Kalimullin Pair and Semicomputability in $\alpha$-Computability Theory

TL;DR

This paper extends semicomputability concepts to $oldsymbol{ extalpha}$-Computability Theory, introducing $oldsymbol{ extalpha}$-Kalimullin pairs and analyzing their definability and structural properties in $oldsymbol{ extalpha}$-enumeration degrees.

Contribution

It defines $oldsymbol{ extalpha}$-Kalimullin pairs, proves their definability under certain conditions, and shows that every nontrivial total $oldsymbol{ extalpha}$-enumeration degree is a join of a maximal $oldsymbol{ extalpha}$-Kalimullin pair.

Findings

$oldsymbol{ extalpha}$-Kalimullin pairs are definable in $oldsymbol{ extalpha}$-enumeration degrees for certain $oldsymbol{ extalpha}$.

Every nontrivial total $oldsymbol{ extalpha}$-enumeration degree is a join of a maximal $oldsymbol{ extalpha}$-Kalimullin pair when $oldsymbol{ extalpha}$ is an infinite regular cardinal.

The results generalize classical semicomputability to the setting of $oldsymbol{ extalpha}$-computability theory.

Abstract

We generalize some results on semicomputability by Jockusch \cite{jockusch1968semirecursive} to the setting of $\alpha$-Computability Theory. We define an $\alpha$-Kalimullin pair and show that it is definable in the $\alpha$-enumeration degrees $\mathcal{D}_{\alpha e}$ if the projectum of $\alpha$ is $\alpha^*=\omega$ or if $\alpha$ is an infinite regular cardinal. Finally using this work on $\alpha$-semicomputability and $\alpha$-Kalimullin pairs we conclude that every nontrivial total $\alpha$-enumeration degree is a join of a maximal $\alpha$-Kalimullin pair if $\alpha$ is an infinite regular cardinal.

Equations

$$V_x:=\{y<\alpha :D_x\subseteq D_y\wedge E_x\subseteq E_y\}$$

$$x\in Y⟺x\in X\wedge D_x\subseteq A⟺x\in X\wedge E_x\subseteq B$$

$$R_e:Y\ne {\Phi }_e(X\oplus U).$$

$$X_s\subseteq X_{s+1}$$

$$M_{s+1}\subseteq M_s$$

$$X_{s+1}-X_s\subseteq M_{s+1}$$

$$\forall D,E\in L_{\alpha }[D\subseteq A\wedge E\subseteq B⟹\exists x\in M_s[D\subseteq D_x\subseteq A\wedge E\subseteq E_x\subseteq B]]$$

$$X_s\in L_{\alpha }$$

$$N_s\in L_{\alpha }$$

$$I_s\in L_{\alpha }$$

$$M_s:=(\bigcap _{x\in I_s}{V}_x)-N_s=V_z-N_s$$

$$M_s\in {\Delta }_1(L_{\alpha })$$

$$W:=\{⟨a,b⟩:\exists x\in M_s[a\in D_x\wedge b\in E_x\wedge x\in {\Phi }_e((X_s\cup (M_s\cap V_x))\oplus U)]\}.$$

Full text

Kalimullin Pair and Semicomputability in $\alpha$-Computability Theory

Dávid Natingga

Abstract

We generalize some results on semicomputability by Jockusch [4] to the setting of $\alpha$-Computability Theory. We define an $\alpha$-Kalimullin pair and show that it is definable in the $\alpha$-enumeration degrees $\mathcal{D}_{\alpha e}$ if the projectum of $\alpha$ is $\alpha^{*}=\omega$ or if $\alpha$ is an infinite regular cardinal. Finally using this work on $\alpha$-semicomputability and $\alpha$-Kalimullin pairs we conclude that every nontrivial total $\alpha$-enumeration degree is a join of a maximal $\alpha$-Kalimullin pair if $\alpha$ is an infinite regular cardinal.

1 $\alpha$-Computability Theory

$\alpha$-Computability Theory is the study of the definability theory over Gödel’s $L_{\alpha}$ where $\alpha$ is an admissible ordinal. One can think of equivalent definitions on Turing machines with a transfinite tape and time [5] [6] [7] [8] or on generalized register machines [9]. Recommended references for this section are [12], [2], [10] and [3].

Classical Computability Theory is $\alpha$-Computability Theory where $\alpha=\omega$.

1.1 Gödel’s Constructible Universe

Definition 1.1**.**

(Gödel’s Constructible Universe)

Define Gödel’s constructible universe as $L:=\bigcup_{\beta\in\mathrm{Ord}}L_{\beta}$ where $\gamma,\delta\in\mathrm{Ord}$, $\delta$ is a limit ordinal and:

$L_{0}:=\emptyset$,

$L_{\gamma+1}:=\mathrm{Def}(L_{\gamma}):=\{x|x\subseteq L_{\gamma}$ and $x$ is first-order definable over $L_{\gamma}\}$,

$L_{\delta}=\bigcup_{\gamma<\delta}L_{\gamma}$.

1.2 Admissibility

Definition 1.2**.**

(Admissible ordinal[2])

An ordinal $\alpha$ is $\Sigma_{1}$ admissible (admissible for short) iff $\alpha$ is a limit ordinal and $L_{\alpha}$ satisfies $\Sigma_{1}$-collection: $\forall\phi(x,y)\in\Sigma_{1}(L_{\alpha}).L_{\alpha}\models\forall u[\forall x\in u\exists y.\phi(x,y)\implies\exists z\forall x\in u\exists y\in z.\phi(x,y)]$ where $L_{\alpha}$ is the $\alpha$-th level of the Gödel’s Constructible Hierarchy (1.1).

Example 1.3**.**

(Examples of admissible ordinals [2] [17])

• •

$\omega^{CK}_{1}$ - Church-Kleene $\omega_{1}$, the first non-computable ordinal

• •

every stable ordinal $\alpha$ (i.e. $L_{\alpha}\prec_{\Sigma_{1}}L$), e.g. $\delta^{1}_{2}$ - the least ordinal which is not an order type of a $\Delta^{1}_{2}$ subset of $\mathbb{N}$, \nth1 stable ordinal

• •

every infinite cardinal in a transitive model of $\mathrm{ZF}$

1.3 Basic concepts

Definition 1.4**.**

A set $K\subseteq\alpha$ is $\alpha$-finite iff $K\in L_{\alpha}$.

Definition 1.5**.**

($\alpha$-computability and computable enumerability)

• •

A function $f:\alpha\to\alpha$ is $\alpha$-computable iff $f$ is $\Sigma_{1}(L_{\alpha})$ definable.

• •

A set $A\subseteq\alpha$ is $\alpha$-computably enumerable ($\alpha$-c.e.) iff $A\in\Sigma_{1}(L_{\alpha})$.

• •

A set $A\subseteq\alpha$ is $\alpha$-computable iff $A\in\Delta_{1}(L_{\alpha})$ iff $A\in\Sigma_{1}(L_{\alpha})$ and $\alpha-A\in\Sigma_{1}(L_{\alpha})$.

Proposition 1.6**.**

[2] There exists a $\Sigma_{1}(L_{\alpha})$-definable bijection $b:\alpha\to L_{\alpha}$. ∎

Let $K_{\gamma}$ denote an $\alpha$-finite set $b(\gamma)$. The next proposition establishes that we can also index pairs and other finite vectors from $\alpha^{n}$ by an index in $\alpha$.

Proposition 1.7**.**

[10] For every $n$, there is a $\Sigma_{1}$-definable bijection $p_{n}$:$\alpha\to\alpha\times\alpha\times...\times\alpha$ (n-fold product). ∎

Similarly, we can index $\alpha$-c.e., $\alpha$-computable sets by an index in $\alpha$. Let $W_{e}$ denote an $\alpha$-c.e. set with an index $e<\alpha$.

Proposition 1.8**.**

($\alpha$-finite union of $\alpha$-finite sets111From [12] p162.)

$\alpha$-finite union of $\alpha$-finite sets is $\alpha$-finite, i.e. if $K\in L_{\gamma}$, then $\bigcup_{\gamma\in K}K_{\gamma}\in L_{\alpha}$. ∎

1.4 Enumeration reducibility

The generalization of the enumeration reducibility corresponds to two different notions - weak $\alpha$-enumeration reducibility and $\alpha$-enumeration reducibility.

Definition 1.9**.**

(Weak $\alpha$-enumeration reducibility)

$A$ is weakly $\alpha$-enumeration reducible to $B$ denoted as $A\leq_{w\alpha e}B$ iff $\exists\Phi\in\Sigma_{1}(L_{\alpha})$ st $\Phi(B)=\{x<\alpha:\exists\delta<\alpha[\langle x,\delta\rangle\in\Phi\land K_{\delta}\subseteq B]\}$. The set $\Phi$ is called a weak $\alpha$-enumeration operator.

Definition 1.10**.**

($\alpha$-enumeration reducibility)

$A$ is $\alpha$-enumeration reducible to $B$ denoted as $A\leq_{\alpha e}B$ iff $\exists W\in\Sigma_{1}(L_{\alpha})$ st $\forall\gamma<\alpha[K_{\gamma}\subseteq A\iff\exists\delta<\alpha[\langle\gamma,\delta\rangle\in W\land K_{\delta}\subseteq B]]$.

Denote the fact that $A$ reduces to $B$ via $W$ as $A=W(B)$.

Fact 1.11**.**

(Transitivity)

The $\alpha$-enumeration reducibility $\leq_{\alpha e}$ is transitive. But in general the weak $\alpha$-enumeration reducibility is not transitive.

Lemma 1.12**.**

$A\leq_{\alpha e}B\oplus C\land B\in\Sigma_{1}(L_{\alpha})\implies A\leq_{\alpha e}C$ ∎

Fact 1.13**.**

If $A\leq_{w\alpha e}B$ and $B\leq_{\alpha e}C$, then $A\leq_{w\alpha e}C$.

1.5 Properties of $\alpha$-enumeration operator

Fact 1.14**.**

If $A\subseteq\alpha$, then $\Phi_{e}(A)\leq_{w\alpha e}A$.

Fact 1.15**.**

(Monotonicity)

$\forall e<\alpha\forall A,B\subseteq\alpha[A\subseteq B\implies\Phi_{e}(A)\subseteq\Phi_{e}(B)]$. ∎

Proposition 1.16**.**

(Witness property)

If $x\in\Phi_{e}(A)$, then $\exists K\subseteq A[K\in L_{\alpha}\land x\in\Phi_{e}(K)]$.

Proof.

Note $\Phi_{e}(A):=\bigcup\{K_{\gamma}:\exists\delta<\alpha[\langle\gamma,\delta\rangle\in W_{e}\land K_{\delta}\subseteq A\}$. Thus if $x\in\Phi_{e}(A)$, then $\exists\gamma<\alpha$ st $x\in K_{\gamma}$ and so $\exists\delta<\alpha[\langle\gamma,\delta\rangle\in W_{e}\land K_{\delta}\subseteq A]$. Taking $K$ to be $K_{\delta}$ concludes the proof. ∎

1.6 Totality

Definition 1.17**.**

222From [2] p8.

