Proof It is clear that if $\mathcal {L}$ is non-trivial, it does not contain a principal partition regular subclass. Suppose now $\mathcal {L}$ is trivial, that is, $\mathcal {L}$ contains a finite set $X = \{n_1, \dots , n_k\}$ . Then in particular we have $\{n_1\} \cup \dots \cup \{n_k\} \supseteq X$ . It follows that we must have $\{n_i\} \in \mathcal {L}$ for some $i \leq k$ . Then any set X containing $n_i$ is in $\mathcal {L}$ , that is, we have $\{X \in 2^{\omega}\,{:}\,n_i \in X\} \subseteq \mathcal {L}$ .

Proof If $\mathcal {L}$ is non-trivial then every member of $\mathcal {L}$ is infinite and clearly $\mathcal {L} \subseteq \mathcal {U}_2$ . If $\mathcal {L}$ is trivial then by Proposition 2.5, it contains $\{n\}$ for some $n \in \omega $ and thus we do not have $\mathcal {L} \subseteq \mathcal {U}_2$ .

Proof Let $X \in \mathcal {L}$ . By definition, any $Y \supseteq X$ also belongs to $\mathcal {L}$ . Thus $\mathcal {L}$ is closed under finite addition of elements, that is, if $X \in \mathcal {L}$ and F is finite, then $X \cup F \in \mathcal {L}$ . Consider now any $Y \subseteq X$ such that $|X \setminus Y|$ is finite. In particular, $X = Y \cup \{n_0, \dots , n_k\}$ for some elements $n_0, \dots , n_k$ . As $\mathcal {L}$ contains only infinite elements, we must have $Y \in \mathcal {L}$ . Thus $\mathcal {L}$ is closed by finite suppression, that is, if $X \in \mathcal {L}$ and F is finite, then $X \setminus F \in \mathcal {L}$ . We easily conclude that $\mathcal {L}$ is closed under finite changes.

If $\mathcal {L}$ is measurable, by Kolmogorov 0–1 law, $\mathcal {L}$ is either of measure $0$ or of measure $1$ . Suppose for contradiction that $\mathcal {L}$ is of measure $0$ . As $\mathcal {L}$ is measurable, if must be included in some Borel set $\mathcal {A}$ of measure $0$ . Let O be an oracle such that $\mathcal {A}$ is included in a $\Pi ^0_2(O)$ classes effectively of measure $0$ , that is, $\mathcal {A} \subseteq \bigcap _e \mathcal {U}_e$ for some uniformly O-computable sequence of $\Pi ^0_2(O)$ classes such that $\mu (\mathcal {U}_e) < 2^{-e}$ . Then no element of $\mathcal {L}$ is O-Martin-Löf random. Let Z be any O-Martin-Löf random set. We also have that $\overline {Z}$ is O-Martin-Löf random. Also $\omega \subseteq Z \cup \overline {Z}$ . As $\omega \in \mathcal {L}$ we must have $Z \in \mathcal {L}$ or $\overline {Z} \in \mathcal {L}$ , which is a contradiction. Thus $\mathcal {L}$ is not of measure $0$ and therefore it is of measure $1$ .

Proof It is clear that $\bigcup _{i \in I} \mathcal {L}_i$ is not empty. Let $X \in \bigcup _{i \in I} \mathcal {L}_i$ . Let $Y \supseteq X$ . There is some $i \in I$ such that $X \in \mathcal {L}_i$ . As $\mathcal {L}_i$ is partition regular, $Y \in \mathcal {L}_i \subseteq \bigcup _{i \in I} \mathcal {L}_i$ .

Let $X \in \bigcup _{i \in I} \mathcal {L}_i$ . Let $Y_0 \cup \dots \cup Y_k \supseteq X$ . There is some $i \in I$ such that $X \in \mathcal {L}_i$ . As $\mathcal {L}_i$ is partition regular, $Y_j \in \mathcal {L}_i \subseteq \bigcup _{i \in I} \mathcal {L}_i$ for some $j \leq k$ .

Proof Note that by definition, $\mathcal {L}(\mathcal {A}) \subseteq \mathcal {A}$ , as if $X \notin \mathcal {A}$ then itself as a $1$ -cover is not in $\mathcal {A}$ , so $X \notin \mathcal {L}(\mathcal {A})$ . Let us show that $\mathcal {L}(\mathcal {A})$ contains every partition regular class included in $\mathcal {A}$ . Suppose $\mathcal {L} \subseteq \mathcal {A}$ is partition regular. Then given $X \in \mathcal {L}$ , for every k and every $X_0\cup \dots \cup X_k \supseteq X$ we have $X_i \in \mathcal {L} \subseteq \mathcal {A}$ for some $i \leq k$ . It follows that $X \in \mathcal {L}(\mathcal {A})$ and thus that $\mathcal {L} \subseteq \mathcal {L}(\mathcal {A})$ .

Assume $\mathcal {L}(\mathcal {A}) \neq \emptyset $ . Let us show that $\mathcal {L}(\mathcal {A})$ is partition regular. Suppose $X \in \mathcal {L}(\mathcal {A})$ . Let $Y \supseteq X$ . Then for every k, every k-cover of Y is also a k-cover of X. As $X \in \mathcal {L}(\mathcal {A})$ , one element of the k-cover belongs to $\mathcal {A}$ . Thus for every k and every k-cover of Y, one element of the k-cover belongs to $\mathcal {A}$ . It follows that $Y \in \mathcal {L}(\mathcal {A})$ . Let $X \in \mathcal {L}(\mathcal {A})$ and let $Y_0 \cup \dots \cup Y_k \supseteq X$ for some k. Let us show there is some $i \leq k$ such that $Y_i \in \mathcal {L}(\mathcal {A})$ . Suppose for contradiction that this is not the case. In particular for every $i \leq k$ there are sets $Y^i_{0}, \dots, Y^i_{k_i} \supseteq Y_i$ such that $\forall j \leq k_i$ , we have $Y^i_{j} \notin \mathcal {A}$ . In particular the sets $\{Y^i_{j}\}_{i \leq k, j \leq k_i}$ are a finite cover of X such that for every $i \leq k$ and every $j \leq k_i$ we have $Y^i_{j} \notin \mathcal {A}$ . This contradicts that $X \in \mathcal {L}(\mathcal {A})$ . Thus there must exists $i \leq k$ such that $Y_i \in \mathcal {L}(\mathcal {A})$ . So if $\mathcal {L}(\mathcal {A})$ is non-empty, it is partition regular.

Proof For every n, $\omega \in \mathcal {L}_n$ because $\mathcal {L}_n$ is partition regular. It follows that $\omega \in \bigcap _{n \in \omega } \mathcal {L}_n$ . In particular $\bigcap _{n \in \omega } \mathcal {L}_n$ is not empty.

Suppose $X \in \bigcap _{n \in \omega } \mathcal {L}_n$ . Let $Y \supseteq X$ . For every n, since $X \in \mathcal {L}_n$ then $Y \in \mathcal {L}_n$ as $\mathcal {L}_n$ is partition regular. Thus $Y \in \bigcap _{n \in \omega } \mathcal {L}_n$ .

