Proof Proceed by induction on $\eta $ .

Proof Use [Reference Kechris14, Theorem 22.16].

Proof Use the methods of [Reference Kechris14, Section 8.H].

Proof Use the fact that, if Z is non-empty, then there exists an open continuous surjection $f:\omega ^\omega \longrightarrow Z$ (see [Reference Kechris14, Exercise 7.14]).

Proof If $f:Z\longrightarrow Z$ witnesses that $A\leq B$ in Z, then $f\circ \rho :\omega ^\omega \longrightarrow \omega ^\omega $ will witness that $\rho ^{-1}[A]\leq \rho ^{-1}[B]$ in $\omega ^\omega $ . On the other hand, if $f:\omega ^\omega \longrightarrow \omega ^\omega $ witnesses that $\rho ^{-1}[A]\leq \rho ^{-1}[B]$ in $\omega ^\omega $ , then $\rho \circ (f\upharpoonright Z):Z\longrightarrow Z$ will witness that $A\leq B$ in Z.

Proof Assume that condition (2) fails. We will show that condition (1) holds. Fix such that $f^{-1}[A_0]\neq D$ for every continuous $f:\omega ^\omega \longrightarrow A_0\cup A_1$ . First we claim that Player II does not have a winning strategy in the game $\mathsf {EW}(D,A_0,A_1)$ . Assume, in order to get a contradiction, that there exists a winning strategy $\sigma $ for Player II. Given $x\in \omega ^\omega $ , view x as describing the moves of Player I, then define $f(x)=y$ , where y is the response of Player II to x according to the strategy $\sigma $ . It is easy to realize that f contradicts the assumption at the beginning of this proof.

Since the payoff set of the game $\mathsf {EW}(D,A_0,A_1)$ belongs to $\mathsf {b}\mathbf {\Sigma }(\omega ^\omega )$ , the assumption of $\mathsf {Det}(\mathsf {b}\mathbf {\Sigma }(\omega ^\omega ))$ guarantees the existence of a winning strategy $\tau $ for Player I. Given $y\in \omega ^\omega $ , view y as describing the moves of Player II, then define $g(y)=x$ , where x is the response of Player I to y according to the strategy $\tau $ .

Set $C=g^{-1}[\omega ^\omega \setminus D]$ , and observe that $C\in \mathbf {\Gamma }$ because $\mathbf {\Gamma }$ is continuously closed. Notice that, since $\tau $ is a winning strategy for Player I, for every $y\in A_0\cup A_1$ neither of the following conditions holds:

• • $g(y)\in D$ and $y\in A_0$ ,

• • $g(y)\notin D$ and $y\in A_1$ .

Using this observation, one sees that $A_0\subseteq C$ and $C\cap A_1=\varnothing $ .

Proof For the case $Z=\omega ^\omega $ , apply Lemma 4.3 with $A_0=A$ , $A_1=\omega ^\omega \setminus A$ , and $\mathbf {\Gamma }=B\!\!\downarrow \,$ . To obtain the full result from this particular case, use Lemma 4.2 and the remarks preceding it.

Proof For the case $Z=\omega ^\omega $ , proceed as in [Reference Kechris14, proof of Theorem 21.15], using Theorem 3.2 and Proposition 3.3. To obtain the full result from this particular case, use Lemma 4.2 and the remarks preceding it.

Proof First notice that $\mathcal {U}\neq \varnothing $ because $Z\neq \varnothing $ . It follows that $A\neq \varnothing $ and $A\neq Z$ . In particular, $A\cap U\neq Z$ for every $U\in \mathcal {U}$ . So, by Lemma 4.8, $A\cap V\leq A\cap U<A$ whenever $V\in \mathbf {\Delta }^0_1(Z)$ and $V\subseteq U\in \mathcal {U}$ . Therefore, since Z is zero-dimensional, we can assume without loss of generality that $\mathcal {U}$ is a disjoint clopen cover of Z.

We claim that $U\setminus A\leq A$ for every $U\in \mathcal {U}$ . Pick $U\in \mathcal {U}$ . If $A\cap U=\varnothing $ then $U\setminus A=U\in \mathbf {\Delta }^0_1(Z)$ , hence the claim holds because $A\neq \varnothing $ and $A\neq Z$ . On the other hand, if $A\cap U\neq \varnothing $ then

$$ \begin{align*}U\setminus A=(Z\setminus (A\cap U))\cap U\leq Z\setminus (A\cap U)\leq A, \end{align*} $$

where the first reduction holds by Lemma 4.8 and the second reduction follows from $A\nleq A\cap U$ using Lemma 4.4. In conclusion, we can fix $f_U:Z\longrightarrow Z$ witnessing that $U\setminus A\leq A$ in Z for every $U\in \mathcal {U}$ . It is clear that $f=\bigcup \{f_U\upharpoonright U:U\in \mathcal {U}\}$ will witness that $Z\setminus A\leq A$ .

Proof Assume, in order to get a contradiction, that F has the Baire property. Since $2^\omega \setminus F$ is also a flip-set, we can assume without loss of generality that F is non-meager in $2^\omega $ . By [Reference Kechris14, Proposition 8.26], we can fix $n\in \omega $ and $s\in 2^n$ such that $F\cap \mathsf {N}_s$ is comeager in $\mathsf {N}_s$ . Fix $k\in \omega \setminus n$ and let $h:\mathsf {N}_s\longrightarrow \mathsf {N}_s$ be the homeomorphism defined by

$$ \begin{align*}h(x)(i)=\left\{ \begin{array}{ll} x(i) & \textrm{if }i\neq k,\\ 1-x(i) & \textrm{if }i=k \end{array} \right. \end{align*} $$

for $x\in \mathsf {N}_s$ and $i\in \omega $ . Observe that $(\mathsf {N}_s\cap F)\cap h[\mathsf {N}_s\cap F]$ is comeager in $\mathsf {N}_s$ , hence it is non-empty. It is easy to realize that this contradicts the definition of flip-set.