The computable join of sets $A,B\subseteq\alpha$ denoted $A\oplus B$ is defined to be

$A\oplus B:=\{2a:a\in A\}\cup\{2b+1:b\in B\}$.

The computable join satisfies the usual properties of the case $\alpha=\omega$.

The generalization of the Turing reducibility corresponds to two different notions - weak $\alpha$ reducibility and $\alpha$ reducibility.

Definition 1.18**.**

(Total reducibilities)

• •

$A$ is $\alpha$-reducible to $B$ denoted as $A\leq_{\alpha}B$ iff $A\oplus\overline{A}\leq_{\alpha e}B\oplus\overline{B}$.

• •

$A$ is weakly $\alpha$-reducible to $B$ denoted as $A\leq_{w\alpha}B$ iff $A\oplus\overline{A}\leq_{w\alpha e}B\oplus\overline{B}$.

Definition 1.19**.**

(Total set)

A subset $A\subseteq\alpha$ is total iff $A\leq_{\alpha e}\overline{A}$ iff $A\equiv_{\alpha e}A\oplus\overline{A}$.

1.7 $\Sigma_{1}$-projectum

Definition 1.20**.**

(Projectum333Definition 1.19 in [2].)

The $\Sigma_{1}$ projectum (projectum for short) of $\alpha$ is

$\alpha^{*}:=\mathrm{min}\{\gamma\leq\alpha:\exists A\subseteq\gamma[A\in\Sigma_{1}(L_{\alpha})\land A\not\in L_{\alpha}]\}$.

Proposition 1.21**.**

444Theorem 1.20 in [2].

The following ordinals are equal:

i) $\alpha^{*}:=\mathrm{min}\{\gamma<\alpha:\exists A\subseteq\gamma[A\in\Sigma_{1}(L_{\alpha})\land A\not\in L_{\alpha}]\}$

ii) $\mathrm{min}\{\gamma\leq\alpha:\exists$ partial surjection $p_{1}:\gamma\rightharpoonup\alpha\in\Sigma_{1}(L_{\alpha})\}$

iii) $\mathrm{min}\{\gamma\leq\alpha:\exists$ total injection $i:\alpha\rightarrowtail\gamma\in\Sigma_{1}(L_{\alpha})\}$. ∎

Proposition 1.22**.**

(Indexing $\alpha$-c.e. sets with a projectum)

We can index all $\alpha$-c.e. sets just with indices from $\alpha^{*}$. ∎

1.8 Degree Theory

Definition 1.23**.**

(Degrees)

• •

$\mathcal{D}_{\alpha}:=\mathcal{P}\left({\alpha}\right)/\equiv_{\alpha}$ is a set of $\alpha$-degrees.

• •

$\mathcal{D}_{\alpha e}:=\mathcal{P}\left({\alpha}\right)/\equiv_{\alpha e}$ is a set of $\alpha$-enumeration degrees.

Induce $\leq$ on $\mathcal{D}_{\alpha}$ and $\mathcal{D}_{\alpha e}$ by $\leq_{\alpha}$ and $\leq_{\alpha e}$ respectively.

Fact 1.24**.**

(Embedding of the total degrees)

$\langle\mathcal{D}_{\alpha},\leq\rangle$ embeds into $\langle\mathcal{D}_{\alpha e},\leq\rangle$ via $\iota:\mathcal{D}_{\alpha}\hookrightarrow\mathcal{D}_{\alpha e}$, $A\mapsto A\oplus\overline{A}$.

Definition 1.25**.**

(Total degrees)

Let $\iota:\mathcal{D}_{\alpha}\hookrightarrow\mathcal{D}_{\alpha e}$ be the embedding from above. The total $\alpha$-enumeration degrees $\mathcal{TOT}_{\alpha e}$ are the image of $\iota$, i.e. $\mathcal{TOT}_{\alpha e}:=\iota[\mathcal{D}_{\alpha}]$.

Proposition 1.26**.**

(Unboundedness of $\alpha$-enumeration degrees) For every set $A\subseteq\alpha$, there is a set $B\subseteq\alpha$ st $A<_{\alpha e}B$. ∎

1.9 Regularity

Regularity and quasiregularity

Definition 1.27**.**

(Regularity and quasiregularity)

• •

A subset $A\subseteq\alpha$ is $\alpha$-regular iff $\forall\gamma<\alpha.A\cap\gamma\in L_{\alpha}$.

• •

A subset $A\subseteq\alpha$ is $\alpha$-quasiregular iff $\forall\gamma<\mathrm{sup}(A).A\cap\gamma\in L_{\alpha}$.

If clear from the context, we just say regular and quasiregular instead of $\alpha$-regular and $\alpha$-quasiregular respectively.

Fact 1.28**.**

i) $\forall A\subseteq\alpha[A$ regular $\iff\overline{A}$ regular$]$,

ii) $\forall A,B\subseteq\alpha[A$ regular $\land B$ regular $\implies A\oplus B$ regular $]$,

iii) $\forall A,B\subseteq\alpha[A$ regular $\land B$ regular $\implies A\cup B$ regular $]$.

Theorem 1.29**.**

(Sack’s Theorem on regular set existence555Sacks [12], theorem 4.2)

Let $A$ be $\alpha$-computably enumerable. Then there exists a regular, $\alpha$-computably enumerable $B$ of the same $\alpha$-degree as $A$.

Theorem 1.30**.**

(Shore’s Splitting Theorem [15])

Let $B$ be $\alpha$-computably enumerable and regular. Then there exists $\alpha$-computably enumerable $A_{0},A_{1}$ st $B=A_{0}\sqcup A_{1}$ and $B\not\leq_{\alpha}A_{i}(i\in\{0,1\})$. ∎

Megaregularity

Megaregularity of a set $A$ measures the amount of the admissibility of a structure structure $\langle L_{\alpha},A\rangle$, i.e. a structure extended by a predicate with an access to $A$.

Note 1.31**.**

(Formula with a positive/negative parameter)

• •

Let $B$ denote a set, $B^{+}$ its enumeration, $B^{-}$ the enumeration of its complement $\overline{B}$.

• •

Denote by $\Sigma_{1}(L_{\alpha},B)$ the class of $\Sigma_{1}$ formulas with a parameter $B$ or in $L_{\alpha}$.

• •

A $\Sigma_{1}(L_{\alpha},B)$ formula $\phi(\overline{x},B)$ is $\Sigma_{1}(L_{\alpha},B^{+})$ iff $B$ occurs in $\phi(\overline{x},B)$ only positively, i.e. there is no negation up the formula tree above the literal $x\in B$.

• •

Similarly, a $\Sigma_{1}(L_{\alpha},B)$ formula $\phi(\overline{x},B)$ is $\Sigma_{1}(L_{\alpha},B^{-})$ iff $B$ occurs in $\phi(\overline{x},B)$ only negatively.

Definition 1.32**.**

(Megaregularity)

Let $\mathcal{B}\in\{B,B^{-},B^{+}\}$ and add $B$ as a predicate to the language for the structure $\langle L_{\alpha},\mathcal{B}\rangle$.

• •

Then $\mathcal{B}$ is $\alpha$-megaregular iff $\alpha$ is $\Sigma_{1}(L_{\alpha},\mathcal{B})$ admissible iff the structure $\langle L_{\alpha},\mathcal{B}\rangle$ is admissible, i.e. every $\Sigma_{1}(L_{\alpha},\mathcal{B})$ definable function satisfies the replacement axiom: $\forall f\in\Sigma_{1}(L_{\alpha},\mathcal{B})\forall K\in L_{\alpha}.f[K]\in L_{\alpha}$.

• •

$B$ is positively $\alpha$-megaregular iff $B^{+}$ is $\alpha$-megaregular.

• •

$B$ is negatively $\alpha$-megaregular iff $B^{-}$ is $\alpha$-megaregular.

If clear from the context, we just say megaregular instead of $\alpha$-megaregular.

Remark 1.33**.**

(Hyperregularity and megaregularity)

A person familiar with the notion of hyperregularity shall note that a set is megaregular iff it is regular and hyperregular.

Proposition 1.34**.**

Let $\mathcal{B}\in\{B,B^{-},B^{+}\}$ be megaregular and let $A\subseteq\alpha$. Then: $A\in L_{\alpha}$ iff $A\in\Delta_{1}(L_{\alpha},\mathcal{B})$ and $A$ is bounded by some $\beta<\alpha$.

Proof.

$\implies$ direction is clear. For the other direction, assume that $A\in\Delta_{1}(L_{\alpha},\mathcal{B})$ and $A\subseteq\beta<\alpha$ for some $\beta$. WLOG let $A\not=\emptyset$ and let $a\in A$. Define a function $f:\alpha\to\alpha$ by $f(x)=y:\iff x\not\in\beta\land y=a\lor x\in\beta\land[x\in A\land x=y\lor x\not\in A\land y=a]$. Since $A\in\Delta_{1}(L_{\alpha},\mathcal{B})$, the function $f$ is $\Sigma_{1}(L_{\alpha},\mathcal{B})$ definable. By the megaregularity of $\mathcal{B}$, we have that $A=f[\beta]\in L_{\alpha}$ as required. ∎

Corollary 1.35**.**

(Megaregularity closure and degree invariance)

i) If $A\leq_{\alpha e}B$ and $B^{+}$ megaregular, then $A^{+}$ megaregular.

ii) If $A\equiv_{\alpha e}B$, then $[A^{+}$ megaregular iff $B^{+}$ megaregular $]$.

iii) If $A\leq_{\alpha}B$ and $B$ megaregular, then $A$ megaregular.

iv) If $A\equiv_{\alpha}B$, then $[A$ megaregular iff $B$ megaregular $]$.

v) If $A\in\Sigma_{1}(L_{\alpha})$, then $A^{+}$ is megaregular.

vi) If $A\in\Delta_{1}(L_{\alpha})$, then $A$ is megaregular. ∎

Regularity and definability

Proposition 1.36**.**

($\Sigma_{1}$ definability and $\alpha$-enumeration reducibilities correspondence)

We have the following implication diagram:

{A\in\Sigma_{1}(L_{\alpha},B^{+})}$${A\leq_{w\alpha e}B}$${A\leq_{\alpha e}B}if $\scriptstyle B$ regularif $\scriptstyle B^{+}$ megaregularalwaysalways

∎

Notions of regularity by strength

Remark 1.37**.**

We have the following strict separation of the notions where $\alpha$-finiteness is the strongest condition and quasiregularity is the weakest:

$\alpha$-finite $\implies$ $\alpha$-computable $\implies$ megaregular $\implies$ regular $\implies$ quasiregular

1.10 Useful lemmas

Lemma 1.38**.**

666From lemma 6 in [11] on p66.

$A_{0}\cap A_{1}=\emptyset,A_{i\in\{0,1\}}\in\Sigma_{1}(L_{\alpha},A_{0}\sqcup A_{1})\implies A_{0}\sqcup A_{1}\equiv_{\alpha}A_{0}\oplus A_{1}$.

Proof.

