{\alpha} degrees as an automorphism base for the {\alpha}-enumeration degrees

TL;DR

This paper generalizes Selman's Theorem from classical Computability Theory to the setting of lpha-Computability Theory, establishing conditions under which lpha-degrees form an automorphism base, extending known results to broader ordinal contexts.

Contribution

The paper extends Selman's Theorem to lpha-Computability Theory, providing a full generalization for regular cardinals and partial results for admissible ordinals.

Findings

lpha-degrees form an automorphism base under certain conditions

Full generalization for regular cardinals lpha

Partial results for admissible ordinals lpha

Abstract

Selman's Theorem in classical Computability Theory gives a characterization of the enumeration reducibility for arbitrary sets in terms of the enumeration reducibility on the total sets: $A \le_e B \iff \forall X [X \equiv_{e} X \oplus \overline{X} \land B \le_{e} X \oplus \overline{X} \implies A \le_{e} X \oplus. \overline{X} ]$. This theorem implies directly that the Turing degrees are an automorphism base of the enumeration degrees. We lift the classical proof to the setting of the $\alpha$-Computability Theory to obtain the full generalization when $\alpha$ is a regular cardinal and partial results for a general admissible ordinal $\alpha$.

Equations

$$s=2e+1\wedge x={\tau }_{s+1}(|{\tau }_s|)⟹x\in C\wedge x\notin A\vee x\notin C\wedge x\in A$$

$$\forall l<\alpha [f(l)\in A⟺l\in f^{-1}[A]⟺l\in {\Phi }_e(f)]$$

Full text

$\alpha$ degrees as an automorphism base for the $\alpha$-enumeration degrees

Dávid Natingga

Abstract

Selman’s Theorem [10] in classical Computability Theory gives a characterization of the enumeration reducibility for arbitrary sets in terms of the enumeration reducibility on the total sets: $A\leq_{e}B\iff\forall X[X\equiv_{e}X\oplus\overline{X}\land B\leq_{e}X\oplus\overline{X}\implies A\leq_{e}X\oplus.\overline{X}]$. This theorem implies directly that the Turing degrees are an automorphism base of the enumeration degrees. We lift the classical proof to the setting of the $\alpha$-Computability Theory to obtain the full generalization when $\alpha$ is a regular cardinal and partial results for a general admissible ordinal $\alpha$.

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 [3] [4] [5] [6] or on generalized register machines [7]. Recommended references for this section are [9], [1], [8] and [2].

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[1])

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 [1] [12])

• •

$\omega^{CK}_{1}$ - 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**.**

[1] 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**.**

[8] 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 [9] 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.

1.5 Properties of $\alpha$-enumeration operator

Fact 1.12**.**

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

Fact 1.13**.**

(Monotonicity)

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

Proposition 1.14**.**

(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 1.15**.**

222From [1] 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 1.16**.**

(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 1.17**.**

(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 Degree Theory

Definition 1.18**.**

(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 1.19**.**

(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 1.20**.**

(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. $\mathcal{TOT}_{\alpha e}:=\iota[\mathcal{D}_{\alpha}]$.

1.8 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 1.21**.**

(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 1.22**.**

(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.

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

Fact 1.23**.**

(Megaregularity and definability)

• •

$B\in\Sigma_{1}(L_{\alpha})\implies B^{+}$ is megaregular,

• •

$B\in\Delta_{1}(L_{\alpha})\implies B$ is megaregular.

Proposition 1.24**.**

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 1.25**.**

(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 $]$. ∎

Proposition 1.26**.**

(Correspondence between the $\alpha$-enumeration reducibilities)

We have the following implication diagram:

{A\leq_{w\alpha e}B}$${A\leq_{\alpha e}B}if $\scriptstyle B^{+}$ megaregularalways

∎

2 Selman’s theorem

We generalize the theorem of Selman present in [10].

Definition 2.1**.**

(Odd enumeration and $\alpha$-finite part)

• •

Let $B\subseteq\alpha$. The total function $f:\alpha\to\alpha$ is an odd enumeration of $B$ iff $f[\{2\gamma+1<\alpha:\gamma<\alpha\}]=B$.

• •

$B$ odd $\alpha$-finite part is a function $\tau:[0,2s)\to\alpha$ for $s<\alpha$ st $\forall x<\alpha[2x+1\in\mathrm{dom}(\tau)\implies\tau(2x+1)\in B]$.