Proof Using Lemma 4.8, one sees that $V\cap A\leq A$ and $V\setminus A\leq Z\setminus A$ , where both reductions are in Z. On the other hand, since $V\cap A<A$ would contradict the assumption that $V\notin \mathcal {I}(A)$ , we see that $V\cap A\equiv A$ . It follows that $V\setminus A\leq Z\setminus A\equiv A\equiv V\cap A$ . Let $f:Z\longrightarrow Z$ be a function witnessing that $V\setminus A\leq V\cap A$ . Notice that $V\setminus A\neq \varnothing $ , otherwise we would have $V=V\cap A\equiv A$ , contradicting the assumption that $A\notin \mathbf {\Delta }^0_1(Z)$ . So we can fix $z\in V\setminus A$ , and define $g:Z\longrightarrow V$ by setting

$$ \begin{align*}g(x)=\left\{ \begin{array}{ll}x & \textrm{if }x\in V,\\ z & \textrm{if }x\in Z\setminus V. \end{array} \right. \end{align*} $$

Since $V\in \mathbf {\Delta }^0_1(Z)$ , the function g is continuous. Finally, it is straightforward to verify that $g\circ (f\upharpoonright V):V\longrightarrow V$ witnesses that $V\cap A\leq V\setminus A$ in V.

Proof Assume, in order to get a contradiction, that $V\in \mathbf {\Delta }^0_1(Z)\setminus \mathcal {I}(A)$ . Fix a complete metric on Z that induces the given Polish topology. We will recursively construct sets $V_n$ and functions $f_n:V_n\longrightarrow V_n$ for $n\in \omega $ . Before specifying which properties we require from them, we introduce some more notation. Given a set X and a function $f:X\longrightarrow X$ , set $f^0=\mathsf {id}_X$ and $f^1=f$ . Furthermore, given $m,n\in \omega $ such that $m\leq n$ and $z\in 2^\omega $ (or just $z\in 2^{[m,n]}$ ), define

$$ \begin{align*}f_{[m,n]}^z=f_m^{z(m)}\circ\cdots\circ f_n^{z(n)}. \end{align*} $$

We will make sure that the following conditions are satisfied for every $n\in \omega $ , where $\mathsf {diam}(X)$ denotes the diameter of $X\subseteq Z$ :

1. (1) $V_n\in \mathbf {\Delta }^0_1(Z)$ ,

2. (2) $V_n\notin \mathcal {I}(A)$ ,

3. (3) $V_m\supseteq V_n$ whenever $m\leq n$ ,

4. (4) $f_n:V_n\longrightarrow V_n$ witnesses that $V_n\cap A\leq V_n\setminus A$ in $V_n$ ,

5. (5) $\mathsf {diam}(f_{[m,n]}^s[V_{n+1}])\leq 2^{-n}$ whenever $m\leq n$ and $s\in 2^{[m,n]}$ .

Start by setting $V_0=V$ and let $f_0:V_0\longrightarrow V_0$ be given by Lemma 5.3. Now fix $n\in \omega $ , and assume that $V_m$ and $f_m$ have already been constructed for every $m\leq n$ . Fix a partition $\mathcal {U}$ of Z consisting of clopen sets of diameter at most $2^{-n}$ . Given $m\leq n$ and $s\in 2^{[m,n]}$ , define

$$ \begin{align*}\mathcal{V}^s_m=\{(f_{[m,n]}^s)^{-1}[U\cap V_m]:U\in\mathcal{U}\}. \end{align*} $$

Observe that each $\mathcal {V}^s_m\subseteq \mathbf {\Delta }^0_1(V_n)$ because each $f_{[m,n]}^s$ is continuous. Furthermore, it is clear that each $\mathcal {V}^s_m$ consists of pairwise disjoint sets, and that $\bigcup \mathcal {V}^s_m=V_n$ . Since there are only finitely many $m\leq n$ and $s\in 2^{[m,n]}$ , it is possible to obtain a partition $\mathcal {V}\subseteq \mathbf {\Delta }^0_1(V_n)$ of $V_n$ that simultaneously refines each $\mathcal {V}^s_m$ . This clearly implies that any choice of $V_{n+1}\in \mathcal {V}$ will satisfy condition (5). On the other hand since $\mathcal {I}(A)$ is $\sigma $ -additive and $V_n\notin \mathcal {I}(A)$ , it is possible to choose $V_{n+1}\in \mathcal {V}$ such that $V_{n+1}\notin \mathcal {I}(A)$ , thus ensuring that condition (2) is satisfied as well. To obtain $f_{n+1}:V_{n+1}\longrightarrow V_{n+1}$ that satisfies condition (4), simply apply Lemma 5.3. This concludes the construction.

Fix an arbitrary $y_{n+1}\in V_{n+1}$ for $n\in \omega $ . Given $m\in \omega $ and $z\in 2^\omega $ , observe that the sequence $(f_{[m,n]}^z(y_{n+1}):m\leq n)$ is Cauchy by condition (5), hence it makes sense to define

$$ \begin{align*}x_m^z=\underset{n\to\infty}{\mathsf{lim}}f_{[m,n]}^z(y_{n+1}). \end{align*} $$

To conclude the proof, we will show that $F=\{z\in 2^\omega :x_0^z\in A\}$ is a flip-set. Since the function $g:2^\omega \longrightarrow Z$ defined by setting $g(z)=x_0^z$ is continuous and $A\in \mathbf {\Sigma }(Z)$ , Proposition 3.3 and Lemma 5.2 will easily yield a contradiction.