$A_{0}\sqcup A_{1}\leq_{\alpha}A_{0}\oplus A_{1}$ trivially. Let $i\in\{0,1\}$. For $A_{0}\oplus A_{1}\leq_{\alpha}A_{0}\sqcup A_{1}$: $x\in A_{i}$ recognizable by $A_{i}\in\Sigma_{1}(L_{\alpha},A_{0}\sqcup A_{1})$. Also $x\not\in A_{i}$ is recognizable since $x\not\in A_{i}\iff x\in A_{1-i}\lor x\not\in A_{0}\sqcup A_{1}$ by disjointness and both $x\in A_{1-i},x\not\in A_{0}\sqcup A_{1}$ are recognizable from $A_{0}\sqcup A_{1}$. Hence $A_{0}\sqcup A_{1}\equiv_{\alpha}A_{0}\oplus A_{1}$. ∎

The lemma implies that if $A_{0}$, $A_{1}$ are disjoint $\alpha$-incomparable $\alpha$-computably enumerable sets, then $A_{0}\sqcup A_{1}\equiv_{\alpha}A_{0}\oplus A_{1}$ (proposition 3.3 in [16]).

Lemma 1.39**.**

There exists an $\alpha$-computable function $g:\alpha\times\alpha\times\alpha\to\alpha$ st $D_{\eta}:=\{x|g(\eta,x,1)=1\}\in L_{\alpha}$, $E_{\eta}:=\{x|g(\eta,x,2)=1\}\in L_{\alpha}$ and for every pair $(\hat{D},\hat{E})$ of $\alpha$-finite subsets of $\alpha$ there is an index $\eta<\alpha$ st $D_{\eta}=\hat{D}$ and $E_{\eta}=\hat{E}$.

Therefore we can $\alpha$-effectively number the pairs of the $\alpha$-finite subsets of $\alpha$ by the indices of $\alpha$.

Proof.

Note that there are $\alpha$-computable bijections $j:\alpha\to L_{\alpha}$ and $f:\alpha\to\alpha\times\alpha$. Let $\pi_{1}$ and $\pi_{2}$ be the projections. Define $g(\eta,x,k):=[x\in j\circ\pi_{k}\circ f(\eta)]$. Then $g$ is the required $\alpha$-computable function. ∎

Lemma 1.40**.**

Let $i,j,k:\alpha\times\alpha\to\alpha$ be any $\alpha$-computable numberings of $\alpha$-finite subsets of $\alpha$. Then:

i) There is an $\alpha$-computable function $u:\alpha\to\alpha$ st

$\forall\gamma<\alpha.\bigcup_{x\in j(\gamma)}i(x)=k(u(\gamma))$.

ii) There is an $\alpha$-computable function $v:\alpha\times\alpha\to\alpha$ st

$\forall\gamma,\delta<\alpha.k(v(\gamma,\delta))=i(\gamma)\oplus j(\delta)$.

iii) There exist $\alpha$-computable functions $i_{\pi_{1}},i_{\pi_{2}}:\alpha\to\alpha$ st

$\forall l\in\{1,2\}\forall\gamma<\alpha.k(i_{\pi_{l}}(\gamma))=\{x_{l}:\langle x_{1},x_{2}\rangle\in i(\gamma)\}$.

iv) There exists an $\alpha$-computable function $i_{p_{2}}:\alpha\to\alpha$ st

$\forall\gamma<\alpha.k(i_{p_{2}}(\gamma))=i(\gamma)\times j(\gamma)$.

v) There exists an $\alpha$-computable function $w:\alpha\times\alpha\to\alpha$ st if $\gamma,\delta<\alpha$, then $k(w(\gamma,\delta))=\{\langle x,y\rangle:x\in j(\delta)\land y\in j(\gamma)\land y\in i(x)\}$.

vi) There exists a function $t_{i,j}:\alpha\to\alpha\in\Sigma_{1}(L_{\alpha})$ st $\forall\gamma<\alpha.i(\gamma)=j(t_{i,j}(\gamma))$.

vii) Let $K(\gamma):=\bigcup_{x\in j(\gamma)}i(x)$. Then there exists a function $s_{i,j}:\alpha\to\alpha\in\Sigma_{1}(L_{\alpha})$ st $\forall\gamma<\alpha.s_{i,j}(\gamma)=$ $\begin{cases}0&K(\gamma)=\emptyset\\ \mathtt{sup}(K(\gamma))&K(\gamma)\not=\emptyset\end{cases}$. ∎

2 Semicomputability

The goal of this section is to lift the necessary results of Jockusch [4] on semicomputable sets from the level $\omega$ to a level $\alpha$.

Definition 2.1**.**

A set $A\subseteq\alpha$ is $\alpha$-semicomputable iff there exists a total $\alpha$-computable function $s_{A}:\alpha\times\alpha\to\alpha$ called a selector function satisfying:

i)$\forall x,y\in\alpha.s_{A}(x,y)\in\{x,y\}$,

ii)$\forall x,y\in\alpha[\{x,y\}\cap A\not=\emptyset\implies s_{A}(x,y)\in A]$.

Denote by $\mathrm{sc}(L_{\alpha})$ the class of $\alpha$-semicomputable sets.

Fact 2.2**.**

(Semicomputability closure)

i) $A\in\mathrm{sc}(L_{\alpha})\iff\overline{A}\in\mathrm{sc}(L_{\alpha})$,

ii) $A\oplus B\in\mathrm{sc}(L_{\alpha})\implies A\in\mathrm{sc}(L_{\alpha})\land B\in\mathrm{sc}(L_{\alpha})$.

Definition 2.3**.**

(Index set)

An index set for a set $A\subseteq\alpha$ denoted as $A_{I}$ is a set of all indices of $\alpha$-finite subsets of $A$, i.e. $A_{I}:=\{\gamma<\alpha:K_{\gamma}\subseteq A\}$.

Proposition 2.4**.**

(Semicomputability of an index set)

For every set $A\subseteq\alpha$, its index set $A_{I}$ is $\alpha$-semicomputable.

Proof.

Define the selector function of $A_{I}$ as $s_{A_{I}}:=\{\langle\gamma,\delta,\rangle:K_{\gamma}\subseteq K_{\delta}\}$. The function $s_{A_{I}}$ is $\alpha$-computable as required. ∎

Definition 2.5**.**

(Binary ordering)

Define $<_{b}\subseteq\mathcal{P}(\alpha)\times\mathcal{P}(\alpha)$ and $\leq_{b}\subseteq\mathcal{P}(\alpha)\times\mathcal{P}(\alpha)$ to be numerical orderings on the binary representation of the compared sets:

• •

$A<_{b}B:\iff\exists\beta\in\alpha[\beta\not\in A\land\beta\in B\land A\cap\beta=B\cap\beta]$,

• •

$A\leq_{b}B:\iff A<_{b}B\lor A=B$.

Remark 2.6**.**

The restrictions of the orderings $<_{b}$ and $\leq_{b}$ to $\alpha$-finite sets are first-order definable and $\alpha$-computable since an $\alpha$-finite set is bounded.

Proposition 2.7**.**

(Properties of binary ordering)

Let $\lhd\in\{<,\leq\}$, then:

i) $<_{b}$ is a strict total order,

ii) $\leq_{b}$ is a total order,

iii) $([0,1],\lhd_{{}^{*}\mathbb{R}})\cong(\mathcal{P}(\alpha),\lhd_{b}))$ where ${}^{*}\mathbb{R}$ is an appropriate model of the hyperreal numbers,

iv) $A\lhd_{b}B\iff\overline{B}\lhd_{b}\overline{A}$.

Proof.

i), ii), iii) are trivial. To prove iv), use iii) and consider $\mathcal{P}(\alpha)$ as the interval $[0,1]$ from the field of hyperreals, where $0:=\emptyset$ and $1:=\alpha$. Then: $\overline{B}\lhd_{b}\overline{A}\iff 1-B\lhd_{b}1-A\iff-B\lhd_{b}-A\iff A\lhd_{b}B$. ∎

Fact 2.8**.**

(Binary and subset ordering)

i) $A\subset B\implies A<_{b}B$,

ii) $A\subseteq B\implies A\leq_{b}B$,

iii) $A=B\iff A\equiv_{b}B$.

Note 2.9**.**

If $A\leq_{b}C$ and $B\leq_{b}C$, is it true that $A\cup B\leq_{b}C$?

No. Consider $A=011...,B=100...,C=110...$. Then $A\cup B=111...$. Thus $A\leq_{b}C$ and $B\leq_{b}C$, but $\neg A\cup B\leq_{b}C$.

Definition 2.10**.**

Given a set $A$ define $L_{A}:=\{x\in\alpha:K_{x}\leq_{b}A\},R_{A}:=\overline{L_{A}}$.

Remark 2.11**.**

If $A\not\in L_{\alpha}$, then:

• •

$L_{A}=\{x<\alpha:K_{x}<_{b}A\}$ are $\alpha$-finite sets left of $A$,

• •

$R_{A}=\{x<\alpha:A<_{b}K_{x}\}$ are $\alpha$-finite sets right of $A$.

Fact 2.12**.**

(Properties of left/right $\alpha$-finite sets)

Let $A\subseteq\alpha$ and $\beta,\gamma,\delta<\alpha$. Then:

i) $K\in L_{\alpha}\land K_{\delta}=\bigcup_{\gamma\in K}K_{\gamma}\land\delta\in L_{A}\implies K\subseteq L_{A}$,

ii) $\beta\in L_{A}\land\gamma\in R_{A}\land K_{\beta}\cap\delta=K_{\gamma}\cap\delta\implies K_{\beta}\cap\delta\subseteq A$.

Lemma 2.13**.**

For any $A\subseteq\alpha$ the sets $L_{A},R_{A}$ are $\alpha$-semicomputable.

Proof.

$L_{A}$ is $\alpha$-semicomputable since it has an $\alpha$-computable selector function $s:=\{(x,y):K_{x}\leq_{b}K_{y}\}\cup\{(y,x):K_{x}>_{b}K_{y}\}$ by remark 2.6. ∎

Lemma 2.14**.**

Let $A\subseteq\alpha$ be a quasiregular set, then $A\equiv_{\alpha}L_{A}\equiv_{\alpha}R_{A}$.

Proof.

If $A\in\Delta_{1}(L_{\alpha})$, then trivially $A\equiv_{\alpha}L_{A}\equiv_{\alpha}R_{A}$. Hence WLOG assume that $A\not\in L_{\alpha}$ and use 2.11. Also WLOG $A\not\in\Delta_{1}(L_{\alpha})$ and so in the proof implicitly use the property: $\forall x\in A\exists y,z[x<y<\alpha\land x<z<\alpha\land y\not\in A\land z\in A]$.

Note that $\bigcup_{x\in K_{\gamma}}K_{x}\in L_{\alpha}$. Hence for any $\gamma<\alpha$ we have: $K_{\gamma}\subseteq L_{A}\iff\exists\beta<\alpha[K_{\beta}<_{b}A\land\forall x\in K_{\gamma}.K_{x}<_{b}K_{\beta}]$. Thus $L_{A}\leq_{\alpha e}A$ via $W:=\{\langle\gamma,\delta\rangle:\exists\beta<\alpha[K_{\delta}=\{\beta\}\land\forall x\in K_{\gamma}.K_{x}<_{b}K_{\beta}]\}\in\Sigma_{1}(L_{\alpha})$. By symmetry $R_{A}\leq_{\alpha e}A$. Hence $L_{A}\oplus R_{A}\leq_{\alpha e}A$.

