Proof The backward direction is obvious, as by $ \mathsf {SLO}^{\mathsf {W}}_X$ we have $ A_n \le _{\mathsf {W}}^X \neg A $ for all $ n \in \omega $ : if $ (f_n)_{n \in \omega } $ are continuous maps witnessing this, then $ f = \bigcup _{n \in \omega } (f_n \restriction V_n) $ is a continuous map witnessing $ A \le _{\mathsf {W}}^X \neg A $ .

Conversely, let $ X $ be a closed subspace of $ \mathcal {N} $ and $ \rho \colon \mathcal {N} \to X $ be a retraction. If $ A $ is selfdual in $ X $ , then $ A' = \rho ^{-1}(A) $ is selfdual in $ \mathcal {N} $ . Then the set of those $ s \in \omega ^{<\omega } $ such that $ A' \le _{\mathsf {W}}^{\mathcal {N},\boldsymbol {N}_s} A' \cap \boldsymbol {N}_s $ is a nonempty well-founded tree $ \boldsymbol {T}(A') \subseteq \omega ^{<\omega } $ [Reference Andretta1, Lemma 22]. Let $ (V_n)_{n \in \omega } $ be an enumeration of the sets of the form $ \boldsymbol {N}_{s {}^{\smallfrown {}} k} \cap X $ with $ k \in \omega $ and $ s $ a leaf in $ \boldsymbol {T}(A') $ , and set $ A_n = X $ if $ V_n \subseteq A $ and $ A_n = A \cap V_n $ otherwise. It is easy to check that the $ V_n $ ’s and $ A_n $ ’s are as required.

The additional part on nonselfduality of the $ A_n $ ’s follows by iterating the construction on those $ A_n $ which happen to be selfdual in $ X $ . (This process must terminate on each branch after finitely many steps because otherwise we would construct a strictly $ \le _{\mathsf {W}}^X $ -decreasing sequence of subsets of $ X $ , against Proposition 2.3.)

Finally, set $ \alpha = ||A||_{\mathsf {W}}^X $ and $ \alpha _n = ||A_n||_{\mathsf {W}}^X $ , so that in particular $ \alpha _n < \alpha $ . If $ B \subseteq X $ is a set with $ ||B||_{\mathsf {W}}^X = \sup _{n \in \omega } (\alpha _n+1) $ , then $ A_n <_{\mathsf {W}}^X B $ because $ ||A_n||_{\mathsf {W}}^X = \alpha _n < \alpha _n+1 \leq ||B||_{\mathsf {W}}^X $ : if $ f_n $ witnesses $ A_n \le _{\mathsf {W}}^X B $ , then $ A \le _{\mathsf {W}}^X B $ via $ \bigcup _{n \in \omega } (f_n \restriction V_n) $ , thus $ \sup _{n \in \omega } (\alpha _n+1) = ||B||_{\mathsf {W}}^X \geq ||A||_{\mathsf {W}}^X = \alpha $ . The other inequality is obvious.

Proof Without loss of generality, $ X $ is a closed subset of $ Y $ and $ Y $ is a closed subspace of some $ X ' $ which is homeomorphic to $ X $ via some $ h \colon X \to X' $ . Let $ \rho _1 \colon Y \to X $ and $ \rho _2 \colon X' \to Y $ be retractions. We already observed that $ A \mapsto \rho _1^{-1} (A) $ provides an embedding of into , and hence it canonically induces an embedding of into : we claim that the latter is also surjective (hence an isomorphism), i.e., that for every $ B \subseteq Y $ there is $ A \subseteq X $ such that $ B \equiv _{\mathsf {W}}^Y \rho _1^{-1}(A) $ . To see this, it is enough to notice that since $ \rho _1 $ and $ \rho _2 $ are retractions and $ h $ is a homeomorphism, then

$$\begin{align*}B \equiv_{\mathsf{W}}^{Y,X'} \rho_2^{-1}(B) \equiv_{\mathsf{W}}^{X',X} h^{-1}(\rho_2^{-1}(B)) \equiv_{\mathsf{W}}^{X,Y} \rho_1^{-1}(h^{-1}(\rho_2^{-1}(B))). \end{align*}$$

Thus setting $ A = h^{-1}(\rho _2^{-1}(B)) $ we are done.

Proof Let $ A \subseteq X $ be complete for $ \boldsymbol {\Lambda } $ . If $ B \in (\rho ^{-1}(\boldsymbol {\Lambda }))(X) $ then $ \rho ^{-1}(B) \in \rho ^{-1}(\boldsymbol {\Lambda }) $ because $ \rho $ is continuous, thus $ \rho ^{-1}(B) \le _{\mathsf {W}}^{\mathcal {N}} \rho ^{-1}(A) $ by definition of $ \rho ^{-1}(\boldsymbol {\Lambda }) $ , hence $ B \le _{\mathsf {W}}^X A $ , and finally $ B \in \boldsymbol {\Lambda } $ by choice of $ A $ .

Conversely, pick any $ B \in \boldsymbol {\Lambda } $ , let $ f \colon X \to X $ witness $ B \le _{\mathsf {W}}^X A $ , and let $ g \colon \mathcal {N} \to X $ be any continuous function. Then $ f \circ g $ witnesses $ g^{-1}(B) \le _{\mathsf {W}}^{\mathcal {N},X} A $ , and since $ A \equiv _{\mathsf {W}} ^{X,\mathcal {N}} \rho ^{-1}(A) $ we get $ g^{-1}(B) \le _{\mathsf {W}}^{\mathcal {N}} \rho ^{-1}(A) $ , i.e., $ g^{-1}(B) \in \rho ^{-1}(\boldsymbol {\Lambda }) $ . Since $ g $ was arbitrary, this shows that $ B \in (\rho ^{-1}(\boldsymbol {\Lambda }))(X) $ , as desired.