Suppose $X \in \bigcap _{n \in \omega } \mathcal {L}_n$ . Let $Y_0 \cup \dots \cup Y_k \supseteq X$ . Suppose for contradiction that for every $i \leq k$ the set $Y_i$ is not in $\bigcap _{n \in \omega } \mathcal {L}_n$ . For every $i \leq k$ , let $n_i$ be such that $Y_i \notin \mathcal {L}_{n_i}$ . Let n be larger than these numbers. For every $i \leq k$ , since $\mathcal {L}_{n_i} \supseteq \mathcal {L}_n$ , the set $Y_i$ is not in $\mathcal {L}_n$ . As $X \in \mathcal {L}_n$ , it follows that $\mathcal {L}_n$ is not partition regular, which contradicts our hypothesis. Thus for every $X \in \bigcap _{n \in \omega } \mathcal {L}_n$ and for every $Y_0 \cup \dots \cup Y_k \supseteq X$ , there is some $i \leq k$ such that $Y_i \in \bigcap _{n \in \omega } \mathcal {L}_n$ .

Proof By Theorem 2.10, the largest partition regular subclass of $\mathcal {U}$ is defined by

$$ \begin{align*}\mathcal{L}(\mathcal{U}) = \{X \in 2^\omega\ :\ \forall k\ \forall X_0\cup\dots\cup X_k \supseteq X\ \exists i \leq k\ X_i \in \mathcal{U}\}.\end{align*} $$

Since $\mathcal {U}$ is upward-closed, then $\forall X_0\cup \cdots \cup X_k \supseteq X\ \exists i \leq k\ X_i \in \mathcal {U}$ is equivalent to $\forall X_0\cup \dots \cup X_k = X\ \exists i \leq k\ X_i \in \mathcal {U}$ . Since $\mathcal {U}$ is $\Sigma ^0_1$ , then by compactness, the previous formula is equivalent to $\exists n \forall X_0 \cup \dots \cup X_k = X \cap \{0, \dots , n\}\ \exists i \leq k\ X_i \in \mathcal {U}$ , which is a $\Sigma ^0_1$ formula. Hence, $\mathcal {L}(\mathcal {U})$ is $\Pi ^0_2$ .

Proof By Theorem 2.10, the class $\mathcal {U}$ contains a partition regular class iff $\omega \in \mathcal {L}(\mathcal {U})$ , which is a $\Pi ^0_2$ sentence.

Proof Let $\mathcal {U}$ be such a $\Sigma ^{0,Z}_1$ class. Let $X \not \in \mathcal {U}$ . For any n and every string $\sigma $ of length n we have $\sigma 0^{\infty } \in \mathcal {U}$ . For every n, let $f(n) \in \mathbb {N}$ be large enough such that for any string $\sigma $ of length n we have $[\sigma 0^{f(n)}] \subseteq \mathcal {U}$ . Note that the function f is total and Z-computable. It must be that $X\upharpoonright _{n+f(n)}$ contains at least a $1$ at a position greater than n. By repeating this we see that we can Z-computably bound $p_X$ .

Proof Suppose for contradiction that there exists a non-trivial $\mathbf {\Sigma ^0_2}$ partition regular class $\mathcal {C}$ . Let Z be such that $\mathcal {C}$ is $\Sigma ^{0,Z}_2$ . In particular, $2^\omega \setminus \mathcal {C}$ contains all the finite sets, so by Proposition 2.15, it contains all the Z-hyperimmune sets. Consider now any set X such that both X and $\overline {X}$ are Z-hyperimmune. By partition regularity, at least one of them belongs to $\mathcal {C}$ , which is a contradiction.

Proof Suppose $\mathcal {A}$ is upward-closed and contains a partition regular subclass $\mathcal {L} \subseteq \mathcal {A}$ . (a) is trivially satisfied by hypothesis. By partition regularity of $\mathcal {L}$ , $\omega \in \mathcal {L}$ and for every k and every $Y_0 \cup \dots \cup Y_k \supseteq \omega $ , there is some $j \leq k$ such that $Y_j \in \mathcal {L} \subseteq \mathcal {A}$ . So $\mathcal {A}$ is partition large.

Suppose now $\mathcal {A}$ is partition large. By (a), it is upward-closed. We claim that $\mathcal {L}(\mathcal {A})$ is partition regular. By (b) and Theorem 2.10, $\omega \in \mathcal {L}(\mathcal {A})$ , so $\mathcal {L}(\mathcal {A}) \neq \emptyset $ . By definition of $\mathcal {L}(\mathcal {A})$ , it is partition regular, hence $\mathcal {A}$ contains a partition regular subclass.

Proof Let $Y_0 \cup \dots \cup Y_k \supseteq \omega $ . In particular we have $Y_0 \cap X \cup \dots \cup Y_k \cap X \supseteq X$ . As $X \in \mathcal {L}$ we must have $Y_j \cap X \in \mathcal {L}$ for some $j \leq k$ . In particular, since $\mathcal {L}$ is non-trivial, $Y_j \cap X$ is infinite, so $Y_j \cap X \in \mathcal {L}_X$ . Therefore, there is some $j \leq k$ such that $Y_j \cap X \in \mathcal {L} \cap \mathcal {L}_X$ .

Proof Let $\mathcal {L} = \{ X : \mathcal {A} \cap \mathcal {L}_X \mbox { is partition large} \}$ . Since $\mathcal {A}$ is partition large, then by Proposition 2.18, $\mathcal {L}(\mathcal {A})$ is the largest partition regular subclass of $\mathcal {A}$ . Moreover, by Proposition 2.6, $\mathcal {L}(\mathcal {A})$ is non-trivial since $\mathcal {L}(\mathcal {A}) \subseteq \mathcal {U}_2$ . Thus, by Theorem 2.19, for every $X \in \mathcal {L}(\mathcal {A})$ , $\mathcal {L}(\mathcal {A}) \cap \mathcal {L}_X$ is partition large. In particular, $\mathcal {A} \cap \mathcal {L}_X$ is partition large, hence $X \in \mathcal {L}$ . It follows that $\mathcal {L}(\mathcal {A}) \subseteq \mathcal {L}$ .

Let us show that $\mathcal {L}$ is partition regular. First, $\mathcal {A} \cap \mathcal {L}_\omega = \mathcal {A}$ is partition large, so $\omega \in \mathcal {L}$ . Let $X \in \mathcal {L}$ and $Y_0 \cup \dots \cup Y_k \supseteq X$ . In particular, $\mathcal {A} \cap \mathcal {L}_X$ is partition large, so by Proposition 2.18, $\mathcal {L}(\mathcal {A} \cap \mathcal {L}_X)$ is partition regular. Moreover, $X \in \mathcal {L}(\mathcal {A} \cap \mathcal {L}_X)$ , so there is some $i \leq k$ such that $Y_i \in \mathcal {L}(\mathcal {A} \cap \mathcal {L}_X)$ . By Theorem 2.19, $\mathcal {L}(\mathcal {A} \cap \mathcal {L}_X) \cap \mathcal {L}_{Y_i}$ is partition large. Since $\mathcal {L}(\mathcal {A} \cap \mathcal {L}_X) \cap \mathcal {L}_{Y_i} \subseteq \mathcal {A} \cap \mathcal {L}_{Y_i}$ , then $\mathcal {A} \cap \mathcal {L}_{Y_i}$ is partition large, so $Y_i \in \mathcal {L}$ .