Let $\widehat{A}$ denote $A$ or $\overline{A}$. Then $K_{\gamma}\subseteq\widehat{A}\iff\exists\beta_{L},\beta_{R}<\alpha[\forall x\in K_{\gamma}\forall y\leq x[y\in K_{\beta_{L}}\iff y\in K_{\beta_{R}}]\land K_{\gamma}\subseteq\widehat{K_{\beta_{L}}}\land\beta_{L}\in L_{A}\land\beta_{R}\in R_{A}]$ for any $\gamma<\alpha$ using the quasiregularity of $A$ and 2.12ii. Hence define $W:=\{\langle\gamma,\delta\rangle:\exists\beta_{L},\beta_{R}<\alpha[\forall x\in K_{\gamma}\forall y\leq x[y\in K_{\beta_{L}}\iff y\in K_{\beta_{R}}]\land K_{\gamma}\subseteq\widehat{K_{\beta_{L}}}\land K_{\delta}=\{\beta_{L}\}\oplus\{\beta_{R}\}]\}$. Note that $W\in\Sigma_{1}(L_{\alpha})$ and so $\widehat{A}\leq_{\alpha e}L_{A}\oplus R_{A}$ via $W$. Hence $A\oplus\overline{A}\leq_{\alpha e}L_{A}\oplus R_{A}$.

Therefore $A\oplus\overline{A}\equiv_{\alpha e}L_{A}\oplus R_{A}=L_{A}\oplus\overline{L_{A}}=R_{A}\oplus\overline{R_{A}}$ and so $A\equiv_{\alpha}L_{A}\equiv_{\alpha}R_{A}$ as required. ∎

Lemma 2.15**.**

777Adapted from Lemma 5.5 in [4] for $\alpha=\omega$

$B\in\Sigma_{1}(L_{\alpha})\land B>_{\alpha}0\implies$

$\exists A[A$ regular $\land A\equiv_{\alpha}B\land L_{A}\not\in\Pi_{1}(L_{\alpha})\land L_{A}\not\in\Sigma_{1}(L_{\alpha})]$.

Proof.

By 1.29 every $\Sigma_{1}(L_{\alpha})$ set is $\alpha$-equivalent to some regular set, so WLOG assume that $B$ is regular. By Shore’s Splitting Theorem 1.30, $\exists C_{0},D_{0}\in\Sigma_{1}(L_{\alpha})[B=C_{0}\sqcup D_{0}\land C_{0}|_{\alpha}D_{0}$ (incomparable wrt $\alpha$-reducibility) $]$. Using 1.29 again, let $C,D$ be $\alpha$-c.e. regular sets st $C\equiv_{\alpha}C_{0}$ and $D\equiv_{\alpha}D_{0}$. Define $A:=C\oplus\overline{D}$.

Note $A=C\oplus\overline{D}\equiv_{\alpha}C_{0}\oplus\overline{D_{0}}$. Hence $A\equiv_{\alpha}B$ by lemma 1.38 as required.

As $D$ is regular, so $\overline{D}$ is regular. As $C$ and $\overline{D}$ are regular, so $A=C\oplus\overline{D}$ is regular as required.

Next we prove $L_{A}\not\in\Pi_{1}(L_{\alpha})\land L_{A}\not\in\Sigma_{1}(L_{\alpha})$. For suppose to the contrary that $\neg(L_{A}\not\in\Pi_{1}(L_{\alpha})\land L_{A}\not\in\Sigma_{1}(L_{\alpha}))$. Then $L_{A}\in\Sigma_{1}(L_{\alpha})\lor L_{A}\in\Pi_{1}(L_{\alpha})$.

• •

Case $L_{A}\in\Sigma_{1}(L_{\alpha})$:

Note that $\overline{D}\leq_{\alpha e}C\oplus\overline{C}$ via

$W:=\{\langle\gamma,\delta\rangle:\beta=\mathrm{min}\{\epsilon<\alpha:K_{\gamma}\cap\epsilon=K_{\gamma}\}\land\exists\zeta\in L_{A}\forall x<\beta[$

$(2x\in K_{\delta}\iff 2x+1\not\in K_{\delta}\iff 2x\in K_{\zeta})\land$

$(x\in K_{\gamma}\implies 2x+1\in K_{\zeta})\land$

$(2x+1\not\in K_{\zeta}\implies x\in D)]\}$. The set $W$ is $\alpha$-c.e. since $L_{A}$ and $D$ are $\alpha$-c.e. The condition $2x\in K_{\delta}\iff 2x+1\not\in K_{\delta}$ ensures that $K_{\delta}$ contains the initial segment $C\cap\beta$ of $C$. The conditions $2x\in K_{\delta}\iff 2x\in K_{\zeta}$ and $2x+1\not\in K_{\zeta}\implies x\in D$ ensure that $K_{\zeta}$ contains the initial segment $(C\cap\beta)\oplus(\overline{D}\cap\beta)$ of $C\oplus\overline{D}$. Finally, the condition $x\in K_{\gamma}\implies 2x+1\in K_{\zeta}$ verifies that $K_{\gamma}$ is a subset of $\overline{D}$, or more precisely a subset of its initial segment $\overline{D}\cap\beta$.

As $D$ is $\alpha$-c.e., so this gives us $D\leq_{\alpha}C$ which is a contradiction to the case $L_{A}\in\Sigma_{1}(L_{\alpha})$.

• •

Case $L_{A}\in\Pi_{1}(L_{\alpha})$:

Note that $R_{A}=\overline{L_{A}}\in\Sigma_{1}(L_{\alpha})$. Hence similarly $C\leq_{\alpha}D$ using the fact that $R_{A}$ and $C$ are both $\alpha$-c.e. by applying a symmetric argument to the one above. This is a contradiction to the case $L_{A}\in\Pi_{1}(L_{\alpha})$.

So by the two cases $L_{A}\not\in\Pi_{1}(L_{\alpha})\land L_{A}\not\in\Sigma_{1}(L_{\alpha})$.

Therefore given $B>_{\alpha}0$, there is a regular set $A$ st $A\equiv_{\alpha}B\land L_{A}\not\in\Pi_{1}(L_{\alpha})\land L_{A}\not\in\Sigma_{1}(L_{\alpha})$ as required. ∎

Theorem 2.16**.**

Let $B\subseteq\alpha$ be quasiregular and $B>_{\alpha}0$. Then there exists an $\alpha$-semicomputable set $A$ st $A\equiv_{\alpha}B\land A\not\in\Sigma_{1}(L_{\alpha})\land A\not\in\Pi_{1}(L_{\alpha})$.

Proof.

If $\mathrm{deg}_{\alpha}({B})$ is $\alpha$-c.e. degree, then WLOG let $B\in\Sigma_{1}(L_{\alpha})$. Then by lemma 2.15 there is $C$ st $C$ is quasiregular, $B\equiv_{\alpha}C\land L_{C}\not\in\Sigma_{1}(L_{\alpha})\land L_{C}\not\in\Pi_{1}(L_{\alpha})$. By lemma 2.14 and quasiregularity of $C$ we have that $C\equiv_{\alpha}L_{C}$ and so $B\equiv_{\alpha}L_{C}$. Hence $A:=L_{C}$ is the required $\alpha$-semicomputable set by lemma 2.13.

Otherwise $\mathrm{deg}_{\alpha}({B})$ is not an $\alpha$-c.e. degree and so $\forall C\in\mathrm{deg}_{\alpha}({B})[C\not\in\Sigma_{1}(L_{\alpha})\land C\not\in\Pi_{1}(L_{\alpha})]$. Note that $A:=L_{B}\equiv_{\alpha}B$ by the quasiregularity of $B$ and by lemma 2.14 and so $A\not\in\Sigma_{1}(L_{\alpha})\land A\not\in\Pi_{1}(L_{\alpha})$. Finally, $A$ is $\alpha$-semicomputable by lemma 2.13 as required. ∎

3 Kalimullin pair

The goal of this section is to generalize the results of Kalimullin [14] on the definability of a Kalimullin pair to a level $\alpha$.

3.1 Introduction and basic properties

Definition 3.1**.**

Sets $A,B\subseteq\alpha$ are a $\alpha$-$U$-Kalimullin pair denoted by $\mathcal{K}_{U}(A,B)$ iff $\exists W\leq_{\alpha e}U[A\times B\subseteq W\land\overline{A}\times\overline{B}\subseteq\overline{W}]$. If clear, we omit the prefix $\alpha$ and say $U$-Kalimullin pair (or just $U$-$\mathcal{K}$-pair) and denote it by $\mathcal{K}_{U}(A,B)$. Similarly, if $U\in\Sigma_{1}(L_{\alpha})$, then we say that $A,B$ are a Kalimullin pair (or just $\mathcal{K}$-pair) and denote $\mathcal{K}(A,B)$.

The set $W$ is called a witness to the U-Kalimullin pair.

Proposition 3.2**.**

888Proposition 2.2 in [14] for $\alpha=\omega$.

If $A\leq_{\alpha e}U$, then $\forall B\subseteq\alpha.\mathcal{K}_{U}(A,B)$.

Proof.

Take the witness $W:=A\times\alpha$. ∎

Proposition 3.3**.**

If $A$ is $\alpha$-semicomputable, then $\mathcal{K}(A,\overline{A})$.

Proof.

Define the witness $W\in\Sigma_{1}(L_{\alpha})$ to the Kalimullin pair $\mathcal{K}(A,\overline{A})$ to be $W:=\{(x,y)\in\alpha:s_{A}(x,y)=x\}$ where $s_{A}$ is an $\alpha$-computable selector function for an $\alpha$-semicomputable set $A$. ∎

Definition 3.4**.**

$A,B\subseteq\alpha$ are a trivial Kalimullin pair iff $\mathcal{K}(A,B)$ and $A\in\Sigma_{1}(L_{\alpha})\lor B\in\Sigma_{1}(L_{\alpha})$. If $A,B$ are a not a trivial Kalimullin pair, they form a nontrivial Kalimullin pair, denoted by $\mathcal{K}_{\mathrm{nt}}(A,B)$.

Definition 3.5**.**

(Maximal Kalimullin pair)

A Kalimullin pair $\mathcal{K}(A,B)$ is maximal denoted by $\mathcal{K}_{\mathrm{max}}(A,B)$ iff

$\forall C,D[A\leq_{\alpha e}C\land B\leq_{\alpha e}D\land\mathcal{K}(C,D)\implies A\equiv_{\alpha e}C\land B\equiv_{\alpha e}D]$.

Remark 3.6**.**

Note that in the definition of a maximal Kalimullin pair we use $\alpha$-enumeration reducibility instead of a weak $\alpha$-enumeration reducibility since we want that a maximal Kalimullin pair is definable (given that a Kalimullin pair is definable) in the structure $\langle\mathcal{D}_{\alpha e},\leq\rangle$ where $\leq$ is induced by $\leq_{\alpha e}$.

Proposition 3.7**.**

999From [14] and proposition 1.8 in [1].