Proof Without loss of generality, we may assume that $ Y $ (and hence also $ X $ ) is a closed subset of $ \mathcal {N} $ . Let $ A $ and $ \rho $ be as in the statement, and fix a retraction $ \rho ' \colon \mathcal {N} \to Y $ , so that $ \rho \circ \rho ' $ is a retraction of $ \mathcal {N} $ onto $ X $ . Then $ (\rho \circ \rho ')^{-1}(\boldsymbol {\Gamma }(X)) = \boldsymbol {\Gamma } $ by Lemma 3.4 applied to $ \boldsymbol {\Lambda } = \boldsymbol {\Gamma }(X) $ and Theorem 2.9(1). Using this and applying once again Lemma 3.4 with $ \boldsymbol {\Lambda } = \rho ^{-1}(\boldsymbol {\Gamma }(X)) \in \mathrm {Class}_{\mathsf {W}}(Y) $ , it follows that

$$\begin{align*}\rho^{-1}(\boldsymbol{\Gamma}(X)) = ((\rho')^{-1}(\rho^{-1}(\boldsymbol{\Gamma}(X))))(Y) = ((\rho \circ \rho')^{-1}(\boldsymbol{\Gamma}(X))) (Y) = \boldsymbol{\Gamma}(Y) , \end{align*}$$

which proves at once that $ \boldsymbol {\Gamma }(Y) $ is nonselfdual (because so is $ \rho ^{-1}(\boldsymbol {\Gamma }(X)) $ ) and that $ \rho ^{-1}(A) $ is complete for $ \boldsymbol {\Gamma }(Y) $ .

Proof For each $ i \in I $ pick a retraction $ \rho _i \colon X \to X_i $ , so that $ \boldsymbol {\Gamma }_i(X) = \rho _i^{-1}(A_i) {\downarrow }_X $ by Corollary 3.5. Since the identity function witnesses $ A_i \le _{\mathsf {W}}^{X_i,X} A $ and $\rho _i^{-1}(A_i) \equiv _{\mathsf {W}}^{X,X_i} A_i $ , we get $ \boldsymbol {\Gamma }_i(X) \subseteq A {\downarrow }_X = \boldsymbol {\Gamma } $ . Moreover, if $ B \subseteq X $ is such that $ B {\downarrow }_X \supseteq \boldsymbol {\Gamma }_i(X) $ , then $ \rho _i^{-1}(A_i) \le _{\mathsf {W}}^X B $ , and hence $ A_i \le _{\mathsf {W}}^{X_i,X} B $ via some $ f_i \colon X_i \to X $ . It follows that $ \bigcup _{i \in I } f_i $ witnesses $ A \le _{\mathsf {W}}^X B $ , and thus $ \boldsymbol {\Gamma } \subseteq B {\downarrow }_X $ .

For the second part, notice that we can assume without loss of generality that $ X $ is closed in $ \mathcal {N} $ , so that the $ X_i $ ’s are closed in $ \mathcal {N} $ as well. Let $ r_i \colon \mathcal {N} \to X_i $ be retractions. If $ i,j \in I $ are as in the statement, then using Corollary 3.5 again we get

$$\begin{align*}X_i \setminus A_i \equiv_{\mathsf{W}}^{X_i,\mathcal{N}} {\mathcal{N} \setminus r_i^{-1}(A_i)} \le_{\mathsf{W}}^{\mathcal{N}} r_{j}^{-1}(A_{j}) \equiv_{\mathsf{W}}^{\mathcal{N},X_{j}} A_{j}, \end{align*}$$

where the middle reduction exists because . If $ f_i \colon X_i \to X_{j} $ witnesses $ X_i \setminus A_i \le _{\mathsf {W}}^{X_i,X_{j}} A_{j} $ , then $ \bigcup _{i \in I} f_i $ witnesses $ \neg A \le _{\mathsf {W}}^X A $ , as desired.

Fact 3.8. A zero-dimensional second-countable space is not compact if and only if it admits an infinite (countable) clopen partition.

Proof If U is infinite discrete then it is not compact, and if it is closed in $ X $ then $ X $ cannot be compact either.

Conversely, suppose that X is not compact. By Fact 3.8, there exists an infinite clopen partition $(C_n)_{n \in \omega }$ of X. Since $\operatorname {ker}_{\operatorname {CB}}(Z)$ is compact, there exists $m \in \omega $ such that $\operatorname {ker}_{\operatorname {CB}}(Z)\subseteq \bigcup _{n\leq m}C_n$ . For each $n>m$ there is at least one isolated point $ x_n \in C_n $ because $ C_n \cap \operatorname {ker}_{\operatorname {CB}}(X) = \emptyset $ : then $U = \{x_n \mid n> m\}$ is as required.

Proof Since $\alpha < ||X||_{\operatorname {Comp}} \leq ||X||_{\operatorname {CB}}$ we have $D^{\alpha +1}_{\operatorname {CB}}(X) \subsetneq D^{\alpha }_{\operatorname {CB}}(X)$ and the latter is not compact. By Lemma 3.9 there is an infinite discrete clopen (relatively to $ D^{\alpha }_{\operatorname {CB}}(X) $ ) set $U = \{ x_i \mid i \in \omega \} \subseteq D^{\alpha }_{\operatorname {CB}}(X)$ , so that $ U \cap D^{\alpha +1}_{\operatorname {CB}}(X) = \emptyset $ . Let $ (U_n)_{n \in \omega } $ be open sets of $ X $ such that $ U_0 \cap D^{\alpha }_{\operatorname {CB}}(X) = D^{\alpha }_{\operatorname {CB}}(X) \setminus U $ and $ U_{i+1} \cap D^{\alpha }_{\operatorname {CB}}(X) = \{ x_i \} $ . Without loss of generality, we may assume that $ \bigcup _{n \in \omega } U_n = X $ (otherwise we replace $ U_0 $ with $ U_0 \cup (X \setminus D^{\alpha }_{\operatorname {CB}}(X) $ ). Using the generalized reduction property [Reference Kechris17, Theorem 22.16] we obtain a clopen partition $ (C_n)_{n \in \omega } $ of $ X $ such that $ C_n \subseteq U_n $ , so that in particular $ C_n \cap D^{\alpha }_{\operatorname {CB}}(X) = U_n \cap D^{\alpha }_{\operatorname {CB}}(X) $ because $ (U_n \cap D^{\alpha }_{\operatorname {CB}}(X))_{n \in \omega } $ is a clopen partition of $ D^{\alpha }_{\operatorname {CB}}(X) $ . Since $ U \cap D^{\alpha +1}_{\operatorname {CB}}(X) = \emptyset $ , by Lemma 2.11 the clopen partition $(C_n)_{n \in \omega }$ of $ X $ is as desired.