Last, let us show that $\mathcal {L} \subseteq \mathcal {A}$ . Indeed, if $\mathcal {A} \cap \mathcal {L}_X$ is partition large, then since $X \cup \overline {X} = \omega $ , either X or $\overline {X}$ belongs to $\mathcal {A} \cap \mathcal {L}_X$ . However, $\overline {X} \not \in \mathcal {L}_X$ , so $X \in \mathcal {A} \cap \mathcal {L}_X$ . In particular, $X \in \mathcal {A}$ . It follows that $\mathcal {L}$ is a partition regular subclass of $\mathcal {A}$ , so by maximality of $\mathcal {L}(\mathcal {A})$ , $\mathcal {L} \subseteq \mathcal {L}(\mathcal {A})$ .

Proof Let $\mathcal {L} \subseteq \mathcal {A}$ be any non-trivial $\Pi ^0_2$ partition regular subclass of $\mathcal {A}$ . Since X is partition generic in $\mathcal {A}$ , then $X \in \mathcal {L}$ . Since $Y =^{*} X$ , by Theorem 2.7, $Y \in \mathcal {L}$ . Therefore Y is partition generic in $\mathcal {A}$ .

Proof Let A be bi-partition generic. Let X be an infinite subset of A. Then $\mathcal {L}_X$ is a partition regular class such that $\overline {A} \not \in \mathcal {L}_X$ . Since $\overline {A}$ is partition generic, $\mathcal {L}_X$ is not $\Pi ^0_2$ , so X is not computable. Similarly, $\overline {A}$ has no infinite computable subset.

Proof By Theorem 2.7, every non-trivial measurable partition regular class is of measure $1$ . It follows that any Kurtz-random belongs to every $\Pi ^0_2$ partition regular class and thus that any Kurtz-random is bi-partition generic.

Proof Let A be a co-hyperimmune set. Suppose for the contradiction that $A \not \in \mathcal {L}$ for some non-trivial $\Pi ^0_2$ partition regular class $\mathcal {L} \subseteq 2^\omega $ . In particular, there is a partition large $\Sigma ^0_1$ class $\mathcal {U} \supseteq \mathcal {L}$ such that $A \not \in \mathcal {U}$ . Since $\mathcal {U}$ is partition large, for every $t \in \omega $ , there is some $\rho \in 2^{<\omega }$ with $\min \rho> t$ such that $\rho \in \mathcal {U}$ . In particular, for every such $\rho $ , we have $\rho \cap \overline {A} \neq \emptyset $ . Moreover, such a string $\rho $ can be found computably uniformly in t. We can therefore compute an array tracing $\overline {A}$ , contradicting hyperimmunity of $\overline {A}$ .

Proof Immediate by Proposition 2.27.

Proof Consider the following $\Pi ^0_2$ class $\mathcal {L} = \{X\ :\ \forall k\ \exists n\ |X \upharpoonright _{n^2}| \geq nk \}$ . Let us prove it is partition regular: Let $X \in \mathcal {L}$ and $Y_1 \cup \cdots \cup Y_m \supseteq X$ . For $k \in \omega $ there is some n such that $|X \upharpoonright _{n^2}| \geq n m k$ . Then, there is some $e_k \leq m$ such that $|Y_{e_k} \upharpoonright _{n^2}| \geq n k$ . Let $e \leq m$ be such that $e = e_k$ for infinitely many k. We then have for infinitely many k that there exists n such that $|Y_{e} \upharpoonright _{n^2}| \geq n k$ . This is therefore true in particular for every k. Thus $Y_e \in \mathcal {L}$ , so $\mathcal {L}$ is partition regular.

It suffices to construct an effectively co-immune set A such that for every n, $|A \upharpoonright _{n^2}| < n$ for every n. Then $A \not \in \mathcal {L}$ , hence A is not partition generic. To construct such a set, for every e, let $x_e$ be the $e^2$ th element of $W_e$ (in the $<_\omega $ order), if it exists. Assume by convention that $W_0 = \emptyset $ , so $e_0$ does not exist. Let $A = \{ x_e : e \in \omega \}$ . Then A is an infinite set, which is effectively co-immune, as witnessed by the function $n \mapsto n^2$ . Last, for every n, $A \upharpoonright _{n^2} \subseteq \{ x_e : e < n \}$ , so $|A \upharpoonright _{n^2}| < n$ .

Proof Consider the notion of forcing whose conditions are strings $\sigma $ of even length, such that for every $n < |\sigma |/2$ , $\sigma (2n) \neq \sigma (2n+1)$ . The conditions are partially ordered by the suffix relation. Let us show that every sufficiently generic set G is bi-partition generic.

Let $\sigma $ be a condition, and let $\mathcal {U}$ be a non-trivial $\Sigma ^0_1$ partition large class. Let $X_0 = \{2n : 2n> |\sigma | \}$ and $X_1 = \{2n+1 : 2n+1> |\sigma | \}$ . Since $\{0\} \cup \dots \cup \{|\sigma |\} \cup X_0 \cup X_1 \supseteq \omega $ , then either $X_0 \in \mathcal {U}$ , or $X_1 \in \mathcal {U}$ . Say the former case holds as the other case is symmetric. Since $\mathcal {U}$ is $\Sigma ^0_1$ , then there is some k such that $[X_0 \upharpoonright _{2k}] \subseteq \mathcal {U}$ . Let $\tau = \sigma \cup X_0 \upharpoonright _{2k}$ . Note that $\tau $ is a valid condition. Moreover, since $\mathcal {U}$ is closed under superset, then $[\tau ] \subseteq \mathcal {U}$ . Therefore, every sufficiently generic set G for this notion of forcing belongs to every non-trivial $\Sigma ^0_1$ partition large class. By symmetric, so does the complement of G, so G is bi-partition generic.

Proof Consider the set A of Proposition 2.30. It is clearly neither hyperimmune nor co-hyperimmune since $F_0, F_1, \dots $ defined by $F_n = \{2n, 2n+1\}$ is a c.e. array tracing both A and $\overline {A}$ . Furthermore, A is not Kurtz random, since $A \in \{ X : \forall n X(2n) \neq X(2n+1) \}$ which is a $\Pi ^0_1$ class of measure 0.

Proof Suppose for contradiction that there is a non-trivial $\Pi ^0_2$ partition regular class $\mathcal {V} \subseteq \mathcal {A}$ such that $Y_1 \notin \mathcal {V}$ . In particular, $\mathcal {V} \subseteq \mathcal {L}(A)$ , so $Y_0 \notin \mathcal {V}$ . By partition regularity of $\mathcal {V}$ , since $Y_0 \cup Y_1 \supseteq X$ , then $X \notin \mathcal {V}$ , which contradicts partition genericity of X in $\mathcal {A}$ .

Proof We prove the statement by the lemma by induction on k. For $k = 0$ , $Y_0 \supseteq X$ so $Y_0$ is partition generic in $\mathcal {A}$ by upward-closure of partition genericity. Take $\mathcal {U} = 2^\omega $ and we are done.