• •

Let $|\tau|$ denote the order type of $\mathrm{dom}(\tau)$, i.e. $|\tau|:=\mathrm{ot}(\mathrm{dom}(\tau))$.

If $\mathrm{dom}(\tau)$ is an initial segment of $\alpha$, we have $|\tau|=\mathrm{dom}(\tau)$. If $\tau:[0,2s)\to\alpha$ is a function, then $\mathrm{dom}(\tau)=2s$ and so $|\tau|=2s$. If $\tau$ is a $B$ odd $\alpha$-finite part, then there is an odd enumeration $f:\alpha\to\alpha$ of $B$ st $\tau\subseteq f$.

Lemma 2.2**.**

Let $f:\alpha\to\alpha$ be an odd enumeration of $B$. Then $B\leq_{w\alpha e}f$.

Proof.

We have $B\leq_{w\alpha e}f$ via $\Phi:=\{\langle b,\delta\rangle:\exists\gamma<\alpha[2\gamma+1<\alpha\land K_{\delta}=\{(2\gamma+1,b)\}]\}$. ∎

Lemma 2.3**.**

Assume that $A\leq_{w\alpha e}f$. Then $f^{-1}[A]\leq_{w\alpha e}f$.

Proof.

Note $f^{-1}[A]:=\{x<\alpha:f(x)\in A\}$. Let $A\leq_{w\alpha e}f$ via $\Psi\in\Sigma_{1}(L_{\alpha})$. Then $f^{-1}[A]\leq_{w\alpha e}f$ via $\Phi:=\{\langle x,\delta\rangle:\exists y<\alpha[K_{\delta}=K_{\epsilon}\cup\{(x,y)\}\}\land\langle y,\epsilon\rangle\in\Psi]\}\in\Sigma_{1}(L_{\alpha})$. ∎

Definition 2.4**.**

(Weak halting set)

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

Proposition 2.5**.**

333Generalized from [11] for $\alpha=\omega$.

Let $A,B\subseteq\alpha$ and $A\not\leq_{w\alpha e}B$. Assume that $A\oplus B\oplus K(\emptyset)$ is megaregular. Then there exists an odd enumeration $f:\alpha\to\alpha$ of $B$ st $A\not\leq_{w\alpha e}f$ and $B\leq_{\alpha e}f$.

Proof.

Construction

Note that $B\not=\emptyset$ and $B\not\equiv_{\alpha e}\emptyset$ as $A\not\leq_{w\alpha e}B$. We construct a sequence of $B$ odd $\alpha$-finite parts in $\alpha$ many stages st:

$\tau_{0}\subseteq\tau_{1}\subseteq...\subseteq\tau_{s}\subseteq...$

In the end, the desired function $f:\alpha\to\alpha$ is defined as $f=\bigcup_{s<\alpha}\tau_{s}$.

Let $\tau_{0}=\emptyset$. If $s$ is a limit ordinal, then let $\tau_{s}:=\bigcup_{r<s}\tau_{r}$. Now assume that $\tau_{s}$ has been constructed, then at the stage $s$ construct $\tau_{s+1}$:

• •

Stage $s=2e$:

Set $\tau_{s+1}:=\tau_{s}\cdot 0\cdot b$ where $b=\mu y\{y<\alpha:y\in B\land y\not\in\tau_{s}[\{2\gamma+1<|\tau_{s}|:\gamma<\alpha\}]\}$ and $0\cdot b$ is the concatenation of [math] and $b$. E.g. if $\tau=a\cdot b$, then $\tau(0)=a,\tau(1)=b$.

• •

Stage $s=2e+1$:

Use $e$ and $\tau_{s}$ to define set $C$ as

$C:=\{x<\alpha|\exists\rho\supseteq\tau_{s}[\rho$ is a $B$ odd $\alpha$-finite part and $x=\rho(|\tau_{s}|)\land|\tau_{s}|\in\Phi_{e}(\rho)]\}$.

As $C\leq_{w\alpha e}B$, so $C\not=A$. Thus we have two cases:

• –

Case $\exists x<\alpha[x\in C\land x\not\in A]$: Then let $\tau_{s+1}$ be the minimal $\rho$ from $C$. Note $\tau_{s+1}=\tau_{s}\cdot x\cdot b\cdot\sigma$ for some $b\in B$ where $\sigma$ is a $B$ odd $\alpha$-finite part.