Proof Since $ K_{\alpha +1} $ is compact, it is enough to prove by induction on $\alpha $ that there is a continuous injection $i \colon K_{\alpha + 1} \to X \setminus \operatorname {ker}_{\operatorname {CB}}(X)$ .

If $ \alpha = 0 $ then $ K_1 = \{ (0)^{\infty } \} $ and $ X $ has at least one isolated point $ x $ : setting $ i((0)^{\infty }) = x $ we are done.

Assume now that $ \alpha> 0 $ . Since $ ||X||_{\operatorname {CB}} \geq \alpha +1 $ , then $ D^{\alpha }_{\operatorname {CB}}(X) \setminus D^{\alpha +1}_{\operatorname {CB}}(X) \neq \emptyset $ . Pick any $ x \in D^{\alpha }_{\operatorname {CB}}(X) \setminus D^{\alpha +1}_{\operatorname {CB}}(X) $ and fix any $ \subseteq $ -decreasing clopen basis $(V_n)_{n\in \omega }$ of neighborhoods of x with $ V_0 \cap D^{\alpha }_{\operatorname {CB}}(X) = \{ x \} $ (hence also $ V_0 \cap D^{\alpha +1}_{\operatorname {CB}}(X) = \emptyset $ ). Set $ U_n = V_n \setminus V_{n+1} $ , so that each $ U_n $ is a countable clopen set, $ U_n \to x $ , $ ||U_n||_{\operatorname {CB}} \leq \alpha $ because $ U_n \cap D^{\alpha }_{\operatorname {CB}}(X) = \emptyset $ , and $ \limsup _{n \in \omega } ||U_n||_{\operatorname {CB}} = \alpha $ Footnote ¹⁰ because every clopen $ x \in V \subseteq V_0 $ is such that $ ||V||_{\operatorname {CB}} = \alpha +1 $ . (Here we repeatedly use Lemma 2.11.) Let $ (\alpha _n)_{n \in \omega } $ be the constant sequence with value $ \alpha $ if the latter is a successor ordinal, and a sequence of successor ordinals cofinal in $\alpha $ otherwise. By $ \limsup _{n \in \omega } ||U_n||_{\operatorname {CB}} = \alpha $ , there is an injective $ f \colon \omega \to \omega $ such that $ ||U_{f(n)}||_{\operatorname {CB}} \geq \alpha _n $ for all $ n \in \omega $ , so that by inductive hypothesis there is a continuous injection $ i_n \colon K_{\alpha _n} \to U_{f(n)} $ . Then the map $ i \colon K_{\alpha +1} \to V_0 \subseteq X \setminus \operatorname {ker}_{\operatorname {CB}}(X) $ defined by

$$\begin{align*}i(y) = \begin{cases} x, & \text{if } y = (0)^{\infty}, \\ i_n(z), & \text{if } y = (0)^n {}^{\smallfrown{}} (1) {}^{\smallfrown{}} z \text{ with } z \in K_{\alpha_n}, \end{cases} \end{align*}$$

is the desired continuous injection.

Proof Since $ D_{\gamma }(\boldsymbol {\Sigma }^0_1(X)) \subseteq D_{\gamma '}(\boldsymbol {\Sigma }^0_1(X)) $ if $ \gamma \leq \gamma ' $ , it is enough to consider the case $ ||X||_{\operatorname {CB}} = \alpha $ and prove by induction that . Note that $ 1 \leq \alpha < \omega _1 $ and $D^{\alpha }_{\operatorname {CB}}(X)=\emptyset $ because $ X \neq \emptyset $ is countable.

If $\alpha = 1$ then X is discrete, so . If moreover X is simple, then $ X $ is a singleton and .

Suppose now that $\alpha = \beta + 1$ for some $1 \leq \beta < \omega _1$ , and consider an arbitrary $A\subseteq X$ . Let $Y =X \setminus D^{\beta }_{\operatorname {CB}}(X)$ and $A'=Y \setminus A$ . Since $||Y||_{\operatorname {CB}} = \beta $ by Lemma 2.11, we get $A' \in D_{\beta }(\boldsymbol {\Sigma }^0_1(Y))\subseteq D_{\beta }(\boldsymbol {\Sigma }^0_1(X))$ by inductive hypothesis and the fact that Y is open in X, so we can write $ A' = D_{\beta }((U_\gamma )_{\gamma < \beta })$ with $ U_\gamma \in \boldsymbol {\Sigma }^0_1(X) $ . Moreover, $D^{\beta }_{\operatorname {CB}}(X)$ is discrete because $||X||_{\operatorname {CB}} = \beta + 1$ , therefore $A \cap D^{\beta }_{\operatorname {CB}}(X)$ is open in $D^{\beta }_{\operatorname {CB}}(X)$ : let $U \in \boldsymbol {\Sigma }^0_1(X)$ be such that $U \cap D^{\beta }_{\operatorname {CB}}(X) = A \cap D^{\beta }_{\operatorname {CB}}(X)$ , so that the open subset $ U_\beta = U \cup Y $ of $ X $ satisfies $A = U_\beta \setminus A'$ : then $ A = D_{\beta +1}((U_\gamma )_{\gamma < \beta +1}) $ , witnessing $A \in D_{\alpha }(\boldsymbol {\Sigma }^0_1(X))$ . If moreover X is simple, then $D^{\beta }_{\operatorname {CB}}(X)=\{x\}$ is a singleton. If $ x \notin A $ , then $ A \subseteq Y $ and using the induction hypothesis we again obtain $ A \in D_{\beta }(\boldsymbol {\Sigma }^0_1(Y))\subseteq D_{\beta }(\boldsymbol {\Sigma }^0_1(X)) $ . If instead $ x \in A $ , then $ A = X \setminus A' $ , hence . In both cases, .