Assume $A,B\subseteq\alpha\land A\not\in\Sigma_{1}(L_{\alpha})\land B\not\in\Sigma_{1}(L_{\alpha})\land\mathcal{K}(A,B)$ where the witness of $\mathcal{K}(A,B)$ is $W$. Then:

i) $A=\{a:\exists b[b\not\in B\land(a,b)\in W]\}$.

ii) $B=\{b:\exists a[a\not\in A\land(a,b)\in W]\}$.

Proof.

i):

0. Assume $A,B\subseteq\alpha\land A\not\in\Sigma_{1}(L_{\alpha})\land B\not\in\Sigma_{1}(L_{\alpha})\land\mathcal{K}(A,B)$.

1. Define $A_{2}:=\{a:\exists b[b\not\in B\land(a,b)\in W]\}$.

2. Assume $a\in A$.

3. Assume $a\not\in A_{2}$.

4. $\forall b[\neg(b\not\in B\land(a,b)\in W)]$ by 3.

5. $\forall b[b\in B\lor(a,b)\not\in W]$ by 4.

6. $\forall b[(a,b)\in W\implies b\in B]$ by 5.

7. $A\times B\subseteq W$ by 0.

8. $B=\{b:\exists a\in\alpha.(a,b)\in W\}$ by 7.

9. $B\leq_{\alpha e}W$ by 8.

10. $W\in\Sigma_{1}(L_{\alpha})$ by 0.

11. $B\in\Sigma_{1}(L_{\alpha})$ by 9, 10.

12. $\mathrm{false}$ by 0, 11.

13. $a\in A_{2}$ by 3, 12.

14. $A\subseteq A_{2}$ by 2, 13.

15. Assume $a\in A_{2}$.

16. $\exists b[b\not\in B\land(a,b)\in W]$ by 1, 15.

17. Assume $a\not\in A$.

18. $b\not\in B\land a\not\in A$ by 16, 17.

19. $\overline{A}\times\overline{B}\subseteq\overline{W}$ by 0.

20. $(a,b)\in\overline{W}$ by 18, 19.

21. $\mathrm{false}$ by 16, 20.

22. $a\in A$ by 17, 21.

23. $A_{2}\subseteq A$ by 15, 22.

24. $A=A_{2}$ by 14, 23. QED of i).

The proof of ii) is symmetric. ∎

Corollary 3.8**.**

Assume $A,B\subseteq\alpha\land A\not\in\Sigma_{1}(L_{\alpha})\land B\not\in\Sigma_{1}(L_{\alpha})\land\mathcal{K}(A,B)$. Then:

i) $A\leq_{w\alpha e}\overline{B}$ and $B\leq_{w\alpha e}\overline{A}$,

ii) $A\leq_{\alpha e}\overline{B}$ if $B^{-}$ is megaregular, $B\leq_{\alpha e}\overline{A}$ if $A^{-}$ is megaregular.

Proof.

Follows from 3.7. ∎

Lemma 3.9**.**

(Kalimullin pair distributivity)

Suppose that $\bigwedge_{i\in 2}A_{i}\not=\emptyset$. Then

$\bigwedge_{i\in 2}\mathcal{K}(A_{i},B)\iff\mathcal{K}(\bigoplus_{i\in 2}A_{i},B)\iff\mathcal{K}(\prod_{i\in 2}A_{i},B)$

Proof.

Suppose $\bigwedge_{i\in 2}\mathcal{K}(A_{i},B)$. For any $i\in 2$ let $A_{i}\times B\subseteq U_{i}\in\Sigma_{1}(L_{\alpha})$ and $\overline{A}_{i}\times\overline{B}\subseteq\overline{U}_{i}$.

Define $V:=\{(2a+i,b):(a,b)\in U_{i},i\in 2\}$.

Define $W:=\{((a_{0},a_{1}),b):\forall i\in 2.(2a_{i}+i,b)\in V\}$.

Define $U^{*}_{i}:=\{(a_{i},b):\exists(a_{0},a_{1}).((a_{0},a_{1}),b)\in W\}$.

Then $\bigwedge_{i\in 2}A_{i}\times B\subseteq U_{i}\in\Sigma_{1}(L_{\alpha})\land\overline{A}_{i}\times\overline{B}\subseteq\overline{U}_{i}\implies$

$(\bigoplus_{i\in 2}A_{i})\times B\subseteq V\in\Sigma_{1}(L_{\alpha})\land\overline{\bigoplus_{i\in 2}A_{i}}\times\overline{B}\subseteq\overline{V}\implies$

$(\prod_{i\in 2}A_{i})\times B\subseteq W\in\Sigma_{1}(L_{\alpha})\land\overline{\prod_{i\in 2}A_{i}}\times\overline{B}\subseteq\overline{W}\implies$ (by $\bigwedge_{i\in 2}A_{i}\not=\emptyset$)

$\bigwedge_{i\in 2}A_{i}\times B\subseteq U^{*}_{i}\in\Sigma_{1}(L_{\alpha})\land\overline{A}_{i}\times\overline{B}\subseteq\overline{U}^{*}_{i}$. ∎

Lemma 3.10**.**

$\mathcal{K}_{\mathrm{max}}(A,B)\implies\mathcal{K}_{\mathrm{nt}}(A,B)$

Proof.

(Of lemma 3.10)

1. Assume $\mathcal{K}_{\mathrm{max}}(A,B)$.

2. Assume $B\equiv_{\alpha e}0$.

3. Assume $A\equiv_{\alpha e}0$.

4. $\exists C\subseteq\alpha\land C>_{\alpha e}A$ by unboudedness of $\mathcal{D}_{\alpha e}$ 1.26.

5. $L_{C}\oplus R_{C}\equiv_{\alpha e}C$ by definition 2.10.

6. Assume WLOG $L_{C}>_{\alpha e}A$ by 4, 5.

7. $\mathcal{K}_{\mathrm{max}}(L_{C},R_{C})$ by $L_{C}$ $\alpha$-semicomputable and $R_{C}=\overline{L_{A}}$.

8. $L_{C}\geq_{\alpha e}A\land R_{C}\geq_{\alpha e}B$ by 2, 3.

9. $L_{C}\equiv_{\alpha e}A$ by 1, 8.

10. $\mathrm{false}$ by 6, 9.

11. $A>_{\alpha e}0$ by 3, 10.

12. $K_{A}:=\{x\in\alpha:\Phi^{A}_{x}(x)\downarrow\}$.

13. $\overline{K_{A}}\equiv_{\alpha e}K_{A}\oplus\overline{K}_{A}$ by 12.

14. $A\equiv_{\alpha e}K_{A}$ by 12.

15. $\overline{K_{A}}>_{\alpha e}K_{A}$ by 13, 14.

16. $\overline{K_{A}}>_{\alpha e}A$ by 14, 15.

17. $\mathcal{K}(\overline{K_{A}},B)$ by 2.

18. $\overline{K_{A}}\equiv_{\alpha e}A$ by 1, 17.

19. $\mathrm{false}$ by 16, 18.

20. $B>_{\alpha e}0$ by 2, 19.

21. $\mathcal{K}_{\mathrm{nt}}(A,B)$ by 11, 20. ∎

3.2 Definability of an $\alpha$-Kalimullin pair

For this section let $D_{x},E_{x}$ be a pair of $\alpha$-finite sets indexed by $x<\alpha$ according to lemma 1.39. For any $x<\alpha$ define

[TABLE]

Lemma 3.11**.**

Assume that $x\in Y\iff x\in X\land D_{x}\subseteq A$ where $D_{x}$ is an $\alpha$-finite set with an $\alpha$-computable index $x$. Then $Y\leq_{\alpha e}X\oplus A$.

Proof.

Recall $Y\leq_{\alpha e}X\oplus A\iff\exists W\in\Sigma_{1}(L_{\alpha})\forall\gamma<\alpha[K_{\gamma}\subseteq Y\iff\exists\langle\gamma,\delta\rangle\in W.K_{\delta}\subseteq X\oplus A]$. Note $K_{\gamma}\subseteq Y\iff\forall x\in K_{\gamma}.x\in Y\iff\forall x\in K_{\gamma}[x\in X\land D_{x}\subseteq A]\iff K_{\gamma}\subseteq X\land\bigcup_{x\in K_{\gamma}}D_{x}\subseteq A\iff$ (By lemma 1.40i) $K_{\gamma}\subseteq X\land K_{u(\gamma)}\subseteq A\iff$ (By lemma 1.40ii) $K_{v(\gamma,u(\gamma))}\subseteq X\oplus A$. Hence define $W:=\{\langle\gamma,\delta\rangle<\alpha:\delta=v(\gamma,u(\gamma))\}$. As $u,v\in\Sigma_{1}(L_{\alpha})$, so $W\in\Sigma_{1}(L_{\alpha})$. Moreover, $K_{\gamma}\subseteq Y\iff\exists\langle\gamma,\delta\rangle\in W.K_{\delta}\subseteq X\oplus A$. Therefore $Y\leq_{\alpha e}X\oplus A$. ∎

Lemma 3.12**.**

Assume $M_{s}\in\Sigma_{1}(L_{\alpha})$ and $X_{s}\in L_{\alpha}$. Let $W:=\{\langle a,b\rangle:\exists x\in M_{s}[a\in D_{x}\land b\in E_{x}\land x\in\Phi_{e}((X_{s}\cup(M_{s}\cap V_{x}))\oplus U)]\}$. Assume $U^{+}$ is megaregular. Then $W\leq_{\alpha e}U$.

Proof.

Let $S_{e}:=\Phi_{e}((X_{s}\cup(M_{s}\cap V_{x}))\oplus U)$. We first prove $W\leq_{\alpha e}S_{e}$. Note $K_{\gamma}\subseteq W\iff$ $\forall\langle a,b\rangle\in K_{\gamma}.\langle a,b\rangle\in W\iff$ $\forall\langle a,b\rangle\in K_{\gamma}.\exists x\in M_{s}[a\in D_{x}\land b\in E_{x}\land x\in S_{e}]\iff$ $\forall\langle a,b\rangle\in K_{\gamma}.\exists x\in M_{s}[\langle a,b\rangle\in P_{x}\land x\in S_{e}]$ where $i_{P}:\alpha\to\alpha\in\Sigma_{1}(L_{\alpha})$ is a function of lemma 1.40iv and $P_{x}:=K_{i_{P}(x)}$. Define $\phi(\gamma,\delta):\iff\forall y\in K_{\gamma}\exists x\in K_{\delta}.y\in P_{x}$. Define $V:=\{\langle\gamma,\delta\rangle:K_{\delta}\subseteq M_{s}\land\phi(\gamma,\delta)\}$. Then continuing $K_{\gamma}\subseteq W\iff$ $\forall y\in K_{\gamma}\exists x\in M_{s}[y\in P_{x}\land x\in S_{e}]\iff$ $\exists\delta[K_{\delta}\subseteq M_{s}\land K_{\delta}\subseteq S_{e}\land\phi(\gamma,\delta)]$ (Where $K_{\delta}\in L_{\alpha}$ has to exist as an image of an $\alpha$-computable function restricted to an $K_{\gamma}\in L_{\alpha}$ by the admissibility of $\alpha$.) $\iff$ $\exists\langle\gamma,\delta\rangle\in V.K_{\delta}\subseteq S_{e}$. Note $\phi(\gamma,\delta)\iff\exists H[H=w(\gamma,\delta)\land\forall y\in K_{\gamma}\exists x\in K_{\delta}.\langle x,y\rangle\in H]$ where $w:\alpha\times\alpha\to\alpha\in\Sigma_{1}(L_{\alpha})$ with $K_{w(\gamma,\delta)}:=\{\langle x,y\rangle:x\in K_{\delta}\land y\in K_{\gamma}\land y\in P_{x}\}$ is a function of lemma 1.40v. Hence $\phi(\gamma,\delta)\in\Sigma_{1}(L_{\alpha})$. As $M_{s}\in\Delta_{1}(L_{\alpha})$ by $M_{s}\in\Delta_{1}(L_{\alpha})$, so $V\in\Sigma_{1}(L_{\alpha})$. Therefore $W\leq_{w\alpha e}S_{e}$.