Define $A^0=A$ and $A^1=Z\setminus A$ . Given any $m\in \omega $ and $\varepsilon \in 2$ , it is clear from the definition of $f_m^\varepsilon $ and condition (4) that

$$ \begin{align*}x\in A\text{ iff }f_m^\varepsilon(x)\in A^\varepsilon \end{align*} $$

for every $x\in V_m$ . Furthermore, using the continuity of $f_m^\varepsilon $ and the definition of $x_m^z$ , it is easy to see that

$$ \begin{align*}f_m^{z(m)}(x_{m+1}^z)=x_m^z \end{align*} $$

for every $z\in 2^\omega $ and $m\in \omega $ .

Fix $z\in 2^\omega $ and notice that, by the observations in the previous paragraph,

$$ \begin{align*}x_0^z\in A\text{ iff } x_1^z\in A^{z(0)}\text{ iff }\cdots\text{ iff }x_{m+1}^z\in (\cdots(A^{z(0)})^{z(1)}\cdots)^{z(m)} \end{align*} $$

for every $m\in \omega $ . Now fix $w\in 2^\omega $ and $m\in \omega $ such that $z\upharpoonright \omega \setminus \{m\}=w\upharpoonright \omega \setminus \{m\}$ and $z(m)\neq w(m)$ . We need to show that $x_0^z\in A$ iff $x_0^w\notin A$ . For exactly the same reason as above, we have

$$ \begin{align*}x_0^w\in A\text{ iff } x_1^w\in A^{w(0)}\text{ iff }\cdots\text{ iff }x_{m+1}^w\in (\cdots(A^{w(0)})^{w(1)}\cdots)^{w(m)}. \end{align*} $$

Since $z\upharpoonright m=w\upharpoonright m$ and $z(m)\neq w(m)$ , in order to finish the proof, it will be enough to show that $x_{m+1}^z=x_{m+1}^w$ . To see this, observe that

$$ \begin{align*}x_{m+1}^z=\underset{n\to\infty}{\mathsf{lim}}f_{[m+1,n]}^z(y_{n+1})=\underset{n\to\infty}{\mathsf{lim}}f_{[m+1,n]}^w(y_{n+1})=x_{m+1}^w, \end{align*} $$

where the middle equality uses the assumption that $z\upharpoonright \omega \setminus (m+1)=w\upharpoonright \omega \setminus (m+1)$ .

Proof As one can easily check, it will be enough to show that there exists a partition $\mathcal {V}\subseteq \mathbf {\Delta }^0_1(U)$ of U such that for every $V\in \mathcal {V}$ either $A\cap V\in \mathbf {\Delta }^0_1(Z)$ or $A\cap V$ is non-selfdual in Z. If this were not the case, then, using Theorem 5.4, one could recursively construct a strictly $\leq $ -decreasing sequence of subsets of Z, which would contradict Theorem 4.7.

Proof In order to prove that (1) $\rightarrow $ (2), assume that condition (1) holds. Pick an embedding $j:Z\longrightarrow \omega ^\omega $ , then set $A_0=j[A]$ and $A_1=j[Z\setminus A]$ . Using the fact that $\mathbf {\Sigma }$ is a nice topological pointclass and that Z is a Borel space, it is easy to see that $A_0,A_1\in \mathbf {\Sigma }(\omega ^\omega )$ . Therefore, by the assumption $\mathsf {Det}(\mathbf {\Sigma }(\omega ^\omega ))$ , it is possible to apply Lemma 4.3. Notice that it would be sufficient to show that there exists $B\in \mathbf {\Gamma }$ such that $A_0\subseteq B$ and $B\cap A_1=\varnothing $ , as $j^{-1}[B]=A$ would clearly follow. So assume, in order to get a contradiction, that no such $B\in \mathbf {\Gamma }$ exists. Then, by Lemma 4.3, for all there exists a continuous $f:\omega ^\omega \longrightarrow \omega ^\omega $ such that $f[\omega ^\omega ]\subseteq A_0\cup A_1$ and $f^{-1}[A_0]=B$ . In particular, we can fix such a function f when B is such that . Set $g=j^{-1}\circ f:\omega ^\omega \longrightarrow Z$ , and observe that g is continuous because j is an embedding. Then

$$ \begin{align*}B=f^{-1}[A_0]=f^{-1}[j[A]]=g^{-1}[A]\in\mathbf{\Gamma} \end{align*} $$

by condition (1), contradicting the fact that $\mathbf {\Gamma }$ is non-selfdual.

The implication (2) $\rightarrow $ (3) holds because Z is zero-dimensional. The implication (3) $\rightarrow $ (4) is trivial. In order to prove that (4) $\rightarrow $ (1), assume that $f:Z\longrightarrow \omega ^\omega $ and $B\in \mathbf {\Gamma }$ are such that $A=f^{-1}[B]$ . Pick a continuous $g:\omega ^\omega \longrightarrow Z$ . Since $f\circ g:\omega ^\omega \longrightarrow \omega ^\omega $ is continuous and $\mathbf {\Gamma }$ is continuously closed, one sees that

$$ \begin{align*}g^{-1}[A]=g^{-1}[f^{-1}[B]]=(f\circ g)^{-1}[B]\in\mathbf{\Gamma}. \\[-38pt] \end{align*} $$

Proof In order to prove the left-to-right implication, pick $B\in \mathbf {\Gamma }(W)$ . Since Z is zero-dimensional, we can fix an embedding $j:Z\longrightarrow \omega ^\omega $ . Notice that $i=j\upharpoonright W:W\longrightarrow \omega ^\omega $ is also an embedding, hence by condition (2) of Lemma 6.3 there exists $C\in \mathbf {\Gamma }$ such that $i^{-1}[C]=B$ . Let $A=j^{-1}[C]$ , and observe that $A\cap W=B$ . The fact that $A\in \mathbf {\Gamma }(Z)$ follows from condition (3) of Lemma 6.3. In order to prove the right-to-left implication, pick $A\in \mathbf {\Gamma }(Z)$ . Let $i:W\longrightarrow Z$ be the inclusion. It follows from Lemma 6.2.1 that $A\cap W=i^{-1}[A]\in \mathbf {\Gamma }(W)$ .