Finally, suppose that $\alpha $ is limit, and take a strictly increasing sequence $(\alpha _n)_{n \in \omega }$ of nonzero ordinals cofinal in $\alpha $ . For $n\in \omega $ , consider the open sets $V_n^* = X \setminus D^{\alpha _n}_{\operatorname {CB}}(X)$ . Apply the generalized reduction property to the sequence $(V_n^*)_{n \in \omega }$ to obtain a clopen partition $(V_n)_{n \in \omega }$ of $ X $ such that $V_n \subseteq V_n^*$ . Given any $A\subseteq X$ , set $A_n = A \cap V_n$ . Then $A_n \subseteq V_n$ and $||V_n||_{\operatorname {CB}} \le \alpha _n$ by Lemma 2.11, so by induction hypothesis $A_n \in D_{\alpha _n}(\boldsymbol {\Sigma }^0_1(V_n)) \subseteq D_{\alpha _n}(\boldsymbol {\Sigma }^0_1(X))\subseteq D_{\alpha }(\boldsymbol {\Sigma }^0_1(X))$ . Write each $ A_n $ as $ A_n = D_{\alpha }((U^n_\beta )_{\beta <\alpha }) $ with $ U_\beta ^n \in \boldsymbol {\Sigma }^0_1(X) $ . Then $ A = \bigcup _{n \in \omega } (A_n \cap V_n) = D_{\alpha }((\bigcup _{n \in \omega } (U^n_\beta \cap V_n))_{\beta < \alpha })$ , hence $A \in D_{\alpha }(\boldsymbol {\Sigma }^0_1(X))$ .

Proof By definition, $ A_0 = \emptyset $ and $ K_{\alpha +1} \setminus A_\alpha = \operatorname {ptgl}( A_\beta ) $ if $ \alpha = \beta +1 $ . More over, since when $\alpha $ is limit the choice of the sequence $ (\alpha _n)_{n \in \omega } $ in the definition of $ K_{\alpha +1} = \operatorname {ptgl}((K_{\alpha _n+1})_{n\in \omega }) $ is irrelevant (up to homeomorphism), we can assume that the ordinals $ \alpha _n $ are all odd, so that also in the limit case $ K_{\alpha +1} \setminus A_\alpha = \operatorname {ptgl}((A_{\alpha _n})_{n \in \omega }) $ .

Recall the Wadge degrees from (2.1). Since $ D_{\alpha }(\boldsymbol {\Sigma }^0_1) \in \mathrm {Class}_{\operatorname {NSD}}(\mathcal {N}) $ , it is enough to prove by induction that $ A_\alpha \equiv _{\mathsf {W}}^{K_{\alpha +1}, \mathcal {N}} B_\alpha $ for some/any $ B_\alpha \in \mathbf {D}_\alpha $ . If $ \alpha = 0 $ the result is obvious because $ A_0 = \emptyset = B_0 $ . Suppose now that $ \alpha = \beta +1 $ and fix $ B_\beta \in \mathbf {D}_\beta $ . Let $ f \colon K_{\beta +1} \to \mathcal {N} $ and $ g \colon \mathcal {N} \to K_{\beta +1} $ witness $ A_\beta \le _{\mathsf {W}}^{K_{\beta +1},\mathcal {N}} B_\beta $ and $ B_\beta \le _{\mathsf {W}}^{\mathcal {N},K_{\beta +1}} A_\beta $ , respectively. Let $ B_\alpha \in \mathbf {D}_\alpha $ be defined as in (2.1), that is, $ B_\alpha = \{ (0)^n {^{\smallfrown }} (m+1) {^{\smallfrown }} x \mid n,m \in \omega \wedge x \in \neg B_\beta \} $ . Using $ K_{\alpha +1} \setminus A_\alpha = \operatorname {ptgl}( A_\beta ) $ it is straightforward to see that $ \operatorname {ptgl}(f) \colon K_{\alpha +1} \to \operatorname {ptgl}(\mathcal {N}) \subseteq \mathcal {N} $ witnesses $ A_\alpha \le _{\mathsf {W}}^{K_{\alpha +1},\mathcal {N}} B_\alpha $ , and that the map $ \operatorname {ptgl}(g) \circ \rho \colon \mathcal {N} \to K_{\alpha +1} $ , where $ \rho \colon \mathcal {N} \to \operatorname {ptgl}(\mathcal {N}) $ is the retraction defined by $ \rho ((0)^{\infty }) = (0)^{\infty } $ and $ \rho ( (0)^n {^{\smallfrown }} (m+1) {^{\smallfrown }} x) = (0)^n {^{\smallfrown }} (1) {^{\smallfrown }} x $ , witnesses $ B_\alpha \le _{\mathsf {W}}^{\mathcal {N},K_{\alpha +1}} A_\alpha $ . The limit case is similar.

Proof By Proposition 3.14 we know that for $ \alpha = ||X||_{\operatorname {CB}} $ , so it is enough to show that for every $\alpha < ||X||_{\operatorname {CB}}$ .

Since $\alpha +1\leq ||X||_{\operatorname {CB}}$ , by Proposition 3.12 we can suppose that $K_{\alpha +1}$ is, up to homeomorphism, a closed subset of X. By Proposition 3.17 we get . Fix a retraction $ \rho \colon X \to K_{\alpha +1} $ and set $ A^{\prime }_\alpha = \rho ^{-1}(A_\alpha ) $ : since $ A_\alpha \equiv _{\mathsf {W}}^{K_{\alpha +1},X} A^{\prime }_\alpha $ , we get , as desired.

