y$ . From (5) we have that $\{U_{s}:s\in \operatorname{succ}(\emptyset )\}\in {\mathcal{A}}(\unicode[STIX]{x1D6E4}(X))$ . Then consider that the immediate successors of $t_{1}$ is the set $I$ and for every $i\in \operatorname{succ}(t_{1})$ set $U_{i}=A_{i}$ . Finally, consider that $U_{t}=U_{t_{2}}$ for any $t\in \operatorname{succ}(t_{2})$ , and $U_{t}=U_{i}$ for any $t\in \operatorname{succ}(i)$ and $i\in I$ . It is easy to see ((6) should be used) that in this way we can define a pruned tree $T$ (on some set) and a ${\mathcal{A}}(\unicode[STIX]{x1D6E4}(X))$ tree scheme $(U_{t})_{t\in T}$ , which, by hypothesis, has a $\unicode[STIX]{x1D6E4}(X)$ disjoint precise refinement $(G_{t})_{t\in T}$ . To complete the proof it suffices to observe that $\{G_{i}\}_{i\in I}\subseteq \unicode[STIX]{x1D6E4}(X)$ is a disjoint precise refinement of $\{A_{i}\}_{i\in I}$ .◻