Proof First we will prove the existence of $\mathbf {\Gamma }$ . Pick $A\subseteq Z$ such that $\mathbf {\Lambda }=A\!\!\downarrow \,$ . By [Reference Kechris14, Theorem 7.8] and Lemma 6.2.2 we can assume without loss of generality that Z is a closed subspace of $\omega ^\omega $ . Therefore, by [Reference Kechris14, Proposition 2.8], we can fix a retraction $\rho :\omega ^\omega \longrightarrow Z$ . Set $\mathbf {\Gamma }=\rho ^{-1}[A]\!\!\downarrow \,$ in $\omega ^\omega $ , and observe that $\mathbf {\Gamma }$ is non-selfdual by Lemma 4.2. We claim that $\mathbf {\Lambda }=\mathbf {\Gamma }(Z)$ . Since $A=\rho ^{-1}[A]\cap Z\in \mathbf {\Gamma }(Z)$ by Lemma 6.4 and $\mathbf {\Gamma }(Z)$ is continuously closed, one sees that $\mathbf {\Lambda }=A\!\!\downarrow \,\subseteq \mathbf {\Gamma }(Z)$ . To see that the other inclusion holds, pick $B\in \mathbf {\Gamma }(Z)$ . Then $\rho ^{-1}[B]\in \mathbf {\Gamma }$ , hence $\rho ^{-1}[B]\leq \rho ^{-1}[A]$ . It follows from Lemma 4.2 that $B\leq A$ .

Now assume, in order to get a contradiction, that $\mathbf {\Gamma },\mathbf {\Gamma }'\in \mathsf {NSD}_{\mathbf {\Sigma }}(\omega ^\omega )$ are such that $\mathbf {\Gamma }\neq \mathbf {\Gamma }'$ and $\mathbf {\Gamma }(Z)=\mathbf {\Lambda }=\mathbf {\Gamma }'(Z)$ . Notice that is impossible, as an application of Lemma 6.2.3 would contradict the fact that $\mathbf {\Lambda }$ is non-selfdual. Therefore, we can assume without loss of generality that $\mathbf {\Gamma }\subseteq \mathbf {\Gamma }'$ , hence $\mathbf {\Gamma }'\nsubseteq \mathbf {\Gamma }$ . By Lemma 4.4, it follows that . Therefore , which contradicts the fact that $\mathbf {\Gamma }(Z)=\mathbf {\Lambda }$ is non-selfdual.

Proof It will be enough to prove the left-to-right implication, as the other implication is perfectly analogous. So assume that $\mathbf {\Gamma }(Z)\subseteq \mathbf {\Lambda }(Z)$ , and let $B\in \mathbf {\Gamma }(W)$ . Since Z is an uncountable Borel space and W is zero-dimensional, there exists an embedding of W into Z. Hence, using Lemma 6.2.2, we can assume without loss of generality that $W\subseteq Z$ . By Lemma 6.4, there exists $A\in \mathbf {\Gamma }(Z)$ such that $A\cap W=B$ . Since $A\in \mathbf {\Lambda }(Z)$ by our assumption, a further application of Lemma 6.4 shows that $B\in \mathbf {\Lambda }(W)$ .

Proof The inclusion $\subseteq $ holds by Lemma 6.5. In order to prove that the inclusion $\supseteq $ holds, pick $\mathbf {\Gamma }\in \mathsf {NSD}_{\mathbf {\Sigma }}(\omega ^\omega )$ . By Lemma 4.5, it will be enough to show that $\mathbf {\Gamma }(Z)$ is continuously closed and non-selfdual. The fact that $\mathbf {\Gamma }(Z)$ is continuously closed follows from Lemma 6.2.1. Now assume, in order to get a contradiction, that $\mathbf {\Gamma }(Z)$ is selfdual. Then by Lemma 6.2.3, which implies by Theorem 7.1. It follows from Lemma 6.2.4 that , which is a contradiction.

Proof Define $z\in F$ if $\{m\in \omega :\{n\in \omega :\langle m,n\rangle \in z\}\in E_m\}\in D$ . The rest of the proof is a straightforward verification.

Proof This follows from Propositions 8.2 and 8.3 (in case $\eta>\omega $ , use a bijection $\pi :\eta \longrightarrow \omega $ ).

Proof This can be proved by induction on $\xi $ , using Propositions 8.2 and 8.3.

Proof This is a straightforward consequence of Lemma 8.4.

Proof By [Reference Kechris14, Theorem 22.3], we can fix a $2^\omega $ -universal set U for $\mathbf {\Sigma }^0_1(Z)$ . Let $h:2^\omega \longrightarrow (2^\omega )^\omega $ be a homeomorphism, and let $\pi _n:(2^\omega )^\omega \longrightarrow 2^\omega $ be the projection on the n-th coordinate for $n\in \omega $ . Notice that, given any $n\in \omega $ , the function $f_n:2^\omega \times Z\longrightarrow 2^\omega \times Z$ defined by $f_n(x,y)=(\pi _n(h(x)),y)$ is continuous. Let $V_n=f_n^{-1}[U]$ for each n, and observe that each $V_n\in \mathbf {\Sigma }^0_1(2^\omega \times Z)$ . Set $V=\mathcal {H}_D(V_0,V_1,\ldots )$ .