Proof Apply Proposition 2.6 to $ A' $ to get a clopen partition $ (V_n)_{n \in \omega } $ of $ X $ and nonselfdual sets $ (A_n)_{n \in \omega } $ such that $ A_n <_{\mathsf {W}}^X A' $ and $ A' = \bigcup _{n \in \omega } (A_n \cap V_n ) $ . If $ A_n \le _{\mathsf {W}}^X A $ for all $ n \in \omega $ via some continuous map $ f_n \colon X \to X $ , then $ f = \bigcup _{n \in \omega } (f_n \restriction V_n) $ would witness $ A' \le _{\mathsf {W}}^X A $ , a contradiction. Since $ A $ is selfdual, this means that by Wadge’s Lemma there is $ \bar {n} \in \omega $ such that $ A <_{\mathsf {W}}^X A_{\bar {n}} $ . Setting $ A" = A_{\bar {n}} $ we are done.

Proof By Corollary 3.19, all sets $ A $ as in the statement are selfdual. Assume towards a contradiction that there is $ A' \notin \bigcup _{\beta < ||X||_{\operatorname {CB}}} D_{\beta }^*(\boldsymbol {\Sigma }^0_1(X)) $ such that $ A' \not \le _{\mathsf {W}}^X A $ , so that $ A' \in \mathrm {Set}_{\operatorname {SD}}(X) $ and $ A <_{\mathsf {W}}^X A' $ . If $ A' $ is $ \le _{\mathsf {W}}^X $ -minimal with such properties, then we get a contradiction with (the restriction to Borel sets of) Proposition 4.1. Thus $ A \equiv _{\mathsf {W}}^X A' $ for all $ A,A' \notin \bigcup _{\beta < ||X||_{\operatorname {CB}}} D_{\beta }^*(\boldsymbol {\Sigma }^0_1(X)) $ .

Finally, fix $ A $ as in the statement and consider an arbitrary $ B \subseteq X $ . If $ B \in D_{\beta }^*(\boldsymbol {\Sigma }^0_1(X)) $ for some $ \beta < ||X||_{\operatorname {CB}}$ , then $ B <_{\mathsf {W}}^X A $ by Wadge’s Lemma; otherwise, $ B \equiv _{\mathsf {W}}^X A $ by the previous paragraph. In all cases $ B \le _{\mathsf {W}}^X A $ , thus .

Proof Let X be an uncountable Polish zero-dimensional space, and consider any $1 \leq \alpha < \Theta _X = \Theta $ . By Corollary 4.2 we only need to show that if $\alpha \in \mathrm {Ord}_{\mathrm {NSD}}(X)$ , then $\alpha + 1 \in \mathrm {Ord}_{\mathrm {SD}}(X)$ . By Theorem 2.9(1) there is $ \boldsymbol {\Gamma } \in \mathrm {Class}_{\operatorname {NSD}}(\mathcal {N}) $ such that $ \boldsymbol {\Gamma }(X) \in \mathrm {Class}_{\operatorname {NSD}}(X) $ and $ ||\boldsymbol {\Gamma }(X)|| = \alpha $ . Since $ X $ is uncountable then $ \operatorname {ker}_{\operatorname {CB}}(X) $ is uncountable as well, thus we can partition $ X $ into two uncountable clopen sets $ X_0, X_1 $ . By Theorem 2.9(2) we have $ \boldsymbol {\Gamma }(X_0) \in \mathrm {Class}_{\operatorname {NSD}}(X_0) $ and . By Lemma 3.6 the smallest Wadge class containing is selfdual, i.e., $ \alpha +1 \in \mathrm {Ord}_{\mathrm {SD}}(X) $ .

Proof Suppose first that X is simple, so that in particular $||X||_{\operatorname {CB}}=\alpha +1$ for some $ \alpha < \omega _1 $ . By Corollary 3.19 we know that , while the “moreover” part of Proposition 3.14 gives . It follows that we cannot have for $ A \subseteq X $ , because if $ B \subseteq X $ is complete for $ D_{\alpha }(\boldsymbol {\Sigma }^0_1(X)) $ then $ B, \neg B \le _{\mathsf {W}}^X A $ would imply $ B <_{\mathsf {W}}^X A $ by $ B \not \le _{\mathsf {W}}^X \neg B $ , a contradiction.

Suppose now that X is not simple. By Corollary 4.3 it is enough to show that there is some $ A \subseteq X $ such that $ A \notin \bigcup _{\beta <||X||_{\operatorname {CB}}}D_{\beta }^*(\boldsymbol {\Sigma }^0_1(X)) $ . We distinguish two cases.

If $ ||X||_{\operatorname {CB}} = \lambda $ is limit, fix a strictly increasing sequence $ (\alpha _i)_{i \in \omega } $ cofinal in $ \lambda $ . By Lemma 2.10 we can find a clopen partition $ (C_n)_{n \in \omega } $ of $ X $ such that $ ||C_n||_{\operatorname {CB}} < \lambda $ and $ \sup _{n \in \omega } ||C_n||_{\operatorname {CB}}= \lambda $ . Fix an injection $ i \mapsto n_i $ such that $ ||C_{n_i}||_{\operatorname {CB}}> \alpha _i $ , so that $ D_{\alpha _i}(\boldsymbol {\Sigma }^0_1(C_{n_i})) \in \mathrm {Class}_{\operatorname {NSD}}(C_{n_i}) $ for all $ i \in \omega $ by Corollary 3.19. By Lemma 3.6 there is a selfdual Wadge class $ A {\downarrow }_X $ which contains $ \bigcup _{i \in \omega } D_{\alpha _i}(\boldsymbol {\Sigma }^0_1(X))$ , hence $ A \notin \bigcup _{\beta < \lambda } D_{\beta }^*(\boldsymbol {\Sigma }^0_1(X)) $ because all the pointclasses $ D_{\beta }(\boldsymbol {\Sigma }^0_1(X)) $ are distinct and nonselfdual by Corollary 3.19.

