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Working in "MK-Regularity-Limitation of Size + Replacement for sets", call it the Base theory, let's coin the following axiom:

Axiom of Super-Choice:$$\forall \ relation \ R \ \exists F \subset R \ (F: dom(R) \to rng(R))$$ Where: $$R \text { is a relation } \equiv_{df} \forall r \in R \ \exists x,y \ (r=\langle x,y \rangle)\\dom(R)=\{x| \exists y (\langle x,y \rangle \in R)\} \\rng(R)=\{y|\exists x (\langle x,y \rangle \in R)\}$$

I think this would be equivalent to Global Choice over the Base theory.

Now if we weaken the above to:

Axiom of Proper Class Choice:$$\forall \ relation \ R \ \text{ with proper class rows } \\\exists F \subset R \ (F: dom(R) \to rng(R))$$

where: $$ z \text{ is a row of }R \iff \\ \exists x \in dom(R)[z=\{y| \langle x,y \rangle \in R\}]$$

A relation $R$ is with proper class rows, if every row of it is a proper class.

Would that still be equivalent to Global Choice over the Base theory?

 

If not, then if we also add Choice over sets [i.e., every set has a choice function], would that result in proving Global Choice?

Working in "MK-Regularity-Limitation of Size + Replacement for sets", call it the Base theory, let's coin the following axiom:

Axiom of Super-Choice:$$\forall \ relation \ R \ \exists F \subset R \ (F: dom(R) \to rng(R))$$ Where: $$R \text { is a relation } \equiv_{df} \forall r \in R \ \exists x,y \ (r=\langle x,y \rangle)\\dom(R)=\{x| \exists y (\langle x,y \rangle \in R)\} \\rng(R)=\{y|\exists x (\langle x,y \rangle \in R)\}$$

I think this would be equivalent to Global Choice over the Base theory.

Now if we weaken the above to:

Axiom of Proper Class Choice:$$\forall \ relation \ R \ \text{ with proper class rows } \\\exists F \subset R \ (F: dom(R) \to rng(R))$$

where: $$ z \text{ is a row of }R \iff \\ \exists x \in dom(R)[z=\{y| \langle x,y \rangle \in R\}]$$

A relation $R$ is with proper class rows, if every row of it is a proper class.

Would that still be equivalent to Global Choice over the Base theory?

 

If not, then if we also add Choice over sets [i.e., every set has a choice function], would that result in proving Global Choice?

Working in "MK-Regularity-Limitation of Size + Replacement for sets", call it the Base theory, let's coin the following axiom:

Axiom of Super-Choice:$$\forall \ relation \ R \ \exists F \subset R \ (F: dom(R) \to rng(R))$$ Where: $$R \text { is a relation } \equiv_{df} \forall r \in R \ \exists x,y \ (r=\langle x,y \rangle)\\dom(R)=\{x| \exists y (\langle x,y \rangle \in R)\} \\rng(R)=\{y|\exists x (\langle x,y \rangle \in R)\}$$

I think this would be equivalent to Global Choice over the Base theory.

Now if we weaken the above to:

Axiom of Proper Class Choice:$$\forall \ relation \ R \ \text{ with proper class rows } \\\exists F \subset R \ (F: dom(R) \to rng(R))$$

where: $$ z \text{ is a row of }R \iff \\ \exists x \in dom(R)[z=\{y| \langle x,y \rangle \in R\}]$$

A relation $R$ is with proper class rows, if every row of it is a proper class.

Would that still be equivalent to Global Choice over the Base theory?

If not, then if we also add Choice over sets [i.e., every set has a choice function], would that result in proving Global Choice?

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Zuhair Al-Johar
Zuhair Al-Johar
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Working in "MK-Regularity-Limitation of Size + Replacement for sets", call it the Base theory, let's coin the following axiom:

Axiom of Super-Choice:$$\forall \ relation \ R \ \exists X \subset R \ (X: dom(R) \to rng(R))$$$$\forall \ relation \ R \ \exists F \subset R \ (F: dom(R) \to rng(R))$$ Where: $$R \text { is a relation } \equiv_{df} \forall r \in R \ \exists x,y \ (r=\langle x,y \rangle)\\dom(R)=\{x| \exists y (\langle x,y \rangle \in R)\} \\rng(R)=\{y|\exists x (\langle x,y \rangle \in R)\}$$

I think this would be equivalent to Global Choice over the Base theory.