We claim that V is a $2^\omega $ -universal set for $\mathbf {\Gamma }_D(Z)$ . It is clear that $V\in \mathbf {\Gamma }_D(2^\omega \times Z)$ . Furthermore, using Lemma 9.5, one can easily check that $V_x\in \mathbf {\Gamma }_D(Z)$ for every $x\in 2^\omega $ . To complete the proof, fix $A\in \mathbf {\Gamma }_D(Z)$ . Let $A_0,A_1,\ldots \in \mathbf {\Sigma }^0_1(Z)$ be such that $A=\mathcal {H}_D(A_0,A_1,\ldots )$ . Since U is $2^\omega $ -universal, we can fix $z_n\in 2^\omega $ such that $U_{z_n}=A_n$ for every $n\in \omega $ . Set $z=h^{-1}(z_0,z_1,\ldots )$ . It is straightforward to verify that $V_z=A$ .

Proof By Proposition 10.2, we can fix a $2^\omega $ -universal set U for $\mathbf {\Gamma }_D(Z)$ . Fix an embedding $j:2^\omega \longrightarrow Z$ and set $W=j[2^\omega ]$ . Notice that $(j\times \mathsf {id}_Z)[U]\in \mathbf {\Gamma }_D(W\times Z)$ by Lemma 9.5.2. Therefore, by Lemma 9.5.3, there exists $V\in \mathbf {\Gamma }_D(Z\times Z)$ such that $V\cap (W\times Z)=(j\times \mathsf {id}_Z)[U]$ . Using Lemma 9.5 again, one can easily check that V is a Z-universal set for $\mathbf {\Gamma }_D(Z)$ .

Proof Fix a Z-universal set $U\subseteq Z\times Z$ for $\mathbf {\Gamma }_D(Z)$ . Assume, in order to get a contradiction, that $\mathbf {\Gamma }_D(Z)$ is selfdual. Let $f:Z\longrightarrow Z\times Z$ be the function defined by $f(x)=(x,x)$ , and observe that f is continuous. Since , we see that $Z\setminus f^{-1}[U]\in \mathbf {\Gamma }_D(Z)$ . Therefore, since U is Z-universal, we can fix $z\in Z$ such that $U_z=Z\setminus f^{-1}[U]$ . If $z\in U_z$ then $f(z)=(z,z)\in U$ by the definition of $U_z$ , contradicting the fact that $U_z=Z\setminus f^{-1}[U]$ . On the other hand, if $z\notin U_z$ then $f(z)=(z,z)\notin U$ by the definition of $U_z$ , contradicting the fact that $Z\setminus U_z=f^{-1}[U]$ .

Proof Pick $D\subseteq \mathcal {P}(\omega )$ . The fact that $\mathbf {\Gamma }_D(Z)$ is non-selfdual follows from Corollary 10.3 and Lemma 10.4. Therefore, it will be enough to show that $\mathbf {\Gamma }_D(Z)$ is a Wadge class. By Proposition 10.2, we can fix a $2^\omega $ -universal set $U\subseteq 2^\omega \times Z$ for $\mathbf {\Gamma }_D(Z)$ . Fix an embedding $j:2^\omega \times Z\longrightarrow Z$ and set $W=j[2^\omega \times Z]$ . By Lemma 9.5, we can fix $A\in \mathbf {\Gamma }_D(Z)$ such that $A\cap W=j[U]$ . We claim that $\mathbf {\Gamma }_D(Z)=A\!\!\downarrow \,$ . The inclusion $\supseteq $ follows from Lemma 9.5.1. In order to prove the other inclusion, pick $B\in \mathbf {\Gamma }_D(Z)$ . Since U is $2^\omega $ -universal, we can fix $z\in 2^\omega $ such that $B=U_z$ . Consider the function $f:Z\longrightarrow 2^\omega \times Z$ defined by $f(x)=(z,x)$ , and observe that f is continuous. It is straightforward to check that $j\circ f:Z\longrightarrow Z$ witnesses that $B\leq A$ in Z.

Proof Pick $A\subseteq Z$ such that $\mathbf {\Gamma }=A\!\!\downarrow \,$ . By Theorem 11.1, we can fix the minimal $\eta $ such that $1\leq \eta <\omega _1$ and . We will only give the proof in the case $A\in \mathsf {D}_\eta (\mathbf {\Sigma }^0_1(Z))$ , as the other case is perfectly analogous. More specifically, assume that $A=\mathsf {D}_\eta (A_\xi :\xi <\eta )$ , where each $A_\xi \in \mathbf {\Sigma }^0_1(Z)$ and $(A_\xi :\xi <\eta )$ is a $\subseteq $ -increasing sequence. We claim that $\mathbf {\Gamma }=\mathsf {D}_\eta (\mathbf {\Sigma }^0_1(Z))$ . By Lemma 4.4 and the fact that $\mathsf {D}_\eta (\mathbf {\Sigma }^0_1(Z))\in \mathsf {NSD}(Z)$ , it will be enough to show that .

Assume, in order to get a contradiction, that . More specifically, assume that $Z\setminus A=\mathsf {D}_\eta (B_\xi :\xi <\eta )$ , where each $B_\xi \in \mathbf {\Sigma }^0_1(Z)$ and $(B_\xi :\xi <\eta )$ is a $\subseteq $ -increasing sequence. First assume that $\eta $ is a limit ordinal. Define $\mathcal {U}=\{A_\xi :\xi <\eta \}\cup \{B_\xi :\xi <\eta \}$ . It is clear that $\mathcal {U}$ is an open cover of Z. Furthermore, it is easy to realize that for every $U\in \mathcal {U}$ there exists $\xi <\eta $ such that either $A\cap U\in \mathsf {D}_\xi (\mathbf {\Sigma }^0_1(Z))$ or $(Z\setminus A)\cap U\in \mathsf {D}_\xi (\mathbf {\Sigma }^0_1(Z))$ . Using Lemma 4.8 and the fact that Z is zero-dimensional, we can assume without loss of generality that $\mathcal {U}$ consists of clopen subsets of Z.