• –

Case $\exists x<\alpha[x\not\in C\land x\in A]$: Then let $\tau_{s+1}:=\tau_{s}\cdot x\cdot b$ for some $b\in B$.

Verification

By the two cases above we have for all $e,s,x<\alpha$:

[TABLE]

Note that $f=\bigcup_{s<\alpha}\tau_{s}$ is an odd enumeration of $B$. This is ensured by stages $s=2e<\alpha$. If $b\in B$, then a pair $(2\gamma+1,b)$ is added to $f$ for some $2\gamma+1<\alpha$ at the stage $s=2b<\alpha$ at latest.

Moreover, $A\not\leq_{w\alpha e}f$. For suppose not, then $A\leq_{w\alpha e}f$. Hence $f^{-1}[A]\leq_{w\alpha e}f$ by lemma 2.3 and so there is $e<\alpha$ st $f^{-1}[A]=\Phi_{e}(f)$ and thus:

[TABLE]

Consider the stage $s=2e+1$. Let $l=|\tau_{s}|$. Note $\tau_{s}\subseteq f$.

• •

Case 1: $l\in f^{-1}[A]\implies f(l)\in A\implies l\in\Phi_{e}(f)$ using statement 2. By the witness property of an $\alpha$-enumeration operator there exists $B$ odd $\alpha$-finite part $\rho$ of $f$ st $\rho\supseteq\tau_{s}\land l\in\Phi_{e}(\rho)\land\rho(l)=f(l)$. So $f(l)\in C$. Hence $f(l)\in C\cap A$. Also by statement 1 we have $f(l)\in C\land f(l)\not\in A\lor x\not\in C\land f(l)\in A$. This is a contradiction.

• •

Case 2: $l\not\in f^{-1}[A]\implies f(l)\not\in A\implies l\not\in\Phi_{e}(f)$ using statement 2. By the monotonicity of an $\alpha$-enumeration operator there is no $B$ odd $\alpha$-finite part $\rho$ of $f$ st $\rho\supseteq\tau_{s}\land l\in\Phi_{e}(\rho)$. So $f(l)\not\in C$. Hence $f(l)\not\in A\land f(l)\not\in C$. By statement 1 we have $f(l)\in C\land f(l)\not\in A\lor x\not\in C\land f(l)\in A$. This is a contradiction.

Hence in any case $A\not\leq_{w\alpha e}f$ as required.

Let $g:\alpha\to\alpha$ be defined by $g:s\mapsto\gamma$ where $K_{\gamma}=\tau_{s}$. During the construction we use the oracle $A\oplus B\oplus K(\emptyset)$, hence $g\in\Sigma_{1}(L_{\alpha},A\oplus B\oplus K(\emptyset))$. We show that $g$ is well-defined and that $\tau_{s}$ is $\alpha$-finite at a limit stage $s$. Let $s<\alpha$ be a limit stage. Then $g[s]=I\in L_{\alpha}$ since $s\in L_{\alpha}$, $g_{|s}\in\Sigma_{1}(L_{\alpha},A\oplus B\oplus K(\emptyset))$ and by the megaregularity of the oracle $A\oplus B\oplus K(\emptyset))$. Hence $\tau_{s}=\bigcup_{r<s}\tau_{r}=\bigcup_{\gamma\in I}K_{\gamma}$ is $\alpha$-finite as required. Therefore $\forall s<\alpha.\tau_{s}\in L_{\alpha}$ as needed.

Note that $f\in\Delta_{1}(L_{\alpha},A\oplus B\oplus K(\emptyset))$ since the construction of $f$ uses the oracle $A\oplus B\oplus K(\emptyset)$. By that and the megaregularity of $A\oplus B\oplus K(\emptyset)$, also $f$ must by megaregular. By lemma 2.2 we have $B\leq_{w\alpha e}f$. By the megaregularity of $f$, we have $B\leq_{\alpha e}f$ as required. ∎

Theorem 2.6**.**

(Selman’s theorem for admissible ordinals)

Let $A,B\subseteq\alpha$ and let $A\oplus B\oplus K(U)$ be megaregular. Then:

$A\leq_{\alpha e}B\iff\forall X[X\equiv_{\alpha e}X\oplus\overline{X}\land B\leq_{\alpha e}X\oplus\overline{X}\implies A\leq_{\alpha e}X\oplus\overline{X}]$

Part of the following proof is adapted from the classical case present in [11].