Suppose now that $ \operatorname {\mathrm {tp}}(X) = (\beta +1, N) $ with $ 1 < N \leq \omega $ . By Lemma 2.10 again we can find a clopen partition $ (C_n)_{n < N} $ of $ X $ such that $ ||C_n||_{\operatorname {CB}} = \beta +1 $ , so that $ \boldsymbol {\Gamma }_0(C_0) = D_{\beta }(\boldsymbol {\Sigma }^0_1(C_0)) \in \mathrm {Class}_{\operatorname {NSD}}(C_0) $ and by Corollary 3.19. (For the remaining $ 1 < n < N $ the choice of the nonselfdual pointclasses $ \boldsymbol {\Gamma }_n(C_n) $ is irrelevant.) Then Lemma 3.6 allows us to conclude that there is a selfdual $ A \subseteq X $ with , which is again as required by Corollary 3.19.

Proof Let X be a countable Polish space. By Corollary 4.2, we only need to prove that if $\alpha \in \mathrm {Ord}_{\mathrm {NSD}}(X)$ and $\alpha +1<\Theta _X$ then $\alpha +1\in \mathrm {Ord}_{\mathrm {SD}}(X)$ . By Corollary 3.19 there exists a $\beta <||X||_{\operatorname {CB}}$ such that $||D_{\beta }(\boldsymbol {\Sigma }^0_1(X))||=\alpha $ . We distinguish two cases.

Assume first that $ \beta + 1 < ||X||_{\operatorname {CB}} $ . Then $D^{\beta +1}_{\operatorname {CB}}(X) \neq \emptyset $ and $D^{\beta }_{\operatorname {CB}}(X)$ is infinite, which implies that we can split X in two clopen sets $X_0$ and $X_1$ with $||X_i||_{\operatorname {CB}} \geq \beta +1$ for $i=0,1$ . (For example, pick $ x \in D^{\beta }_{\operatorname {CB}}(X) \setminus D^{\beta +1}_{\operatorname {CB}}(X) $ and $ X_0 $ clopen such that $ X_0 \cap D^{\beta }_{\operatorname {CB}}(X) = \{ x \} $ : then $ X_0 $ and $ X_1 =X \setminus X_0 $ are as desired.) By Corollary 3.19 we have that $\boldsymbol {\Gamma }_0(X_0) = D_{\beta }(\boldsymbol {\Sigma }^0_1(X_0)) \in \mathrm {Class}_{\operatorname {NSD}}(X_0) $ and , so by Lemma 3.6 their supremum is selfdual and $\alpha +1\in \mathrm {Ord}_{\mathrm {SD}}(X)$ .

Assume now $ ||X||_{\operatorname {CB}} = \beta +1 $ . If $ X $ were simple, then $ \Theta _X = ||D_{\beta }(\boldsymbol {\Sigma }^0_1(X))|| +1 = \alpha +1 $ by Proposition 3.14, contradicting the choice of $\alpha $ . Thus $ X $ is not simple and Proposition 4.5 together with Corollary 4.3 show that at level $ \alpha + 1 $ we have the selfdual Wadge class , so $ \alpha +1 \in \mathrm {Ord}_{\mathrm {SD}}(X) $ again and we are done.

Proof Let $ A \in \mathrm {Set}_{\operatorname {SD}}(X) $ be such that $ ||A||_{\mathsf {W}}^X = \alpha $ . By Proposition 2.6, there are sets $ (A_n)_{n \in \omega } $ such that $ \alpha = \sup _{n \in \omega } (||A_n||_{\mathsf {W}}^X+1) $ : this already shows that $ \alpha $ has countable cofinality.

Proof If $ X $ is countable, then all of (i)–(iii) are false, so without loss of generality we may assume that $ X $ is uncountable.

We start with (i) $ \Rightarrow $ (ii). Since $ \operatorname {ker}_{\operatorname {CB}}(X) $ is not compact, by Fact 3.8 there is an infinite clopen partition $(C^{\prime }_n)_{n \in \omega }$ of $\operatorname {ker}_{\operatorname {CB}}(Z)$ , and since $ \operatorname {ker}_{\operatorname {CB}}(X) $ is perfect each $ C^{\prime }_n $ is uncountable. Thus it is enough to set $ C_n = \rho ^{-1}(C^{\prime }_n) $ where $ \rho \colon X \to \operatorname {ker}_{\operatorname {CB}}(X) $ is any retraction.

Assume now that (ii) holds. Since the $ C_n $ ’s are uncountable Polish spaces, they contain a closed set $ F_n \subseteq C_n $ homeomorphic to $ \mathcal {C} $ : then $ F = \bigcup _{n \in \omega } F_n $ is still closed and homeomorphic to $ \omega \times \mathcal {C} $ , hence (iii) holds.

Finally, we prove (iii) $ \Rightarrow $ (i). Since $ F $ is perfect, then $ F \subseteq \operatorname {ker}_{\operatorname {CB}}(X) $ . But then $ \operatorname {ker}_{\operatorname {CB}}(X) $ cannot be compact, otherwise so would be $ F $ and $ \omega \times \mathcal {C} $ , which is false.

Proof It is enough to prove (i) $ \Rightarrow $ (iii) and (the contrapositive of) (ii) $ \Rightarrow $ (i).