Now if we weaken the above to:

Axiom of Proper Class Choice:$$\forall \ relation \ R \ \text{ with proper class rows } \\\exists X \subset R \ (X: dom(R) \to rng(R))$$$$\forall \ relation \ R \ \text{ with proper class rows } \\\exists F \subset R \ (F: dom(R) \to rng(R))$$

where: $$ z \text{ is a row of }R \iff \\ \exists x \in dom(R)[z=\{y| \langle x,y \rangle \in R\}]$$

A relation $R$ is with proper class rows, if every row of it is a proper class.

Would that still be equivalent to Global Choice over the Base theory?

If not, then if we also add Choice over sets [i.e., every set has a choice function], would that result in proving Global Choice?

Working in "MK-Regularity-Limitation of Size + Replacement for sets", call it the Base theory, let's coin the following axiom:

Axiom of Super-Choice:$$\forall \ relation \ R \ \exists X \subset R \ (X: dom(R) \to rng(R))$$ Where: $$R \text { is a relation } \equiv_{df} \forall r \in R \ \exists x,y \ (r=\langle x,y \rangle)\\dom(R)=\{x| \exists y (\langle x,y \rangle \in R)\} \\rng(R)=\{y|\exists x (\langle x,y \rangle \in R)\}$$

I think this would be equivalent to Global Choice over the Base theory.

Now if we weaken the above to:

Axiom of Proper Class Choice:$$\forall \ relation \ R \ \text{ with proper class rows } \\\exists X \subset R \ (X: dom(R) \to rng(R))$$

where: $$ z \text{ is a row of }R \iff \\ \exists x \in dom(R)[z=\{y| \langle x,y \rangle \in R\}]$$

A relation $R$ is with proper class rows, if every row of it is a proper class.

Would that still be equivalent to Global Choice over the Base theory?

If not, then if we also add Choice over sets [i.e., every set has a choice function], would that result in proving Global Choice?

Working in "MK-Regularity-Limitation of Size + Replacement for sets", call it the Base theory, let's coin the following axiom:

Axiom of Super-Choice:$$\forall \ relation \ R \ \exists F \subset R \ (F: dom(R) \to rng(R))$$ Where: $$R \text { is a relation } \equiv_{df} \forall r \in R \ \exists x,y \ (r=\langle x,y \rangle)\\dom(R)=\{x| \exists y (\langle x,y \rangle \in R)\} \\rng(R)=\{y|\exists x (\langle x,y \rangle \in R)\}$$

I think this would be equivalent to Global Choice over the Base theory.

Now if we weaken the above to:

Axiom of Proper Class Choice:$$\forall \ relation \ R \ \text{ with proper class rows } \\\exists F \subset R \ (F: dom(R) \to rng(R))$$

where: $$ z \text{ is a row of }R \iff \\ \exists x \in dom(R)[z=\{y| \langle x,y \rangle \in R\}]$$

A relation $R$ is with proper class rows, if every row of it is a proper class.

Would that still be equivalent to Global Choice over the Base theory?

If not, then if we also add Choice over sets [i.e., every set has a choice function], would that result in proving Global Choice?

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Zuhair Al-Johar
Zuhair Al-Johar
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Is Proper Class Choice equivalent to Global Choice?

Working in "MK-Regularity-Limitation of Size + Replacement for sets", call it the Base theory, let's coin the following axiom:

Axiom of Super-Choice:$$\forall \ relation \ R \ \exists X \subset R \ (X: dom(R) \to rng(R))$$ Where: $$R \text { is a relation } \equiv_{df} \forall r \in R \ \exists x,y \ (r=\langle x,y \rangle)\\dom(R)=\{x| \exists y (\langle x,y \rangle \in R)\} \\rng(R)=\{y|\exists x (\langle x,y \rangle \in R)\}$$

I think this would be equivalent to Global Choice over the Base theory.

Now if we weaken the above to:

Axiom of Proper Class Choice:$$\forall \ relation \ R \ \text{ with proper class rows } \\\exists X \subset R \ (X: dom(R) \to rng(R))$$

where: $$ z \text{ is a row of }R \iff \\ \exists x \in dom(R)[z=\{y| \langle x,y \rangle \in R\}]$$

A relation $R$ is with proper class rows, if every row of it is a proper class.

Would that still be equivalent to Global Choice over the Base theory?

If not, then if we also add Choice over sets [i.e., every set has a choice function], would that result in proving Global Choice?