Suppose now that the property holds for $k-1$ . Suppose $Y_k$ is not partition generic in $\mathcal {A}$ . Thus there is a non-trivial $\Pi ^0_2$ partition regular class $\bigcap _{e \in C} \mathcal {U}_e \subseteq \mathcal {A}$ such that $Y_k \not \in \bigcap _{e \in C} \mathcal {U}_e$ . In particular, there is some $e \in \mathcal {C}$ such that $Y_k \notin \mathcal {U}_e$ . Note that $\bigcap _{e \in C} \mathcal {U}_e \subseteq \mathcal {A} \cap \mathcal {U}_e$ , so $\mathcal {A} \cap \mathcal {U}_e$ is partition large. By Theorem 2.32, $Y_0 \cup \dots \cup Y_{k-1}$ is partition generic in $\mathcal {U}_e \cap \mathcal {A}$ . By induction hypothesis on $\mathcal {U}_e \cap \mathcal {A}$ and $Y_0 \cup \dots \cup Y_{k-1}$ , there is a $\Sigma ^0_1$ class $\mathcal {V}$ such that $\mathcal {V} \cap \mathcal {U}_e \cap \mathcal {A}$ is partition large, together with some $i < k$ such that $Y_i$ is partition generic in $\mathcal {V} \cap \mathcal {U}_e \cap \mathcal {A}$ . The property therefore holds with the $\Sigma ^0_1$ class $\mathcal {V} \cap \mathcal {U}_e$ .

Proof First, $\omega $ is partition generic in $\mathcal {U}_2 = \{ X : |X| \geq 2 \}$ . Suppose $X \in \mathcal {P}$ and let $Y_0 \cup \dots \cup Y_k \supseteq X$ . Let $\mathcal {U} \subseteq 2^\omega $ be a non-trivial $\Sigma ^0_1$ partition large class in which X is partition generic. By Proposition 2.33, there is a $\Sigma ^0_1$ class $\mathcal {V}$ such that $\mathcal {U} \cap \mathcal {V}$ is partition large, together with some $i \leq k$ such that $Y_i$ is partition generic in $\mathcal {U} \cap \mathcal {V}$ . Note that $\mathcal {U} \cap \mathcal {V} \subseteq \mathcal {U}$ , hence is non-trivial. It follows that $Y_i \in \mathcal {P}$ .

Proof First, let us show that $\mathcal {L}_X \subseteq \mathcal {L}(\mathcal {U}_{X,2})$ . Suppose that $Y \in \mathcal {L}_X$ . By definition, $|Y \cap X| = \infty $ , so $Y \in \mathcal {U}_{X,2}$ . So $\mathcal {L}_X$ is a partition regular subclass of $\mathcal {U}_{X,2}$ . Since $\mathcal {L}(\mathcal {U}_{X,2})$ is the largest partition regular subclass of $\mathcal {U}_{X,2}$ , then $\mathcal {L}_X \subseteq \mathcal {L}(\mathcal {U}_{X,2})$ .

Then, let us show that $\mathcal {L}(\mathcal {U}_{X,2}) \subseteq \mathcal {L}_X$ . Let $Y \in \mathcal {L}(\mathcal {U}_{X,2})$ . We claim that $|Y \cap X| = \infty $ . Indeed, otherwise, consider the $|Y \cap X|$ -cover of $Y \cap X$ made of singletons. By partition regularity of $\mathcal {L}(\mathcal {U}_{X,2})$ , one of the parts belongs to $\mathcal {U}_{X,2}$ , but $\mathcal {U}_{X,2}$ contains no singleton, contradiction. Therefore $|Y \cap X| = \infty $ , so $Y \in \mathcal {L}_X$ .

Proof $\mathcal {U}_{X,2}$ is clearly non-trivial, upward-closed, and $\Sigma ^0_1(X)$ . By Lemma 2.35, $\mathcal {L}(\mathcal {U}_{X,2}) = \mathcal {L}_X$ . Since X is infinite, $X \in \mathcal {L}_X$ , so $\mathcal {L}(\mathcal {U}_{X,2}) \neq \emptyset $ . By Proposition 2.18, $\mathcal {U}_{X,2}$ is large.

Proof Let $\mathcal {V}$ be a partition large subclass of $\mathcal {U}_{X,2}$ . Suppose for the contradiction that $X \not \in \mathcal {V}$ . By largeness of $\mathcal {V}$ , $\overline {X} \in \mathcal {V} \subseteq \mathcal {U}_{X,2}$ . Then $|\overline {X} \cap X| \geq 2$ , contradiction.

Proof By Lemma 2.36, $\mathcal {U}_{H,2}$ is a non-trivial $\Sigma ^0_1(A)$ partition large class. In particular, $\mathcal {L}(\mathcal {U}_{H,2})$ is a non-trivial $\Pi ^0_2(H)$ partition generic class. However, $A \not \in \mathcal {U}_{H,2} \supseteq \mathcal {L}(\mathcal {U}_{H,2})$ , so A is not H-partition generic.

Proof Let A be the bi-partition generic set of Proposition 2.30. Let $H \subseteq A$ . By Lemma 2.38, $\overline {A}$ is not H-partition generic. Let $P = \{ 2n+1 : 2n \in H \} \cup \{2n : 2n+1 \in H \}$ . Then $P \subseteq \overline {A}$ . By Lemma 2.38, $\overline {A}$ is not P-partition generic, hence not H-partition generic. The case $H \subseteq \overline {A}$ is symmetric.

Proof Let $H = A \cap B$ if it is infinite, otherwise $H = \overline {A} \cap B$ . By Lemma 2.36, $\mathcal {U}_{H,2}$ is a non-trivial $\Sigma ^0_1(H)$ partition large class. We claim that B is H-partition generic in $\mathcal {U}_{H,2}$ . By Lemma 2.37, H is H-partition generic in $\mathcal {U}_{H,2}$ . Since $B \supseteq H$ , then A is also H-partition generic in $\mathcal {U}_{H,2}$ .

Proof A split pair is a pair of strings $\rho _0, \rho _1$ such that there is an input $x \in \omega $ for which $\Phi _e^{\sigma \cup \rho _0}(x)\downarrow \neq \Phi _e^{\sigma \cup \rho _1}(x)\downarrow $ . We have two cases.

Case 1: the following class is large:

$$ \begin{align*}\mathcal{A} = \{ Y \in \mathcal{U} : Y\ \mathit{contains\ a\ split\ pair} \}. \end{align*} $$

Since $\mathcal {A}$ is a non-trivial $\Pi ^0_2$ large subclass of $\mathcal {U}$ and X is partition generic in $\mathcal {U}$ , then $X \in \mathcal {A}$ . Let $\rho _0, \rho _1 \subseteq X$ be a split pair, with witness x. In particular, there is some $i < 2$ such that $\Phi _e^{\sigma \cup \rho _i}(x)\neq B(x)$ . Let $Y = X \setminus \{0, \dots , |\rho _i|\}$ . By Proposition 2.23, Y is partition generic in $\mathcal {U}$ . Then $(\sigma \cup \rho _i, Y, \mathcal {U})$ is an extension of $(\sigma , X, \mathcal {U})$ forcing $\Phi _e^G(x) \neq B(x)$ .