Proof.

$\Rightarrow$ direction is by the transitivity of $\leq_{\alpha e}$.

For the $\Leftarrow$ direction assume that $\forall X[X\equiv_{\alpha e}X\oplus\overline{X}\land B\leq_{\alpha e}X\oplus\overline{X}\implies A\leq_{\alpha e}X\oplus\overline{X}]$. We want to show that $A\leq_{\alpha e}B$. Assume not, then $A\not\leq_{\alpha e}B$ and so $A\not\leq_{w\alpha e}B$ as $A\oplus B\oplus K(U)$ is megaregular. Then by 2.5 and the megaregularity of $A\oplus B\oplus K(U)$ there exists a total function $f$ st $A\not\leq_{w\alpha e}f$, but $B\leq_{\alpha e}f$. But $f$ is total and so $A\leq_{\alpha e}f$ which is a contradiction to the statement $A\not\leq_{w\alpha e}f$. Hence $A\leq_{\alpha e}B$ as required. ∎

Corollary 2.7**.**

(Selman’s theorem for regular cardinals)

Let $\alpha$ be a regular cardinal. Then for any $A,B\subseteq\alpha$ we have:

$A\leq_{\alpha e}B\iff\forall X[X\equiv_{\alpha e}X\oplus\overline{X}\land B\leq_{\alpha e}X\oplus\overline{X}\implies A\leq_{\alpha e}X\oplus\overline{X}]$

Proof.

If $\alpha$ is a regular cardinal, then every subset of $\alpha$ is megaregular. Hence $A\oplus B\oplus K(U)$ is megaregular. The remaining proof of the corollary follows from 2.6. ∎

3 $\mathcal{TOT}_{\alpha e}$ as an automorphism base for $\mathcal{D}_{\alpha e}$

Definition 3.1**.**

(Automorphism base)

TFAE:

• •

The subset $B\subseteq\mathrm{dom}(\mathcal{M})$ is an automorphism base of the model $\mathcal{M}$.

• •

$\forall f,g\in\mathrm{Aut}(\mathcal{M})[f_{|B}=g_{|B}\implies f=g]$

• •

$\forall f\in\mathrm{Aut}(\mathcal{M})[f_{|B}=1_{|B}\implies f=1]$

Theorem 3.2**.**

Assume $\alpha$ is a regular cardinal. The total degrees $\mathcal{TOT}_{\alpha e}$ are an automorphism base for $\mathcal{D}_{\alpha e}$.

Proof.

0. Assume $\alpha$ is a regular cardinal.

1. $\forall a,b\in\mathcal{D}_{\alpha e}[a\leq b\iff\forall x\in\mathcal{TOT}_{\alpha e}(b\leq x\implies a\leq x)]$ by [math] and 2.7.

2. $\forall a,b\in\mathcal{D}_{\alpha e}[a=b\iff\forall x\in\mathcal{TOT}_{\alpha e}(b\leq x\iff a\leq x)]$ by 1.

3. Assume $f\in\mathrm{Aut}(\mathcal{D}_{\alpha e})$.

4. $\forall a,b\in\mathcal{D}_{\alpha e}[a\leq b\iff f(a)\leq f(b)]$ by 3.

5. Assume $\forall x\in\mathcal{TOT}_{\alpha e}.f(x)=x$.

6. Assume $y\in\mathcal{D}_{\alpha e}$.

7. $\forall x\in\mathcal{TOT}_{\alpha e}.(f(y)\leq f(x)\iff y\leq x)$ by 4.

8. $\forall x\in\mathcal{TOT}_{\alpha e}.(f(y)\leq x\iff y\leq x)$ by 5, 7.

9. $f(y)=y$ by 2, 8.

10. $\forall y\in\mathcal{D}_{\alpha e}.f(y)=y$ by 6, 9.

11. $\mathcal{TOT}_{\alpha e}$ is an automorphism base for $\mathcal{D}_{\alpha e}$ by 3, 5, 10. ∎

4 Acknowledgements

The author would like to thank Mariya Soskova for the explanation of the proof 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.