We claim that $A\cap U<A$ for every $U\in \mathcal {U}$ . This will conclude the proof by Proposition 5.1, as it will contradict the fact that A is non-selfdual. Pick $U\in \mathcal {U}$ . If there exists $\xi <\eta $ such that $A\cap U\in \mathsf {D}_\xi (\mathbf {\Sigma }^0_1(Z))$ , then the claim holds by the minimality of $\eta $ . If $(Z\setminus A)\cap U\in \mathsf {D}_\xi (\mathbf {\Sigma }^0_1(Z))$ , then

by Lemma 4.8. Hence the claim follows from the minimality of $\eta $ again.

Finally, assume that $\eta =\zeta +1$ is a successor ordinal. Notice that $\zeta \neq 0$ , otherwise A would be a clopen subset of Z such that $\varnothing \subsetneq A\subsetneq Z$ , contradicting the fact that A is non-selfdual. So it makes sense to set $C=\mathsf {D}_\zeta (A_\xi :\xi <\zeta )$ and $D=\mathsf {D}_\zeta (B_\xi :\xi <\zeta )$ . Observe that $A=A_\zeta \setminus C$ and $Z\setminus A=B_\zeta \setminus D$ . Since $A_\zeta \cup B_\zeta =Z$ , using the fact that Z is zero-dimensional it is possible to find $U,V\in \mathbf {\Delta }^0_1(Z)$ such that $U\subseteq A_\zeta $ , $V\subseteq B_\zeta $ , $U\cup V=Z$ and $U\cap V=\varnothing $ . Notice that and $A\cap V=D\cap V\in \mathsf {D}_\zeta (\mathbf {\Sigma }^0_1(Z))$ by Lemma 4.8. As above, this contradicts the fact that A is non-selfdual.

Proof The result is trivial if $\mathcal {A}=\varnothing $ , so assume that $\mathcal {A}\neq \varnothing $ . Let $\mathcal {A}=\{A_n:n\in \omega \}$ be an enumeration. We will proceed by induction on $\xi $ . First assume that $\xi =2$ . Pick $A_{(n,i)}\in \mathbf {\Pi }^0_1(Z)$ for $n,i\in \omega $ such that $A_n=\bigcup _{i\in \omega }A_{(n,i)}$ . Since Z is zero-dimensional, it is possible to pick $A_{(n,i,j)}\in \mathbf {\Delta }^0_1(Z)$ for $n,i,j\in \omega $ such that $Z\setminus A_{(n,i)}=\bigcup _{j\in \omega }A_{(n,i,j)}$ and $A_{(n,i,j)}\cap A_{(n,i,j')}=\varnothing $ whenever $j\neq j'$ . Define $f_{(n,i)}:Z\longrightarrow \omega $ for $n,i\in \omega $ by setting

$$ \begin{align*}f_{(n,i)}(x)=\left\{ \begin{array}{ll} 0 & \textrm{if }x\in A_{(n,i)},\\ j+1 & \textrm{if }x\in A_{(n,i,j)}. \end{array} \right. \end{align*} $$

Fix a bijection $\langle \cdot ,\cdot \rangle :\omega \times \omega \longrightarrow \omega $ and define $f:Z\longrightarrow \omega ^\omega $ by setting $f(x)(\langle n,i\rangle )=f_{(n,i)}(x)$ for $n,i\in \omega $ . It is clear that condition (3) holds.

To show that condition (1) holds, pick $x_m\in Z$ for $m\in \omega $ such that $(x_m,f(x_m))\to (x,y)\in Z\times \omega ^\omega $ . We need to show that $y=f(x)$ . So fix $k=\langle n,i\rangle \in \omega $ . By the definition of f, in order to see that $y(k)=f(x)(k)$ , we need to show that $y(k)=f_{(n,i)}(x)$ . But $f_{(n,i)}(x_m)=f(x_m)(k)$ is eventually constant (with value $y(k)$ ), hence all but finitely many $x_m$ belong to $A_{(n,i)}$ , or there exists $j\in \omega $ such that all but finitely many $x_m$ belong to $A_{(n,i,j)}$ . Since $x_m\to x$ and these sets are all closed, it follows that $f_{(n,i)}(x)=y(k)$ .

To show that condition (2) holds for a given $n\in \omega $ , pick $x\in A_n$ . We need to find $V\in \mathbf {\Sigma }^0_1(F)$ such that $(x,f(x))\in V\subseteq f^\ast [A_n]$ . Pick $i\in \omega $ such that $x\in A_{(n,i)}$ , then set $U=\{y\in \omega ^\omega :y(\langle n,i\rangle )=0\}$ , and observe that $U\in \mathbf {\Delta }^0_1(\omega ^\omega )$ . Using the definition of f, it is easy to realize that $V=(U\times Z)\cap F$ is as required.

Now assume that $\xi \geq 3$ and that the theorem holds for all $\xi '$ such that $2\leq \xi '<\xi $ . Pick $A_{(n,i)}\in \mathbf {\Pi }^0_{\xi (n,i)}(Z)$ for $n,i\in \omega $ such that $A_n=\bigcup _{i\in \omega }A_{(n,i)}$ , where $1\leq \xi (n,i)<\xi $ . Then pick $A_{(n,i,j)}\in \mathbf {\Delta }^0_{\xi (n,i)}$ for $n,i,j\in \omega $ such that $Z\setminus A_{(n,i)}=\bigcup _{j\in \omega }A_{(n,i,j)}$ and $A_{(n,i,j)}\cap A_{(n,i,j')}=\varnothing $ whenever $j\neq j'$ .