Case 2: there is some $k \in \omega $ such that the following class is non-empty:

$$ \begin{align*}\mathcal{C} = \{ Z_0 \oplus \dots \oplus Z_{k-1} : Z_0 \cup \dots \cup Z_{k-1} = \omega \wedge \forall i < k\ Z_i\, \not \in \mathcal{U} \mbox{ or } Z_i \mbox{ contains no split pair} \}. \end{align*} $$

Note that $\mathcal {C}$ is a $\Pi ^0_1$ class. By the cone avoidance basis theorem [Reference Jockusch and Soare7], there is an element $Z_0 \oplus \dots \oplus Z_{k-1} \in \mathcal {C}$ such that $B \not \leq _T Z_0 \oplus \dots \oplus Z_{k-1}$ . By Proposition 2.33, there is a $\Sigma ^0_1$ class $\mathcal {V}$ such that $\mathcal {V} \cap \mathcal {U}$ is large and some $i < k$ such that $Z_i \cap X$ is partition generic in $\mathcal {V} \cap \mathcal {U}$ . Note that in particular, $Z_i \cap X \in \mathcal {V} \cap \mathcal {U}$ , so $Z_i \in \mathcal {U}$ , hence $Z_i$ contains no split pair.

The condition $(\sigma , Z_i \cap X, \mathcal {V} \cap \mathcal {U})$ is a valid extension of $(\sigma , X, \mathcal {U})$ .

We claim that $(\sigma , Z_i \cap X, \mathcal {V} \cap \mathcal {U})$ forces $\Phi ^G_e \neq B$ . Suppose for the contradiction that $\Phi ^{G_{\mathcal {F}}}_e = B$ for some generic filter $\mathcal {F}$ . Let $f : \omega \to 2$ be the partial $Z_i$ -computable function which on input x, searches for some $\rho \subseteq Z_i$ such that $\Phi ^{\sigma \cup \rho }_e(x)\downarrow $ . If such a $\rho $ is found, then $f(x) = \Phi ^{\sigma \cup \rho }_e(x)\downarrow $ . Since $\Phi ^{G_{\mathcal {F}}}_e = B$ , then f is total. Since $Z_i$ contains no split pair, then $f = B$ , which contradicts the assumption that $B \not \leq _T Z_i$ .

Proof Let $\mathcal {F}$ be a sufficiently generic filter containing $(\emptyset , A, \mathcal {U})$ . By Lemma 3.1, $G_{\mathcal {F}}$ is infinite. By construction, $G_{\mathcal {F}} \subseteq A$ . Last, by Lemma 3.2, $B \not \leq _T G_{\mathcal {F}}$ .

Proof By Corollary 2.34, either A or $\overline {A}$ is partition generic in a non-trivial $\Sigma ^0_1$ large class. Apply Theorem 3.3.

Proof For every k, let $S_k$ be the set of pairs $(x, v) \in \omega \times 2$ such that for every k-cover $Z_0 \sqcup \dots \sqcup Z_{k-1} \supseteq \omega $ , there is some $j < k$ such that $Z_j \in \mathcal {U}$ , and some $\rho \subseteq Z_j$ such that $\Phi _e^{\sigma \cup \rho }(x)\downarrow = v$ . Note that the set $S_k$ is $\Sigma ^0_1$ uniformly in k. We have two cases:

Case 1: for every $k \in \omega $ , there is some $x \in \omega $ such that $(x, \Phi _x(x)) \in S_k$ . Then the following class is large:

$$ \begin{align*}\{Y \in \mathcal{U} : \exists x\ \exists \rho \subseteq Y\ \Phi^{\sigma \cup \rho}_e(x)\downarrow = \Phi_x(x) \}. \end{align*} $$

Since X is partition generic, then it belongs to this class, so there is some x and some $\rho \subseteq X$ such that $\Phi ^{\sigma \cup \rho }_e(x)\downarrow = \Phi _x(x)$ . Let $Y = X \setminus \{0, \dots , |\rho |\}$ . Note that Y is again partition generic in $\mathcal {U}$ , so $(\sigma \cup \rho , Y, \mathcal {U})$ is an extension forcing $\Phi ^G_e(x)\downarrow = \Phi _x(x)$ .

Case 2: there is some $k \in \omega $ such that for every $x \in \omega $ , if $\Phi _x(x)\downarrow $ , then $(x, \Phi _x(x)) \not \in S_k$ . Then there must be some x such that $(x, 0), (x, 1) \not \in S_k$ , otherwise we would compute a DNC ${}_2$ function by waiting, for each x, for some $v < 2$ such that $(x, v)$ is enumerated in $S_k$ . Let x be such that $(x, 0), (x, 1) \not \in S_k$ . By definition of $S_k$ , there are two k-covers of $\omega $ , $Z^0_0 \cup \dots \cup Z^0_{k-1} = \omega $ and $Z^1_0 \cup \dots \cup Z^1_{k-1} = \omega $ such that for every $i < 2$ and every $j < k$ such that $Z^i_j \in \mathcal {U}$ , and every $\rho \subseteq Z^i_j$ such that $\Phi _e^{\sigma \cup \rho }(x)\downarrow $ , then $\Phi _e^{\sigma \cup \rho }(x) \neq i$ .

Then $\langle Z^0_s \cap Z^1_t : s, t < k \rangle $ is a $k^2$ -cover of $\omega $ such that for every $s, t < k$ , either $Z^0_s \cap Z^1_t \not \in \mathcal {U}$ , or for every $\rho \subseteq Z^i_j$ , $\Phi _e^{\sigma \cup \rho }(x)\uparrow $ . By Proposition 2.33, there is a non-trivial large $\Sigma ^0_1$ class $\mathcal {V} \subseteq \mathcal {U}$ and some $s, t < k$ such that $Z^0_s \cap Z^1_t \cap X$ is partition regular in $\mathcal {V}$ . The condition $(\sigma , Z^0_s \cap Z^1_t \cap X, \mathcal {V})$ is an extension forcing $\Phi ^G_e(x)\uparrow $ .

Proof Let $\mathcal {F}$ be a sufficiently generic filter containing $(\emptyset , A, \mathcal {U})$ . By Lemma 3.1, $G_{\mathcal {F}}$ is infinite. By construction, $G_{\mathcal {F}} \subseteq A$ . Last, by Lemma 3.5, $G_{\mathcal {F}}$ is of non-PA degree.

Proof By Corollary 2.34, either A or $\overline {A}$ is partition generic in a non-trivial $\Sigma ^0_1$ large class. Apply Theorem 3.6.