Note $V_{x}\in\Sigma_{1}(L_{\alpha})$. By the assumptions $M_{s}\in\Sigma_{1}(L_{\alpha})$ and $X_{s}\in L_{s}$ it is true that $M_{s}\in\Sigma_{1}(L_{\alpha})$ and $X_{s}\in\Sigma_{1}(L_{\alpha})$. Thus $(X_{s}\cup(M_{s}\cap V_{x}))\in\Sigma_{1}(L_{\alpha})$. Hence $S_{e}\leq_{w\alpha e}(X_{s}\cup(M_{s}\cap V_{x}))\oplus U\leq_{\alpha e}U$ by 1.14 and lemma 1.12 respectively. Hence $S_{e}\leq_{w\alpha e}U$ by 1.13.

As $U^{+}$ is megaregular, so $S_{e}\leq_{\alpha e}U$ by 1.36. Hence $W\leq_{w\alpha e}U$ by 1.13. Finally, $W\leq_{\alpha e}U$ by the megaregularity of $U^{+}$ again. ∎

Lemma 3.13**.**

Let $I\in L_{\alpha}$. Then exists an index $z<\alpha$ which is $\alpha$-computable from $I$ st $V_{z}=\bigcap_{x\in I}V_{x}$.

Proof.

Define $f(I)=z:\iff D_{z}=\bigcup_{x\in I}D_{x}\land E_{z}=\bigcup_{x\in I}E_{x}$. By lemma 1.40i the function $f$ is total and $\alpha$-computable. Also $\bigcap_{x\in I}V_{x}=\{y<\alpha:\bigcup_{x\in I}D_{x}\subseteq D_{y}\land\bigcup_{x\in I}E_{x}\subseteq E_{y}\}=V_{f(I)}=V_{z}$ as required. ∎

Lemma 3.14**.**

Let $D\subseteq A\subseteq\alpha$ and $E\subseteq B\subseteq\alpha$ satisfying $A,B\not\in\Sigma_{1}(L_{\alpha})$ and $D,E\in L_{\alpha}$. Define $Z:=Z_{D,E}:=\{x<\alpha:D\subseteq D_{x}\subseteq A\land E\subseteq E_{x}\subseteq B\}$. Then:

i) $Z\equiv_{\alpha e}A\oplus B$,

ii) $\overline{Z}\leq_{w\alpha e}\overline{A}\oplus\overline{B}$,

iii) $Z\not\in\Sigma_{1}(L_{\alpha})$,

iv) $Z$ is unbounded if $A\oplus B$ is megaregular.

Proof.

i) First note that for all $\alpha$-finite sets $K_{\gamma},K_{\delta}$ there is some $x<\alpha$ st $D_{x}=K_{\gamma},E_{x}=K_{\delta}$. Hence if we require that $D_{x}$ (or $E_{x}$) is fixed to some $\alpha$-finite set $K\in L_{\alpha}$, still the remaining sets $E_{x}$ (or $D_{x}$) include all $\alpha$-finite sets. Note $A\leq_{\alpha e}Z$ via $W:=\{\langle\gamma,\delta\rangle:\exists x<\alpha[D\cup K_{\gamma}\subseteq D_{x}\land K_{\delta}=\{x\}]\}\in\Sigma_{1}(L_{\alpha})$. Similarly, $B\leq_{\alpha e}Z$. Consequently, $A\oplus B\leq_{\alpha e}Z$. Define $I_{D,A}:=\{x<\alpha:D\subseteq D_{x}\subseteq A\}$. Define $I_{E,B}:=\{x<\alpha:E\subseteq E_{x}\subseteq B\}$. $I_{D,A}\leq_{\alpha e}A$ via $W_{A}:=\{\langle\gamma,\delta\rangle:\forall x\in K_{\gamma}.D\subseteq D_{x}\land\bigcup_{x\in K_{\gamma}}D_{x}=K_{\delta}\}\in\Sigma_{1}(L_{\alpha})$. Similarly $I_{E,B}\leq_{\alpha e}B$. Note that $Z=I_{D,A}\cap I_{E,B}$. Thus $Z\leq_{\alpha e}I_{D,A}\oplus I_{E,B}\leq_{\alpha e}A\oplus B$. Therefore $A\oplus B\equiv_{\alpha e}Z$.

ii) Note that $\overline{I_{D,A}}\leq_{w\alpha e}\overline{A}$ via $\Phi^{vw}:=\{\langle x,\delta\rangle:\exists y<\alpha[y\not\in D_{x}\land y\in D\land K_{\delta}=\emptyset\lor y\in D_{x}\land K_{\delta}=\{y\}\}$. Similarly, $\overline{I_{E,B}}\leq_{w\alpha e}\overline{B}$. Hence $\overline{Z}=\overline{I_{D,A}}\cup\overline{I_{E,B}}\leq_{w\alpha e}\overline{A}\oplus\overline{B}$ as required.

iii) If $Z\in\Sigma_{1}(L_{\alpha})$, then $Z\in\Sigma_{1}(L_{\alpha})$ and $A\in\Sigma_{1}(L_{\alpha})$, $B\in\Sigma_{1}(L_{\alpha})$ which contradicts the assumption. Hence $Z\not\in\Sigma_{1}(L_{\alpha})$.

iv) From ii) and megaregularity of $A\oplus B$, we have $\overline{Z}\leq_{\alpha e}\overline{A}\oplus\overline{B}$. Note $\overline{A\oplus B}=\overline{A}\oplus\overline{B}$. Combining this with i) it yields $Z\leq_{\alpha}A\oplus B$. Hence $Z\in\Delta_{1}(L_{\alpha},A,B)$. If $Z$ was bounded, then by 1.34 using the megaregularity of $A\oplus B$, $Z$ is $\alpha$-finite. This contradicts iii). Hence $Z$ has to be unbounded. ∎

Definition 3.15**.**

(Weak halting set)

The weak halting set is defined as $K(A):=\{x<\alpha:x\in\Phi_{x}(A)\}$.

Theorem 3.16**.**

101010Theorem 2.5 in [14] for $\alpha=\omega$.

Let $A,B,U\subseteq\alpha$. Let one of the conditions hold:

i) the projectum of $\alpha$ is $\alpha^{*}=\omega$ and $U^{+}$ is megaregular.

ii) $A\oplus B\oplus K(U)$ is megaregular.

Suppose $\neg\mathcal{K}_{U}(A,B)$. Then $\exists X,Y\subseteq\alpha[Y\leq_{\alpha e}X\oplus A\land Y\leq_{\alpha e}X\oplus B\land Y\not\leq_{w\alpha e}X\oplus U]$.

The following proof is a generalization of the proof for the case when $\alpha=\omega$ in [14].

Proof.

We perform a construction in $\alpha^{*}$ stages and define sets $X,Y$ st $\forall x<\alpha$:

[TABLE]

which guarantees $Y\leq_{\alpha e}X\oplus A$ and $Y\leq_{\alpha e}X\oplus B$ by lemma 3.11.

Index the requirements and $\alpha$-enumeration operators by indices in $\alpha^{*}$ using 1.22. Aim to meet for all $e<\alpha^{*}$ the requirements

[TABLE]

At each stage $s<\alpha^{*}$ of the construction aim to define an $\alpha$-finite set $X_{s}$ and an $\alpha$-computable set $M_{s}$ so that for all $s<\alpha^{*}$ they satisfy:

[TABLE]

Pre-construction

By statement 9, the set $M_{s}$ is defined at every stage $s<\alpha^{*}$ by the sets $N_{s}$ and $I_{s}$. Since the set $I_{s}$ is $\alpha$-finite at the stage $s$ by statement 8, so by lemma 3.13 there is an index $z$ which is $\alpha$-computable from $I_{s}$ and $V_{z}=\bigcap_{x\in I_{s}}V_{x}$. Hence the equality $(\bigcap_{x\in I_{s}}V_{x})-N_{s}=V_{z}-N_{s}$ holds at every stage $s$ where $I_{s}\in L_{\alpha}$. Consequently also the set $V_{z}$ is $\alpha$-computable at such stage $s$.

Since the set $N_{s}$ is $\alpha$-finite by statement 7 and $V_{z}$ is $\alpha$-computable at the stage $s$, so the set $M_{s}$ has to be $\alpha$-computable at the stage $s$, hence statement 10 holds.

When proving at the stage $s<\alpha^{*}$ that statement 5 holds, we use the fact that $A$ and $B$ are not $\alpha$-finite by 3.2 since $\neg\mathcal{K}_{U}(A,B)$. This given $\alpha$-finite sets $D,E$, enables us to find arbitrarily large $\alpha$-finite supersets of $D,E$ contained in $A$ and $B$ respectively.

Constructing $X$

The set $X$ will be constructed in $\alpha^{*}$-many stages.

• •

Stage $s=0$. Set $X_{0}:=\emptyset$, $N_{0}:=\emptyset$, $I_{0}:=\emptyset$. Observe statement 5 is true for $M_{0}=\alpha$. Clearly, statements 6, 7 and 8 are satisfied.

• •

Stage $s+1=3e>0$, $3e$ is a successor ordinal. Define $X_{s+1}:=X_{s}$, $N_{s+1}:=N_{s}$, $I_{s+1}:=I_{s}$. Since the sets $X_{s+1}$, $N_{s+1}$, $I_{s+1}$ are the same as the sets $X_{s}$, $N_{s}$, $I_{s}$ and statements 2, 3, 4, 5, 6, 7 and 8 hold at the stage $s$ by IH, they hold at the stage $s+1$ too.

• •

Stage $s+1=3e+1$. By induction hypothesis let $X_{s}$, $N_{s}$, $I_{s}$ be given and $\alpha$-finite by statements 6, 7 and 8. Define $X_{s+1}:=X_{s}$, $N_{s+1}:=N_{s}\cup\{e\}$, $I_{s+1}:=I_{s}$. Trivially, statements 6, 7 and 8 hold at the stage $s+1$ by IH at the stage $s$.