Set $\mathcal {A}_{(n,i)}=\{A_{(n,i,j)}:j\in \omega \}\cup \{Z\setminus A_{(n,i,j)}:j\in \omega \}$ for $(n,i)\in \omega \times \omega $ . By the inductive hypothesis, for each $(n,i)$ we can fix a $\xi (n,i)$ -refining function $g_{(n,i)}:Z\longrightarrow \omega ^\omega $ for $\mathcal {A}_{(n,i)}$ . This means that the following conditions will hold, where $G_{(n,i)}=g_{(n,i)}^\ast [Z]$ denotes the graph of $g_{(n,i)}$ :

1. (4) $G_{(n,i)}$ is closed in $Z\times \omega ^\omega $ ,

2. (5) $g_{(n,i)}^\ast [A_{(n,i,j)}]\in \mathbf {\Delta }^0_1(G_{(n,i)})$ for every $j\in \omega $ ,

3. (6) For all $k\in \omega $ there exists $\xi _k$ such that $1\leq \xi _k<\xi (n,i)$ and $(g_{(n,i)})_k^{-1}(j)\in \mathbf {\Pi }^0_{\xi _k}(Z)$ for every $j\in \omega $ .

Fix a bijection $\langle \cdot ,\cdot ,\cdot \rangle :\omega \times \omega \times \omega \longrightarrow \omega $ and define $g:Z\longrightarrow \omega ^\omega $ by setting $g(x)(\langle n,i,j\rangle )=g_{(n,i)}(x)(j)$ for every $x\in Z$ and $n,i,j\in \omega $ . Denote by $G=g^\ast [Z]$ the graph of g. Using condition $(4)$ and arguments as in the case $\xi =2$ , one can show that G is closed in $Z\times \omega ^\omega $ . Using condition $(5)$ , it is easy to see that each $g^\ast [A_{(n,i,j)}]\in \mathbf {\Delta }^0_1(G)$ . Since $G\setminus g^\ast [A_{(n,i)}]=\bigcup _{j\in \omega }g^\ast [A_{(n,i,j)}]$ , by proceeding as in the proof of the case $\xi =2$ it is possible to obtain $h:G\longrightarrow \omega ^\omega $ that satisfies the following conditions, where $H=h^\ast [G]$ denotes the graph of h:

1. (7) H is closed in $G\times \omega ^\omega $ ,

2. (8) $h^\ast [g^\ast [A_n]]\in \mathbf {\Sigma }^0_1(H)$ for every $n\in \omega $ ,

3. (9) $h_{\langle n,i\rangle }^{-1}(0)=g^\ast [A_{(n,i)}]$ and $h_{\langle n,i\rangle }^{-1}(j+1)=g^\ast [A_{(n,i,j)}]$ for every $n,i,j\in \omega $ .

Finally, define $f:Z\longrightarrow \omega ^{\omega +\omega }$ by setting $f_k(x)=g_k(x)$ and $f_{\omega +k}(x)=h_k(x,g(x))$ for every $x\in Z$ and $k\in \omega $ . Also set $F=f^\ast [Z]$ . Using conditions $(6)$ and $(9)$ , it is easy to check that condition (3) will hold. By identifying $\omega ^{\omega +\omega }$ with $\omega ^\omega \times \omega ^\omega $ in the obvious way, F can be identified with H. Therefore, condition (2) holds by condition $(8)$ . Furthermore, since condition $(7)$ holds and $G\times \omega ^\omega $ is closed in $Z\times \omega ^\omega \times \omega ^\omega $ , it follows that condition (1) holds.

Proof The case $\xi =1$ is trivial, so assume that $\xi \geq 2$ . Then the desired result follows from Theorem 12.2, by setting $W=F$ and $f=f^\ast $ .

Proof Let $\mathcal {U}$ be a countable base for $(Z,\tau )$ . Let f and W be given by applying Corollary 12.3 with $\mathcal {A}=\mathcal {B}\cup \mathcal {U}$ , then define

$$ \begin{align*}\sigma=\{f^{-1}[U]:U\in\mathbf{\Sigma}^0_1(W)\}. \end{align*} $$

It is straightforward to verify that $\sigma $ satisfies all of the desired properties.

Proof By [Reference Kechris14, Lemma 13.3], it will be enough to consider the case $\mathcal {A}=\{A\}$ . We will proceed by induction on $\xi $ . The case $\xi =0$ is Corollary 12.4. Now assume that $\xi>0$ and the result holds for every $\xi '<\xi $ . Write $A=\bigcup _{n\in \omega }(Z\setminus A_n)$ , where each $A_n\in \mathbf {\Sigma }^0_{1+\eta +\xi _n}(Z,\tau )$ for suitable $\xi _n<\xi $ . By the inductive assumption, for each n, we can fix a zero-dimensional Polish topology $\sigma _n$ on the set Z such that $\tau \subseteq \sigma _n\subseteq \mathbf {\Sigma }^0_{1+\eta }(Z,\tau )$ and $A_n\in \mathbf {\Sigma }^0_{1+\xi }(Z,\sigma _n)$ . Using [Reference Kechris14, Lemma 13.3] again, it is easy to check that the topology $\sigma $ on Z generated by $\bigcup _{n\in \omega }\sigma _n$ is as desired.

Proof The space W is simply the set Z with the finer topology given by Theorem 12.5, while $f=\mathsf {id}_Z$ .

Proof This is proved like Proposition 9.3 (in fact, the same D will work).

Proof This is proved by combining Propositions 9.4 and 8.3.

Proof This is proved by combining Proposition 13.5 and Theorem 10.5.

Proof This is a straightforward consequence of Proposition 8.4.