Proof For every k, let $S_k$ be the set of strings $\mu \in 2^{<\omega }$ such that for every k-cover $Z_0 \sqcup \dots \sqcup Z_{k-1} \supseteq \omega $ , there is some $j < k$ such that $Z_j \in \mathcal {U}$ , and some $\rho \subseteq Z_j$ such that $\Phi _e^{\sigma \cup \rho }(|\mu |)\downarrow = \mu $ . Note that the set $S_k$ is $\Sigma ^0_1$ uniformly in k. We have two cases:

Case 1: for every $k \in \omega $ , there is some $\mu \in S_k$ such that $[\mu ] \cap \mathcal {C} = \emptyset $ . Then the following class is large:

$$ \begin{align*}\{ \!Y \in \mathcal{U} : \exists \mu\ \exists \rho \subseteq Y\ \Phi^{\sigma \cup \rho}_e(|\mu|)\downarrow = \mu \wedge [\mu] \cap \mathcal{C} = \emptyset \} \end{align*} $$

is large. Since X is partition generic, then it belongs to this class, so there is some $\mu \in 2^{<\omega }$ and some $\rho \subseteq X$ such that $\Phi ^{\sigma \cup \rho }_e(|\mu |)\downarrow = \mu $ and $[\mu ] \cap \mathcal {C} = \emptyset $ . Let $Y = X \setminus \{0, \dots , |\rho |\}$ . Note that Y is again partition generic in $\mathcal {U}$ , so $(\sigma \cup \rho , Y, \mathcal {U})$ is an extension forcing $\Phi ^G_e(|\mu |)\downarrow = \mu $ .

Case 2: there is some $k \in \omega $ such that for every $\mu \in S_k$ , $[\mu ] \cap \mathcal {C} \neq \emptyset $ . Then there must be some n such that $S_k$ contains no string of length n. Indeed, otherwise, one would compute a 1-trace of $\mathcal {C}$ , contradicting our hypothesis. For every string $\mu $ of length n, let $Z^\mu _0 \cup \dots \cup Z^\mu _{k-1} = \omega $ be a k-cover such that for every $j < k$ such that $Z^\mu _j \in \mathcal {U}$ , and every $\rho \subseteq Z^\mu _j$ such that $\Phi _e^{\sigma \cup \rho }(n)\downarrow $ , then $\Phi _e^{\sigma \cup \rho }(n) \neq \mu $ .

Let $\langle P_j : j < k^{2^n} \rangle $ be the $k^{2^n}$ -cover of $\omega $ refining all the k-covers above. Then for every $j < k^{2^n}$ and every $\rho \subseteq P_j$ , $\Phi _e^{\sigma \cup \rho }(n)\uparrow $ . By Proposition 2.33, there is a non-trivial large $\Sigma ^0_1$ class $\mathcal {V} \subseteq \mathcal {U}$ and some $j < k^{2^n}$ such that $P_j \cap X$ is partition regular in $\mathcal {V}$ . The condition $(\sigma , P_j \cap X, \mathcal {V})$ is an extension forcing $\Phi ^G_e(n)\uparrow $ .

Proof Let $\mathcal {F}$ be a sufficiently generic filter containing $(\emptyset , A, \mathcal {U})$ . By Lemma 3.1, $G_{\mathcal {F}}$ is infinite. By construction, $G_{\mathcal {F}} \subseteq A$ . Last, by Lemma 3.10, none of the classes $\mathcal {C}_n$ admits a $G_{\mathcal {F}}$ -computable 1-trace.

Proof By Corollary 2.34, either A or $\overline {A}$ is partition generic in a non-trivial $\Sigma ^0_1$ large class. Apply Theorem 3.11.

Proof By Proposition 2.26, every Kurtz random is bi-partition generic. Apply Theorem 3.11.

Proof For every $m,c$ , let $\mathcal {C}_{m,c} = \{ X : \forall k K( X \upharpoonright _k ) \geq k/m-c\}$ . A set X has positive effective Hausdorff dimension iff $X \in \mathcal {C}_{m,c}$ for some $m,c \in \omega $ . For every $m,c \in \omega $ , $\mathcal {C}_{m,c}$ has no computable 1-trace (see Corollary 1.4 in [Reference Kjos-Hanssen and Liu10]). By Corollary 3.13, there is an infinite subset $H \subseteq A$ such that none of the classes $\mathcal {C}_{m,c}$ admits an H-computable 1-trace. In particular, H does not compute any set of positive effective Hausdorff dimension.

Proof Let $\bigcap _n \mathcal {V}_n$ be a $\Pi ^0_2$ partition regular class such that $\overline {A} \not \in \bigcap _n \mathcal {V}_n$ . In particular, there is some $m \in \omega $ such that $\overline {A} \not \in \mathcal {V}_m$ . Define a $\Delta ^0_2$ decreasing sequence of Mathias conditions $(\sigma _0, X_0) \geq (\sigma _1, X_1) \geq \cdots $ such that for every $s \in \omega $ :

1. (1) $\sigma _s \subseteq A$ ; $X_s$ is low; $X_s \in \bigcap _n \mathcal {V}_n$ ;

2. (2) $\sigma _{s+1} \in \mathcal {V}_e$ if $s = 2e$ ;

3. (3) $(\sigma _{s+1}, X_{s+1}) \Vdash \Phi ^G_e(e)\downarrow $ or $(\sigma _{s+1}, X_{s+1}) \Vdash \Phi ^G_e(e)\uparrow $ if $s = 2e+1$ .

Start with the condition $(\emptyset , \omega )$ .

Satisfying (2): Given a condition $(\sigma _s, X_s)$ at a stage $s = 2e$ , search $\emptyset '$ -computably for some finite $\rho \subseteq X \cap A$ such that $\rho \in \mathcal {V}_s$ . We claim that such a $\rho $ must exist. Indeed, since X belongs to the partition regular class $\bigcap _n \mathcal {V}_n$ , then either $X \cap A$ or $X \cap \overline {A}$ belongs to $\bigcap _n \mathcal {V}_n$ . However, $\overline {A} \not \in \mathcal {V}_m$ , so by upward-closure of partition regular classes, $X \cap \overline {A} \not \in \bigcap _n \mathcal {V}_n$ , hence $X \cap A \in \bigcap _n \mathcal {V}_n$ . The condition $(\sigma _s \cup \rho , X_s - \{0, \dots , \max \rho \})$ satisfies (2).

Satisfying (3): Given an condition $(\sigma _s, X_s)$ at stage $s = 2e+1$ , consider the $\Pi ^{0,X_s}_1$ class $\mathcal {C}$ of all B such that $\overline {B} \not \in \mathcal {V}_m$ and $(\sigma _s, X_s \cap B) \Vdash \Phi ^G_e(e)\uparrow $ . Decide $\emptyset '$ -computably if $\mathcal {C}$ is empty or not. If $\mathcal {C} = \emptyset $ , then in particular $A \not \in \mathcal {C}$ , so $(\sigma _s, X_s \cap A) \not \Vdash \Phi ^G_e(e)\uparrow $ . Search $\emptyset '$ -computably for some $\rho \subseteq X_s \cap A$ such that $\Phi ^{\sigma _s \cup \rho }_e(e)\downarrow $ . The condition $(\sigma _s \cup \rho , X_s - \{0, \dots , \max \rho \})$ forces $\Phi ^G_e(e)\downarrow $ , hence satisfies (3). If $\mathcal {C} \neq \emptyset $ , then by the low basis theorem, there is some $B \in \mathcal {C}$ of low degree. In particular, $\overline {B} \not \in \mathcal {V}_m$ so $\overline {B} \cap X_s \not \in \mathcal {V}_m$ . However, $X_s \in \bigcap _n \mathcal {V}_n$ , so by partition regularity of $\bigcap _n \mathcal {V}_n$ , $B \cap X_s \in \bigcap _n \mathcal {V}_n$ . The condition $(\sigma _s, X_s \cap B)$ forces $\Phi ^G_e(e)\uparrow $ , hence satisfies (3).