Note $M_{s+1}=M_{s}-\{e\}$ by statement 9. We claim that the set $M_{s+1}$ satisfies statement 5. Let $D,E\in L_{\alpha}\land D\subseteq A\land E\subseteq B$. By IH on $M_{s}$ there is $x\in M_{s}$ st $[D\subseteq D_{x}\subseteq A\land E\subseteq E_{x}\subseteq B]$. Note $D_{x}\in L_{\alpha}$, but by 3.2 $A\not\in L_{\alpha}$, hence $D_{x}\subset A$. Let $z\in A-D_{x}$. Then $\hat{D}:=D_{x}\cup\{z\}\in L_{\alpha}$. By IH on $M_{s}$ there is $y\in M_{s}$ st $\hat{D}\subseteq D_{y}\subseteq A\land E\subseteq E_{y}\subseteq B$. If $x\not=e$, then $x\in M_{s+1}:=M_{s}-\{e\}$. Otherwise $x=e\not=y$ and $y\in M_{s+1}\land D\subseteq D_{x}\subset\hat{D}\subseteq D_{y}\subseteq A\land E\subseteq E_{y}\subseteq B$. Therefore in any case the set $M_{s+1}$ satisfies statement 5.

• •

Stage $s+1=3e+2$. Aim to find $x\in M_{s}$ st one of the two following statements is true:

1: $D_{x}\not\subseteq A\land E_{x}\not\subseteq B\land x\in\Phi_{e}((X_{s}\cup(M_{s}\cap V_{x}))\oplus U)$,

2: $D_{x}\subseteq A\land E_{x}\subseteq B\land x\not\in\Phi_{e}((X_{s}\cup(M_{s}\cap V_{x}))\oplus U)$.

First we prove the existence of such $x\in M_{s}$. Assume that $\forall x\in M_{s}$ the statement 2 is false. Define

[TABLE]

Then $W\leq_{\alpha e}U$ by the regularity of $U^{+}$, lemma 3.12, statement 10 and statement 6.

We prove $A\times B\subseteq W$. Let $(a,b)\in A\times B$. By statement 5 for $M_{s}$ it follows $\exists x\in M_{s}[a\in D_{x}\subseteq A\land b\in E_{x}\subseteq B]$. Since statement 2 is false, we have $x\in\Phi_{e}((X_{s}\cup(M_{s}\cap V_{x}))\oplus U)$. Thus $(a,b)\in W$. Since $\neg\mathcal{K}_{U}(A,B)$, there is a pair $(a,b)\in\overline{A}\times\overline{B}$ st $(a,b)\in W$. Thus there is $x\in M_{s}$ st $a\in D_{x},b\in E_{x}$ and $x\in\Phi_{e}((X_{s}\cap(M_{s}\cap V_{x}))\oplus U)$. Hence $D_{x}\not\subseteq A$, $E_{x}\not\subseteq B$ and statement 1 is true for $x\in M_{s}$. Therefore there is $x\in M_{s}$ st statement 1 or statement 2 is true. Choose such an element $x\in M_{s}$ using the oracle $A\oplus B\oplus K(U)$.

Case 1: If statement 1 is true for $x$, then $x\in\Phi_{e}((X_{s}\cup(M_{s}\cap V_{x})\oplus U)$. By 1.16 and 1.15 there is $F\subseteq X_{s}\cup(M_{s}\cap V_{x})$ st $F\in L_{\alpha}\land x\in\Phi_{e}(F\oplus U)$. Thus define $X_{s+1}:=X_{s}\cup F$, $N_{s+1}:=N_{s}$, $I_{s+1}:=I_{s}$. Note that$M_{s+1}:=M_{s}$. The set $F$ is $\alpha$-finite, by IH $X_{s}$ is $\alpha$-finite and so the union $X_{s+1}=X_{s}\cup F$ is $\alpha$-finite satisfying statement 6. Statements 7 and 8 are true by IH.

Case 2: Otherwise if statement 2 is true for $x$, then define $X_{s+1}:=X_{s}\cup\{x\}$, $N_{s+1}:=N_{s}$, $I_{s+1}:=I_{s}\cup\{x\}$. Trivially, the sets $X_{s+1},N_{s+1},I_{s+1}$ are $\alpha$-finite using IH, hence satisfying statements 6, 7 and 8. Note $M_{s+1}=M_{s}\cap V_{x}$ by statement 9. $M_{s+1}$ satisfies statement 5: if $D\subseteq A,E\subseteq B,D\in L_{\alpha},E\in L_{\alpha}$, then by the hypothesis on $M_{s}$, there is $y\in M_{s}$ st $D\cup D_{x}\subseteq D_{y}\subseteq A$ and $E\cup E_{x}\subseteq E_{y}\subseteq B$. Therefore $y\in M_{s}\cap V_{x}=M_{s+1}$.

Note in both cases $X_{s+1}-X_{s}\subseteq M_{s+1}$ statement 4 being satisfied.

• •

Stage $s=3e>0$, $3e$ is a limit ordinal. If $\alpha^{*}=\omega$, then this stage does not arise. Hence assume that $A\oplus B\oplus K(U)$ is megaregular.

Define $X_{s}:=\bigcup_{r<s}X_{r},N_{s}:=\bigcup_{r<s}N_{r},I_{s}:=\bigcup_{r<s}I_{r}$. We claim that these sets are $\alpha$-finite.

Define a partial function $f:\alpha\rightharpoonup\alpha$ on the ordinals smaller than $s$ by $f(r)=\{\gamma<\alpha:K_{\gamma}=X_{r}\}$. Note that by IH for all $r<s$, the set $X_{r}$ is $\alpha$-finite using statement 6. Also during the construction we only use the oracle $A\oplus B\oplus K(U)$. Thus the index $f(r)$ of an $\alpha$-finite set $X_{r}$ is also $A\oplus B\oplus K(U)$-computable. Consequently, the function $f$ is $\Sigma_{1}(L_{\alpha},A\oplus B\oplus K(U))$ definable. As $s<\alpha^{*}$, so $s$ as a limit ordinal is an $\alpha$-finite set. Therefore by the megaregularity of $A\oplus B\oplus K(U)$, the set $f[s]$ is also $\alpha$-finite. But then $X_{s}=\bigcup_{\gamma\in f[s]}K_{\gamma}$ is $\alpha$-finite by 1.8. So statement 6 holds at the stage $s$ as required. Applying similar reasoning, using the veracity of statements 7 and 8 for all $r<s$ by IH, we conclude that statements 7 and 8 hold at the stage $s$ too.

Note $M_{s}:=\bigcap_{r<s}M_{r}$ by statement 9. We prove that statement 5 holds at the stage $s$. Note that $M_{s}=V_{z}-N_{s}$ by statement 9 for some $z<\alpha$ satisfying both $D_{z}\subseteq A$ and $E_{z}\subseteq B$. Fix $\alpha$-finite sets $D$ and $E$ st $D\subseteq A$ and $E\subseteq B$. WLOG let $D_{z}\subseteq D$ and $E_{z}\subseteq E$. Define $Z:=\{x<\alpha:D\subseteq D_{x}\subseteq A\land E\subseteq E_{x}\subseteq B\}$. As $\neg\mathcal{K}(A,B)$ by the assumption, so $A\not\in\Sigma_{1}(L_{\alpha})$ and $B\not\in\Sigma_{1}(L_{\alpha})$ by 3.2. Note that $A\oplus B$ is megaregular. Hence $Z$ is unbounded by lemma 3.14. On the other hand $N_{s}\subseteq s$. Thus $Z-N_{s}\not=\emptyset$. Note $\{x\in M_{s}:D\subseteq D_{x}\subseteq A\land E\subseteq E_{x}\subseteq B\}=\{x\in V_{z}-N_{s}:D\subseteq D_{x}\subseteq A\land E\subseteq E_{x}\subseteq B\}=Z-N_{s}\not=\emptyset$. Therefore $\forall D,E\in L_{\alpha}[D\subseteq A\land E\subseteq B\implies\exists x\in M_{s}[D\subseteq D_{x}\subseteq A\land E\subseteq E_{x}\subseteq B]]$ and so the statement statement 5 is satisfied at the limit stage $s$.

Finally, define $X:=\bigcup_{s<\alpha^{*}}X_{s}$.

Defining $Y$

To define $Y$ first prove $\forall z\in X[D_{z}\subseteq A\iff E_{z}\subseteq B]$: Let $z\in X$. Then there is a stage $s+1=3e+2$ st $z\in X_{s+1}-X_{s}$. In case 2 $D_{z}\subseteq A$ and $E_{z}\subseteq B$. In case 1 there is $x$ st $X_{s+1}-X_{s}\subseteq V_{x},D_{x}\not\subseteq A$ and $E_{x}\not\subseteq B$. As $z\in X_{s+1}-X_{s}\subseteq V_{x}$, so $D_{x}\subseteq D_{z}$ and $E_{x}\subseteq E_{z}$. Thus $D_{z}\not\subseteq A$ and $E_{z}\not\subseteq B$. Define the set

$Y:=\{z\in X:D_{z}\subseteq A\}=\{z\in X:E_{z}\subseteq B\}$.

Final verification

Note $Y\leq_{\alpha e}X\oplus A$ and $Y\leq_{\alpha e}X\oplus B$ as proved under statement 1.

We prove $Y\not\leq_{w\alpha e}X\oplus U$ by showing $Y\not=\Phi_{e}(X\oplus U)$ for an arbitrary $e<\alpha^{*}$. Consider a stage $s+1=3e+2$. In case 1 $X_{s+1}=X_{s}\cup F$ and there is $x$ st $x\in\Phi_{e}(F\oplus U),D_{x}\not\subseteq A$ and $E_{x}\not\subseteq B$. Hence $x\in\Phi_{e}(X\oplus U)-Y$. In case 2 there is $x$ st $X_{s+1}=X_{s}\cup\{x\},M_{s+1}=M_{s}\cap V_{x},D_{x}\subseteq A,E_{x}\subseteq B$ and $x\not\in\Phi_{e}((X_{s}\cup M_{s+1})\oplus U)$.

Let $z\in X$. Then $\exists t.z\in X_{t+1}-X_{t}\subseteq M_{t+1}$ by statement 4. If $t\geq s$, then $z\in M_{s+1}$ by statement 3. If $t<s$, then $z\in X_{s}$ by statement 2. Hence $z\in X_{s}\cup M_{s+1}$ and thus $X\subseteq X_{s}\cup M_{s+1}$.

Hence $x\in Y-\Phi_{e}(X\oplus U)$ by 1.15. Therefore in both cases $Y\not=\Phi_{e}(X\oplus U)$ and so $Y\not\leq_{w\alpha e}X\oplus U$. ∎

Theorem 3.17**.**

111111From theorem 2.6 in [14] for $\mathcal{D}_{T}$.