Assume first that $ \operatorname {ker}_{\operatorname {CB}}(X) $ is not compact, so that by Lemma 4.8 there is an infinite clopen partition $ (C_n)_{n \in \omega } $ of $ X $ into uncountable pieces. Given any $ \alpha \in \Theta _X \in \mathsf {Cof}_\omega $ , pick a strictly increasing sequence of ordinals $ (\alpha _n)_{n \in \omega } $ cofinal in $\alpha $ such that $ \alpha _n \in \mathrm {Ord}_{\mathrm {NSD}}(X) $ for all $ n \in \omega $ (which exists by the alternating property), and using Theorem 2.9(1) let $ \boldsymbol {\Gamma }_n \in \mathrm {Class}_{\operatorname {NSD}}(\mathcal {N}) $ be such that $ ||\boldsymbol {\Gamma }_n(X)||_X = \alpha _n $ . By Theorem 2.9(2) and uncountability of $ C_n $ , we have $ \boldsymbol {\Gamma }_n(C_n) \in \mathrm {Class}_{\operatorname {NSD}}(C_n) $ as well, hence by Lemma 3.6 the supremum $ \boldsymbol {\Gamma } \in \mathrm {Class}_{\mathsf {W}}(X) $ of the pointclasses $ \boldsymbol {\Gamma }_n(X) $ is selfdual, i.e., $ \alpha \in \mathrm {Ord}_{\mathrm {SD}}(X) $ .

Assume now that $ \operatorname {ker}_{\operatorname {CB}}(X) $ is compact and let $ \beta = ||X||_{\operatorname {CB}} $ . Consider any $ \beta + \omega < \alpha < \omega _1 $ : we want to show that if $ \alpha \in \mathrm {Ord}_{\mathrm {SD}}(X) $ then $ \alpha $ is a successor ordinal, so that any limit ordinal between $ \beta +\omega $ and $ \omega _1 $ witnesses the failure of (ii). Let $ A \subseteq X $ be a selfdual set such that $ ||A||_{\mathsf {W}}^X = \alpha $ . By Proposition 2.6 there exists a clopen partition $(V_n)_{n \in \omega }$ of $ X $ and sets $(A_n)_{n \in \omega }$ such that $A_n <_{\mathsf {W}}^X A = \bigcup _{n \in \omega } (A_n \cap V_n)$ and $ \alpha = \sup _{n \in \omega }(\alpha _n+1) $ for $ \alpha _n= ||A_n||_{\mathsf {W}}^X $ . Recall also from Remark 2.7(1) that we can assume that for each $ n \in \omega $ either $ A_n = X $ , or else $ A_n \subseteq V_n $ . Since $ \operatorname {ker}_{\operatorname {CB}}(X) $ is compact, there is $ N \in \omega $ such that $ V_n \cap \operatorname {ker}_{\operatorname {CB}}(X) = \emptyset $ for all $ n \geq N $ . Fix any $ n \geq N $ . Then $ ||V_n||_{\operatorname {CB}} \leq \beta $ by Lemma 2.11, hence if $ A_n \subseteq V_n $ then $ A_n \in D_{\beta }(\boldsymbol {\Sigma }^0_1(V_n)) \subseteq D_{\beta }(\boldsymbol {\Sigma }^0_1(X)) $ by Proposition 3.14, which means that $ \alpha _n < \beta +\omega $ because by Theorem 2.9 and the alternating property $ ||D_{\beta }(\boldsymbol {\Sigma }^0_1(X))||_X \leq ||D_{\beta }(\boldsymbol {\Sigma }^0_1(\mathcal {N}))||_{\mathcal {N}} = 2 \beta +1 $ . If instead $ A_n = X $ , then $ \alpha _n = 1 $ . Hence $ \alpha _n + 1 < \beta +\omega $ for all $ n \geq N $ . Since $ \alpha> \beta +\omega $ , this means that necessarily $ \alpha = \sup _{n \in \omega } (\alpha _n+1) = \sup _{n < N } (\alpha _n+1) = \max _{n < N} (\alpha _n+1) $ , and thus $\alpha $ is a successor ordinal.

Proof (1) Suppose towards a contradiction that $ \alpha> ||X||_{\operatorname {Comp}} $ but $ \alpha \in \mathrm {Ord}_{\mathrm {SD}}(X) $ : we adapt the proof of Theorem 4.9 to show that $\alpha $ needs to be a successor ordinal, contradicting $ \alpha \in \mathsf {Cof}_\omega $ . Given any $ A \in \mathrm {Set}_{\operatorname {SD}}(X) $ such that $ ||A||_{\mathsf {W}}^X = \alpha $ , use Proposition 2.6 to get a clopen partition $(V_n)_{n \in \omega }$ of $ X $ and sets $(A_n)_{n \in \omega }$ such that $A_n <_{\mathsf {W}}^X A = \bigcup _{n \in \omega } (A_n \cap V_n)$ and $ \alpha = \sup _{n \in \omega } (\alpha _n+1) $ for $ \alpha _n= ||A_n||_{\mathsf {W}}^X $ . Since by definition $D^{||X||_{\operatorname {Comp}}}_{\operatorname {CB}}(X)$ is compact, there exists $N \in \omega $ such that $V_n \subseteq X \setminus D^{||X||_{\operatorname {Comp}}}_{\operatorname {CB}}(X)$ for all $n \geq N$ . Using that either $ A_n = X $ or $ A_n \subseteq V_n $ , this implies that $ A_n \in D_{||X||_{\operatorname {Comp}}}(\boldsymbol {\Sigma }^0_1(X)) $ for all $ n \geq N $ . Arguing as in Theorem 4.9 (and using that $ \alpha> ||X||_{\operatorname {Comp}} $ is limit), we get that $ ||D_{||X||_{\operatorname {Comp}}}(\boldsymbol {\Sigma }^0_1(X))||_X \leq ||D_{||X||_{\operatorname {Comp}}}(\boldsymbol {\Sigma }^0_1(\mathcal {N}))||_{\mathcal {N}} = 2 \cdot ||X||_{\operatorname {Comp}}+1 < \alpha $ , hence $ \sup _{n \geq N} (\alpha _n+1) < \alpha $ . Thus $ \alpha = \sup _{n \in \omega }(\alpha _n+1) = \max _{n < N} (\alpha _n+1) $ is a successor ordinal, a contradiction.