This completes the construction. Let $G = \bigcup _s \sigma _s$ . In particular, by (1), $G \subseteq A$ . By (2) $G \in \bigcap _n \mathcal {V}_n$ , hence G is infinite. By (3), $G' \leq _T \emptyset '$ . This completes the proof.

Proof Suppose A is bi-partition generic relative to every low set. By Proposition 2.25, A is bi-immune relative to every low set, hence has no low infinite subset in it or its complement.

Suppose A is not bi-partition generic relative to a low set L. By symmetry, say $\overline {A}$ is not partition generic relative to L. By Theorem 3.15 relativized to L, there is an infinite subset $H \subseteq A$ such that $H' \leq _T L' \leq _T \emptyset '$ , hence A has an infinite low subset.

Proof Immediate by the previous corollary and the existence of a $\Delta ^0_2$ set with no low infinite subset in either it or its complement [Reference Downey, Hirschfeldt, Lempp and Solomon2].

Proof It is clear that $\bigcup _{i \in I} \mathcal {L}_i$ is not empty. Let $(X_0,X_1) \in \bigcup _{i \in I} \mathcal {L}_i$ . Let $Y_0 \supseteq X_0$ and $Y_1 \supseteq X_1$ . There is some $i \in I$ such that $(X_0,X_1) \in \mathcal {L}_i$ . As $\mathcal {L}_i$ is $\mathfrak {H}$ -regular, $(Y_0,Y_1) \in \mathcal {L}_i \subseteq \bigcup _{i \in I} \mathcal {L}_i$ .

Let $(X_0,X_1) \in \bigcup _{i \in I} \mathcal {L}_i$ . Let $Y_0 \cup \dots \cup Y_k \supseteq X$ and $\sigma \in 2^{<\omega }$ . There is some $i \in I$ such that $(X_0,X_1) \in \mathcal {L}_i$ . As $\mathcal {L}_i$ is $\mathfrak {H}$ -regular, $(Y_j,X_1 \setminus \sigma ) \in \mathcal {L}_i \subseteq \bigcup _{i \in I} \mathcal {L}_i$ for some $j \leq k$ .

Proof Suppose $\mathcal {A}$ is upward-closed and contains an $\mathfrak {H}$ -regular subclass $\mathcal {L} \subseteq \mathcal {A}$ . Upward closure is trivially satisfied by hypothesis. By $\mathfrak {H}$ -regularity of $\mathcal {L}$ , $(\omega ,\omega ) \in \mathcal {L}$ and for every k-cover $Y_0, \dots , Y_{k-1}$ of $\omega $ , and every $\sigma \in 2^{<\omega }$ , there is some $j < k$ such that $(\! Y_j, \omega \setminus \sigma ) \in \mathcal {L} \subseteq \mathcal {A}$ . So $\mathcal {A}$ is $\mathfrak {H}$ -large.

Suppose now $\mathcal {A}$ is $\mathfrak {H}$ -large. It is upward-closed by definition. We claim that $\mathcal {L}_{\mathfrak {H}}(\mathcal {A})$ is $\mathfrak {H}$ -regular. By definition of $\mathfrak {H}$ -largeness and Theorem 4.6, $(\omega ,\omega ) \in \mathcal {L}_{\mathfrak {H}}(\mathcal {A})$ , so $\mathcal {L}_{\mathfrak {H}}(\mathcal {A}) \neq \emptyset $ . By definition of $\mathcal {L}_{\mathfrak {H}}(\mathcal {A})$ , it is $\mathfrak {H}$ -regular, hence $\mathcal {A}$ contains an $\mathfrak {H}$ -regular subclass.

Proof Let $\mathcal {L} \subseteq \mathcal {A}$ be any non-trivial $\Pi ^0_2 \mathfrak {H}$ -regular subclass of $\mathcal {A}$ . Since X is $\mathfrak {H}(B)$ -generic in $\mathcal {A}$ , then $(X,\overline {B}) \in \mathcal {L}$ . Since $Y \supseteq X$ , then by upward-closure of $\mathcal {L}$ , $(Y, \overline {B}) \in \mathcal {L}$ . Therefore Y is $\mathfrak {H}(B)$ -generic in $\mathcal {A}$ .

Proof Let $\mathcal {L} \subseteq \mathcal {A}$ be any non-trivial $\Pi ^0_2 \mathfrak {H}$ -regular subclass of $\mathcal {A}$ . Since X is $\mathfrak {H}(B)$ -generic in $\mathcal {A}$ , then $(X,\overline {B}) \in \mathcal {L}$ . Since $Y =^{*} X$ , then there is a finite set F such that $Y \cup F \supseteq X$ . By $\mathfrak {H}$ -regularity of $\mathcal {L}$ , either $(Y,\overline {B})$ or $(F,\overline {B})$ belongs to $\mathcal {L}$ . Since $\mathcal {L}$ is non-trivial, $(F,\overline {B}) \notin \mathcal {L}$ , so $(Y,\overline {B}) \in \mathcal {L}$ . Therefore Y is $\mathfrak {H}(B)$ -generic in $\mathcal {A}$ .

Proof Suppose for contradiction that there is a non-trivial $\Pi ^0_2 \mathfrak {H}$ -regular class $\mathcal {V} \subseteq \mathcal {A}$ such that $(Y_1,\overline {B}) \notin \mathcal {V}$ . In particular, $\mathcal {V} \subseteq \mathcal {L}_{\mathfrak {H}}(A)$ , so $(Y_0,\overline {B}) \notin \mathcal {V}$ . By partition $\mathfrak {H}$ -regularity of $\mathcal {V}$ , since $Y_0 \cup Y_1 \supseteq X$ , then $(X,\overline {B}) \notin \mathcal {V}$ , which contradicts $\mathfrak {H}(B)$ -genericity of X in $\mathcal {A}$ .

Proof We proceed by induction on k. For $k = 0$ , $Y_0 \supseteq X$ , so by Proposition 4.10, $Y_0$ is $\mathfrak {H}(B)$ -generic in $\mathcal {A}$ . Take $\mathcal {U} = \mathcal {A}$ and we are done.

Suppose now that the property holds for $k-1$ . Suppose $Y_k$ is not $\mathfrak {H}(B)$ -generic in $\mathcal {A}$ . Thus there is a non-trivial $\Pi ^0_2 \mathfrak {H}$ -regular class $\bigcap _{e \in C} \mathcal {U}_e \subseteq \mathcal {A}$ such that $(Y_k,\overline {B}) \not \in \bigcap _{e \in C} \mathcal {U}_e$ . In particular, there is some $e \in \mathcal {C}$ such that $(Y_k,\overline {B}) \notin \mathcal {U}_e$ . Note that $\bigcap _{e \in C} \mathcal {U}_e \subseteq \mathcal {A} \cap \mathcal {U}_e$ , so $\mathcal {A} \cap \mathcal {U}_e$ is $\mathfrak {H}$ -large. By Theorem 4.12, $Y_0 \cup \dots \cup Y_{k-1}$ is $\mathfrak {H}(B)$ -generic in $\mathcal {U}_e \cap \mathcal {A}$ . By induction hypothesis on $\mathcal {U}_e \cap \mathcal {A}$ and $Y_0 \cup \dots \cup Y_{k-1}$ , there is a non-trivial $\mathfrak {H}$ -large $\Sigma ^0_1$ class $\mathcal {V} \subseteq \mathcal {U}_e \cap \mathcal {A}$ , together with some $i < k$ such that $Y_i$ is $\mathfrak {H}(B)$ -generic in $\mathcal {V}$ .