The statements i) - iv) are equivalent and imply v). Moreover if the projectum of $\alpha$ is $\alpha^{*}=\omega$ and $U^{+}$ is megaregular or $A\oplus B\oplus K(U)$ is megaregular, then all the statements i) - v) are equivalent.

i) $\mathcal{K}_{U}(A,B)$, i.e. $\exists W\leq_{\alpha e}U.A\times B\subseteq W\land\overline{A}\times\overline{B}\subseteq\overline{W}$,

ii) $\exists f(x,y)\in\Delta_{1}(L_{\alpha}).\forall X\subseteq\alpha.$ $\forall x,y\in\alpha.\Phi_{x}(A\oplus X)\cap\Phi_{y}(B\oplus X)$ $\subseteq\Phi_{f(x,y)}(X\oplus U)$ $\subseteq\Phi_{x}(A\oplus X)\cup\Phi_{y}(B\oplus X)$,

iii) $\exists f(x,y)\in\Delta_{1}(L_{\alpha})\forall x,y<\alpha[\Phi_{x}(A)=\Phi_{y}(B)\implies\Phi_{f(x,y)}(U)=\Phi_{x}(A)]$,

iv) $\forall V_{1},V_{2}[V_{1}\leq_{\alpha e}A\land V_{2}\leq_{\alpha e}B\implies\exists W\leq_{\alpha e}U.V_{1}\cap V_{2}\subseteq W\subseteq V_{1}\cup V_{2}]$,

v) $\forall X\subseteq\alpha.\mathrm{deg}_{\alpha e}({X\oplus U})=\mathrm{deg}_{\alpha e}({A\oplus X\oplus U})\wedge\mathrm{deg}_{\alpha e}({B\oplus X\oplus U})$.

Proof.

The implications ii) $\implies$ iii), ii) $\implies$ iv), ii) $\implies$ v), iv) $\implies$ i) are trivial. It remains to prove the implications i) $\implies$ ii) and iii) $\implies$ i).

i) $\implies$ ii):

Assume $\exists W\leq_{\alpha e}U.A\times B\subseteq W\land\overline{A}\times\overline{B}\subseteq\overline{W}$ and let $W=\Phi(U)$ for some $\alpha$-enumeration operator $\Phi$.

Define $f$ st for any $X\subseteq\alpha$, $x,y\in\alpha$:

$\Phi_{f(x,y)}(X\oplus V):=\{z\in\alpha:\exists D,E\in L_{\alpha}[z\in\Phi_{x}(D\oplus X)\cap\Phi_{y}(E\oplus X)\land D\times E\subseteq\Phi(V)\}$.

Then $f$ is $\alpha$-computable and satisfies the condition ii).

iii) $\implies$ i): Suppose that $A$ and $B$ satisfy the condition iii) with $f$ being computable. Define a computable function $g$ st for every $Y\subseteq\alpha$ and $y<\alpha$:

\Phi_{g(y)}(Y)=\begin{cases}\alpha&if y\in Y,\\ \emptyset&otherwise.\end{cases}

Then $A,B$ are a $U$-Kalimullin pair with a witness $W=\{(m,n):\Phi_{f(g(m),g(n))}(U)\not=\emptyset\}$.

Hence i) $\iff$ ii) $\iff$ iii) $\iff$ iv) $\implies$ v) for any $\alpha$. Note v) $\implies$ i) is the contrapositive of 3.16. Therefore i) $\iff$ ii) $\iff$ iii) $\iff$ iv) $\iff$ v) if $\alpha^{*}=\omega$ and $U^{+}$ is megaregular or $A\oplus B\oplus K(\emptyset)$ is megaregular. ∎

The statement i) iff v) establishes the definability of a $U$-Kalimullin pair.

Proposition 3.18**.**

121212Proposition 1.7 in [1] for $\alpha=\omega$.

Let $B\subseteq\alpha$. The set of all $A$ st $\mathcal{K}(A,B)$ is closed downwards under $\alpha$-enumeration reducibility as well as closed under join.

Proof.

Suppose $\mathcal{K}(A_{0},B)$ and $A_{1}\leq_{\alpha e}A_{0}$. Hence $\exists W_{0}\in\Sigma_{1}(L_{\alpha}).A_{0}\times B\subseteq W_{0}\land\overline{A}_{0}\times\overline{B}\subseteq\overline{W}_{0}$. Let $V_{1}:=A_{1}\times\alpha$, $V_{2}:=\alpha\times B$. As $A_{1}\leq_{\alpha e}A_{0}$, so $V_{1}\leq_{\alpha e}A_{0}\land V_{2}\leq_{\alpha e}B$. Hence by theorem 3.17 (i implies iv), $\exists W_{1}\in\Sigma_{1}(L_{\alpha})$ st $V_{1}\cap V_{2}\subseteq W_{1}\subseteq V_{1}\cup V_{2}$. Therefore $V_{1}\cap V_{2}=A_{1}\times B\subseteq W_{1}$. Also $W_{1}\subseteq V_{1}\cup V_{2}\iff\overline{V}_{1}\cap\overline{V}_{2}\subseteq\overline{W}_{1}$ and so $\overline{V}_{1}\cap\overline{V}_{2}=(\overline{A}_{1}\times\alpha)\cap(\alpha\times\overline{B})=\overline{A}_{1}\times\overline{B}\subseteq\overline{W}_{1}$. Hence $\mathcal{K}(A_{1},B)$.

Let $\mathcal{K}(A_{0},B)\land\mathcal{K}(A_{1},B)$. If $A_{i}=\emptyset$ for $i\in 2$ then $A_{0}\oplus A_{1}\equiv_{\alpha e}A_{1-i}$ and so $\mathcal{K}(A_{0}\oplus A_{1},B)$. Otherwise $\mathcal{K}(A_{0}\oplus A_{1},B)$ by lemma 3.9. ∎

Corollary 3.19**.**

(Definability of a Kalimullin Pair)131313The case for $\alpha=\omega$ proved in [14].

Let $\alpha^{*}=\omega$ or assume $V=L$ and let $\alpha$ be an infinite regular cardinal. Then:

$\forall a,b\in\mathcal{D}_{\alpha e}[\mathcal{K}(a,b)\iff\forall x\in\mathcal{D}_{\alpha e}.(a\vee x)\wedge(b\vee x)=x]$.

Proof.

Note that if $U\in\Sigma_{1}(L_{\alpha})$, then $U^{+}$ is megaregular by 1.35. Thus the statement above follows from $(i\iff v)$ in 3.17 and from the $\mathcal{K}$-pair being a degree theoretic property by its invariance under the $\alpha e$-reducibility by proposition 3.18. ∎

Corollary 3.20**.**

(Definability of an $U$-Kalimullin Pair)

Assume $V=L$ and let $\alpha$ be an infinite regular cardinal. Then:

$\forall a,b,u\in\mathcal{D}_{\alpha e}[\mathcal{K}_{u}(a,b)\iff\forall x\in\mathcal{D}_{\alpha e}.(a\vee x\vee u)\wedge(b\vee x\vee u)=x\vee u]$.

Proof.

Note that since $\alpha$ is an infinite regular cardinal, so $A\oplus B\oplus K(U)$ is megaregular. Thus the statement above follows from $(i\iff v)$ in 3.17 and from the $\mathcal{K}$-pair being a degree theoretic property by its invariance under the $\alpha e$-reducibility by proposition 3.18. ∎

3.3 Maximal Kalimullin pair and total degrees

Proposition 3.21**.**

(Maximality of semicomputable megaregular Kalimullin pairs)141414Generalized from Maximal $\mathcal{K}$-pairs in [1] for $\alpha=\omega$.

Let $A\subseteq\alpha$ and let $A^{+}$ and $A^{-}$ be both megaregular. If $\mathcal{K}(A,\overline{A})\land A\not\in\Sigma_{1}(L_{\alpha})\land A\not\in\Pi_{1}(L_{\alpha})$, then $\mathcal{K}_{\mathrm{max}}(A,\overline{A})$.

Proof.

Suppose $\mathcal{K}(A,\overline{A})$ and $\mathcal{K}(C,D)$, $A\leq_{\alpha e}C$, $\overline{A}\leq_{\alpha e}D$. By 3.18 $\mathcal{K}(A,D)$. By 3.8 and megaregularity of $A^{-}$ we have $D\leq_{\alpha e}\overline{A}$. Similarly, $\mathcal{K}(\overline{A},C)$ and thus $C\leq_{\alpha e}\overline{\overline{A}}=A$ by the megaregularity of $A^{+}$. ∎

Corollary 3.22**.**

Let $\alpha^{*}=\omega$ or assume $V=L$ and let $\alpha$ be an infinite regular cardinal. Then every nontrivial total megaregular degree is a join of a maximal $\mathcal{K}$-pair, i.e.

$\forall a\in\mathcal{TOT}_{\alpha e}^{\mathrm{mr}}-\{0\}\exists b,c\in\mathcal{D}_{\alpha e}[(a=b\vee c)\land\mathcal{K}_{\mathrm{max}}(b,c)]$.

Proof.

Since $\alpha^{*}=\omega$ or $\alpha$ is an infinite regular cardinal, thus a (maximal) Kalimullin pair is definable by 3.19.

Suppose $a\in\mathcal{TOT}_{\alpha e}-\{0\}$ and $a$ is a megaregular degree (at least one or equivalently every set in $a$ is megaregular). Then by theorem 2.16, there is $A\subseteq\alpha$ st $A$ $\alpha$-semicomputable, $A\not\in\Sigma_{1}(L_{\alpha})$, $\overline{A}\not\in\Sigma_{1}(L_{\alpha})$ and $A\oplus\overline{A}\in a$ by the totality of $a$. As $A$ $\alpha$-semicomputable, so $\mathcal{K}(A,\overline{A})$ by proposition 3.3. $\mathcal{K}(A,\overline{A})$ is nontrivial since $A\not\in\Sigma_{1}(L_{\alpha})$ and $\overline{A}\not\in\Sigma_{1}(L_{\alpha})$. Thus by 3.21 and the megaregularity of $A$ we have $\mathcal{K}_{\mathrm{max}}(A,\overline{A})$. ∎

By inspecting whether a degree which is not quasiregular could be a join of a maximal Kalimullin pair, one may establish the following:

Proposition 3.23**.**

If $\mathrm{deg}_{\alpha}({B})$ is not a quasiregular degree, then there is $C$ st $0<_{\alpha}C<_{\alpha}B$ and $\mathcal{K}_{\mathrm{max}}(C,\overline{C})$.

Proof.

Since $\mathrm{deg}_{\alpha}({B})$ is not a quasiregular degree, then $D$ is not quasiregular for any $D\equiv_{\alpha}B$. So $B$ is not quasiregular.

Let $\beta<\alpha$ be the least ordinal st $B\cap\beta\not\in L_{\alpha}$. Define $A:=B\cap\beta$. Then $A\subset B$ by $B$ not being quasiregular. By the minimality of $\beta$, the set $A$ is quasiregular. $A$ is bounded, but not $\alpha$-finite, hence $A$ cannot be $\alpha$-computable. Thus $A>_{\alpha}\emptyset$. By 2.16 there is $\alpha$-semicomputable set $C$ st $A\equiv_{\alpha}C$, $C\not\in\Sigma_{1}(L_{\alpha})$ and $C\not\in\Pi_{1}(L_{\alpha})$. As $C$ is $\alpha$-semicomputable, so $\mathcal{K}(C,\overline{C})$. By 3.21 we have that $\mathcal{K}_{\mathrm{max}}(C,\overline{C})$. ∎

4 Acknowledgements

The author would like to thank Mariya Soskova and Hristo Ganchev for the explanation of the proof of the Kalimullin pair definability in the classical case, i.e. $\alpha=\omega$.

The author was supported by Hausdorff Research Institute for Mathematics during Hausdorff Trimester Program Types, Sets and Constructions.