(2) Suppose now that $\alpha <||X||_{\operatorname {Comp}}$ , so that in particular X is not compact. Then by Lemma 3.10 there is a clopen partition $(C_n)_n$ of X such that $D^{\alpha +1}_{\operatorname {CB}}(X)\subseteq C_0$ and $||C_{n+1}||_{\operatorname {CB}}=\alpha +1$ for all $n\in \omega $ . For every $ n \geq 1 $ and $ \beta < \alpha $ , Corollary 3.19 implies that $ D_{\beta }(\boldsymbol {\Sigma }^0_{1}(C_n)) \in \mathrm {Class}_{\operatorname {NSD}}(C_n) $ . Moreover, $ \beta \leq ||D_{\beta }(\boldsymbol {\Sigma }^0_{1}(X))|| \leq 2\beta +1 $ : if $ X $ is countable this follows from the alternating property and Corollary 3.19 together with $ ||X||_{\operatorname {CB}} \geq ||X||_{\operatorname {Comp}}> \alpha > \beta $ , while if $ X $ is uncountable we can use the alternating property and Lemma 3.13. Since $\alpha $ is limit, then $ 2\beta +1 < \alpha $ and hence $ \beta \leq ||D_{\beta }(\boldsymbol {\Sigma }^0_{1}(X))|| < \alpha $ . Therefore taking any strictly increasing sequence $ (\alpha _n)_{n \geq 1} $ cofinal in $ \alpha $ we have $ D_{\alpha _n}(\boldsymbol {\Sigma }^0_1(C_n)) \in \mathrm {Class}_{\operatorname {NSD}}(C_n) $ , $ ||D_{\alpha _n}(\boldsymbol {\Sigma }^0_{1}(X))||_{X} < \alpha $ , and $ \sup _{n \geq 1} ||D_{\alpha _n}(\boldsymbol {\Sigma }^0_{1}(X))|| = \alpha $ . Setting $ \boldsymbol {\Gamma }_0 = \{ \mathcal {N} \} $ and $ \boldsymbol {\Gamma }_n = D_{\alpha _n}(\boldsymbol {\Sigma }^0_1) $ for $ n \geq 1 $ , and applying Lemma 3.6, we then get $ \alpha \in \mathrm {Ord}_{\mathrm {SD}}(X) $ , as desired.

(3) The proof of the desired equivalence is a variation of the arguments for parts (1) and (2). Assume first that $ \alpha \in \mathrm {Ord}_{\mathrm {SD}}(X) $ . Given any $ A \in \mathrm {Set}_{\operatorname {SD}}(X) $ such that $ ||A||_{\mathsf {W}}^X=\alpha $ , let $ (V_n)_{n \in \omega } $ and $ (A_n)_{n \in \omega } $ be as in Proposition 2.6. Since $ D^{\alpha }_{\operatorname {CB}}(X) $ is compact, there is $ N \in \omega $ such that $ V_n \cap D^{\alpha }_{\operatorname {CB}}(X) = \emptyset $ for all $ n \geq N $ , which also means $ ||V_n||_{\operatorname {CB}} \leq \alpha $ by Lemma 2.11. If $ ||V_n||_{\operatorname {CB}} = \alpha $ for some $ n \geq N $ , then we apply Lemma 2.10 to all such countable spaces $ V_n $ and get the desired partition. (Recall that $\alpha $ is limit.) So we can assume that $ ||V_n||_{\operatorname {CB}} + 1 \leq \alpha $ for all $ n \geq N $ and we only need to show that for every $ \beta < \alpha $ there is $ n \geq N $ with $ ||V_n||_{\operatorname {CB}} \geq \beta $ . Assume not: since we can assume that either $ A_n = X $ or $ A_n \subseteq V_n $ , by Proposition 3.14 we would have $ A_n \in D_{||V_n||_{\operatorname {CB}}}(\boldsymbol {\Sigma }^0_1(V_n)) \subseteq D_{\beta }(\boldsymbol {\Sigma }^0_1(V_n)) \subseteq D_{\beta }(\boldsymbol {\Sigma }^0_1(X)) $ for all $ n \geq N $ , and since the usual computation gives $ ||D_{\beta }(\boldsymbol {\Sigma }^0_1(X))||_X \leq 2 \beta +1 $ and $ 2 \beta +1 < \alpha $ because $\alpha $ is limit, then $ \alpha = \sup _{n \in \omega } (||A_n||_{\mathsf {W}}^X+1) = \max _{n < N} (||A_n||_{\mathsf {W}}^X+1) $ , contradicting the fact that $ \alpha $ is limit.

Conversely, let $ (V_n)_{n \in \omega } $ and $ N \in \omega $ be as in part (3), and fix a sequence $ (\alpha _k)_{k \in \omega } $ cofinal in $ \alpha $ . Since $\alpha $ is limit, the assumption on $ (V_n)_{n \in \omega } $ implies that for each $ k \in \omega $ we can find $ n_k \geq N $ such that $ \alpha _k < ||V_{n_k}||_{\operatorname {CB}} $ and further ensure that the map $ k \mapsto n_k $ is injective. By Corollary 3.19 the class $ \boldsymbol {\Gamma }_{n_k}(V_{n_k}) = D_{\alpha _k}(\boldsymbol {\Sigma }^0_1(V_{n_k})) $ is nonselfdual in $ V_{n_k} $ , and as in the proof of part (2) one can show that $ \alpha _{k} \leq ||D_{\alpha _k}(\boldsymbol {\Sigma }^0_1(X))|| \leq 2 \alpha _k +1 < \alpha $ . Setting $ \boldsymbol {\Gamma }_n = \{ \mathcal {N} \} $ for all $ n \in \omega $ which are not of the form $ n_k $ and applying Lemma 3.6 we get $ \alpha \in \mathrm {Ord}_{\mathrm {SD}}(X) $ , as desired.