Proof If $\mathcal {A}=\varnothing $ the desired result is trivial, so assume that $\mathcal {A}\neq \varnothing $ . Let $\mathcal {A}=\{A_m:m\in \omega \}$ be an enumeration. Given $m\in \omega $ , fix $B_{(m,n)}\in \mathbf {\Sigma }^0_{1+\xi }(Z)$ for $n\in \omega $ such that $A_m=\mathcal {H}_D(B_{(m,0)},B_{(m,1)},\ldots )$ . Define $\mathcal {B}=\{B_{(m,n)}:m,n\in \omega \}$ . By Corollary 12.3, we can fix a zero-dimensional Polish space W and a $\mathbf {\Sigma }^0_{1+\xi }$ -measurable bijection $f:Z\longrightarrow W$ such that $f[B]\in \mathbf {\Sigma }^0_1(W)$ for every $B\in \mathcal {B}$ . It remains to observe that

$$ \begin{align*}f[A_m]=f[\mathcal{H}_D(B_{(m,0)},B_{(m,1)},\ldots)]=\mathcal{H}_D(f[B_{(m,0)}],f[B_{(m,1)}],\ldots)\in\mathbf{\Gamma}_D(W) \end{align*} $$

for every $m\in \omega $ , where the second equality follows from Proposition 8.4.2.

Proof The inclusion $\mathbf {\Gamma }_D(Z)^{(\xi )}\subseteq \mathbf {\Gamma }_D^{(\xi )}(Z)$ follows from Lemma 13.7.2. In order to prove the other inclusion, pick $A\in \mathbf {\Gamma }_D^{(\xi )}(Z)$ . By Lemma 13.8, we can fix a zero-dimensional Polish space W and a $\mathbf {\Sigma }^0_{1+\xi }$ -measurable bijection $f:Z\longrightarrow W$ such that $f[A]\in \mathbf {\Gamma }_D(W)$ . Since $2^\omega $ embeds in Z and W is zero-dimensional, using Lemma 9.5.2 we can assume without loss of generality that W is a subspace of Z, so that $f:Z\longrightarrow Z$ . By Lemma 9.5.3, we can fix $B\in \mathbf {\Gamma }_D(Z)$ such that $B\cap W=f[A]$ . It is easy to check that $A=f^{-1}[B]$ , which concludes the proof.

Proof This follows from Lemma 13.9 and Proposition 13.5.

Proof Pick $A\in \mathbf {\Gamma }(W)$ . To see that $f^{-1}[A]\in \mathbf {\Gamma }^{(\xi )}(Z)$ , we have to show that $g^{-1}[f^{-1}[A]]\in \mathbf {\Gamma }^{(\xi )}$ for every continuous $g:\omega ^\omega \longrightarrow Z$ . So pick such a g. Since $A\in \mathbf {\Gamma }(W)$ , by condition (3) of Lemma 6.3, we can fix an embedding $j:W\longrightarrow \omega ^\omega $ and $B\in \mathbf {\Gamma }$ such that $A=j^{-1}[B]$ . The proof is concluded by observing that

$$ \begin{align*}g^{-1}[f^{-1}[A]]=g^{-1}[f^{-1}[j^{-1}[B]]]=(j\circ f\circ g)^{-1}[B]\in\mathbf{\Gamma}^{(\xi)} \end{align*} $$

by the definition of expansion, since $j\circ f\circ g$ is $\mathbf {\Sigma }^0_{1+\xi }$ -measurable by Lemma 18.1.

Proof In order to prove the inclusion $\subseteq $ , pick $A\in \mathbf {\Gamma }(Z)^{(\xi )}$ . We have to show that $g^{-1}[A]\in \mathbf {\Gamma }^{(\xi )}$ for every continuous function $g:\omega ^\omega \longrightarrow Z$ . So pick such a g. By the definition of expansion, there exists a $\mathbf {\Sigma }^0_{1+\xi }$ -measurable function $f:Z\longrightarrow Z$ and $B\in \mathbf {\Gamma }(Z)$ such that $f^{-1}[B]=A$ . By condition (3) of Lemma 6.3, there exists an embedding $j:Z\longrightarrow \omega ^\omega $ and $C\in \mathbf {\Gamma }$ such that $B=j^{-1}[C]$ . Using Lemma 18.1, one sees that $j\circ f\circ g:\omega ^\omega \longrightarrow \omega ^\omega $ is a $\mathbf {\Sigma }^0_{1+\xi }$ -measurable function. Therefore $g^{-1}[A]=(j\circ f\circ g)^{-1}[C]\in \mathbf {\Gamma }^{(\xi )}$ by the definition of expansion.

In order to prove the inclusion $\supseteq $ , pick $A\in \mathbf {\Gamma }^{(\xi )}(Z)$ . By [Reference Kechris14, Theorem 7.8], we can fix an embedding $i:Z\longrightarrow \omega ^\omega $ such that $i[Z]$ is closed in $\omega ^\omega $ . Therefore, by [Reference Kechris14, Proposition 2.8], there exists a retraction $\rho :\omega ^\omega \longrightarrow i[Z]$ . Observe that $i[A]\in \mathbf {\Gamma }^{(\xi )}(i[Z])$ by Lemma 6.2.2. Set $A'=\rho ^{-1}[i[A]]$ , and observe that $A'\in \mathbf {\Gamma }^{(\xi )}$ . Therefore, there exist a $\mathbf {\Sigma }^0_{1+\xi }$ -measurable function $f:\omega ^\omega \longrightarrow \omega ^\omega $ and $B'\in \mathbf {\Gamma }$ such that $f^{-1}[B']=A'$ . Since Z is uncountable, we can fix an embedding $j:\omega ^\omega \longrightarrow Z$ . By Lemmas 6.2.2, 6.2.4 and 6.4, there exists $B\in \mathbf {\Gamma }(Z)$ such that $B\cap j[\omega ^\omega ]=j[B']$ . The proof is concluded by observing that $A=(j\circ f\circ i)^{-1}[B]$ , and that $j\circ f\circ i$ is $\mathbf {\Sigma }^0_{1+\xi }$ -measurable by Lemma 18.1.