Proof Let $\mathcal {L} \subseteq 2^\omega \times 2^\omega $ be a non-trivial $\Pi ^0_2 \mathfrak {H}$ -regular class. We can assume $\mathcal {L} = \bigcap _n \mathcal {U}_n$ for some decreasing sequence of $\Sigma ^0_1 \mathfrak {H}$ -large classes. Fix $n \in \omega $ and let $W \subseteq 2^{<\omega } \times 2^{<\omega }$ be a c.e. set such that $\mathcal {U}_n = \{ (X,Y) : \exists (\rho _0, \rho _1) \in W\ \rho _0 \subseteq X \wedge \rho _1 \subseteq Y \}$ . Let P be a cofinite set.

Fix some $s \in \omega $ . Since $\mathcal {U}_n$ is $\mathfrak {H}$ -large, then $(\omega , \{s+1, s+2, \dots \}) \in \mathcal {U}_n$ . Therefore, there is some $(\rho _0, \rho _1) \in W$ such that $\min \rho _1> s$ . Let $F_s = \rho _1$ . Note that $F_s$ can be found computably uniformly in s. Since B is hyperimmune, there is some s such that $F_s \subseteq \overline {B}$ . It follows that $(\omega , \overline {B}) \in \mathcal {U}_n$ . Since this holds for every n, then $(\omega , \overline {B}) \in \bigcap _n \mathcal {U}_n = \mathcal {L}$ .

Proof Let $\mathcal {L} \subseteq 2^\omega \times 2^\omega $ be a non-trivial $\Pi ^0_2 \mathfrak {H}$ -regular class. By Proposition 4.14, $(\omega , \overline {B}) \in \mathcal {L}$ . Since $\mathcal {L}$ is $\mathfrak {H}$ -regular, then for every k, every $Z_0 \cup \dots \cup Z_{k-1} = \omega $ , there is some $j < k$ such that $(Z_j, \overline {B}) \in L$ . Thus the class $\mathcal {L}_0 = \{ X : (X, \overline {B}) \in \mathcal {L} \}$ is partition large. Moreover, $\mathcal {L}_0$ is $\Pi ^0_2(B)$ , so since A is partition generic relative to B, $A \in \mathcal {L}_0$ . It follows that $(A,\overline {B}) \in \mathcal {L}$ .

Proof Consider the notion of forcing $(\sigma , X, \mathcal {V})$ where $(\sigma , X)$ is as Mathias condition, and $\mathcal {V} \subseteq \mathcal {U}$ is an $\mathfrak {H}$ -large $\Sigma ^0_1$ class within which X is $\mathfrak {H}(B)$ -generic. A condition $(\tau , Y, \mathcal {W})$ extends another condition $(\sigma , X, \mathcal {V})$ if $(\tau , Y)$ Mathias extends $(\sigma , X)$ and $\mathcal {W} \subseteq \mathcal {V}$ .

Any filter $\mathcal {F}$ induces a subset $G_{\mathcal {F}} = \bigcup _{(\sigma , X, \mathcal {V}) \in \mathcal {F}} \sigma $ of A. The first property we prove is that $G_{\mathcal {F}}$ is an infinite set.

In the following lemma, we interpret $\Phi _e^G$ as a partial G-c.e. array $\{ \Phi _e^G(n) : n \in \omega \}$ . An array $\{F_n : n \in \omega \}$ intersects a set C if $\forall n\ F_n \cap C \neq \emptyset $ .

Proof We have two cases.

Case 1: the following class is $\mathfrak {H}$ -large:

$$ \begin{align*}\mathcal{A} = \{ (Y,C) \in \mathcal{V} : \exists \rho \subseteq Y \exists n\ \Phi_e^{\sigma \cup \rho}(n) \subseteq C \}. \end{align*} $$

Since $\mathcal {A}$ is a non-trivial $\Pi ^0_2 \mathfrak {H}$ -large subclass of $\mathcal {U}$ and X is $\mathfrak {H}(B)$ -generic in $\mathcal {U}$ , then $(X,\overline {B}) \in \mathcal {A}$ . Therefore there is some $\rho \subseteq X$ and some n such that $\Phi _e^{\sigma \cup \rho }(n) \subseteq \overline {B}$ . Let $Y = X \setminus \{0, \dots , |\rho |\}$ . By Proposition 4.11, Y is $\mathfrak {H}(B)$ -generic in $\mathcal {V}$ . Therefore $(\sigma \cup \rho , Y, \mathcal {V})$ is an extension of $(\sigma , X, \mathcal {V})$ forcing $\Phi _e^G(n) \subseteq \overline {B}$ .

Case 2: there is a co-finite set P and some k such that the following class is non-empty:

$$ \begin{align*}\mathcal{C} = \left\{ Z_0 \oplus \dots \oplus Z_{k-1} : \begin{array}{l} Z_0 \cup \dots \cup Z_{k-1} = \omega \wedge \\ \forall i < k\ (Z_i,P) \not \in \mathcal{V} \vee \forall n \forall \rho \subseteq Z_i \Phi_e^{\sigma \cup \rho}(n)\uparrow \vee \Phi_e^{\sigma \cup \rho}(n) \not \subseteq P \end{array} \!\!\!\!\right\}\!. \end{align*} $$

Note that $\mathcal {C}$ is a non-empty $\Pi ^0_1$ class. Pick any element $Z_0 \oplus \dots \oplus Z_{k-1} \in \mathcal {C}$ . By Proposition 4.13, there is an $\mathfrak {H}$ -large $\Sigma ^0_1$ subclass $\mathcal {W} \subseteq \mathcal {V}$ and some $i < k$ such that $Z_i \cap X$ is $\mathfrak {H}(B)$ -generic in $\mathcal {W}$ . In particular, $(Z_i \cap X, \overline {B}) \in \mathcal {L}_{\mathfrak {H}}(W) \subseteq \mathcal {W}$ , so $(Z_i \cap X, \overline {B} \cap P) \in W \subseteq \mathcal {V}$ and by upward-closure of V, $(Z_i, P) \in \mathcal {V}$ , so for every n and every $\rho \subseteq Z_i$ , $\Phi _e^{\sigma \cup \rho }(n)\uparrow \vee \Phi _e^{\sigma \cup \rho }(n) \not \subseteq P$ . The condition $(\sigma , Z_i \cap X, \mathcal {W})$ is an extension of $(\sigma , X, \mathcal {V})$ forcing $\Phi _e^{G}$ not to be a c.e. array.

We are now ready to prove Theorem 4.16. Let $\mathcal {F}$ be a sufficiently generic filter containing $(\emptyset , A, \mathcal {U})$ . By Lemma 4.17, $G_{\mathcal {F}}$ is infinite. By construction, $G_{\mathcal {F}} \subseteq A$ . Last, by Lemma 4.18, B is $G_{\mathcal {F}}$ -hyperimmune.