A question about the definition of 'set'

[pwsnafu]

A is STILL a set⇔∃z(z∈A∨A=∅)

Why is "still" in all caps?
I personally do not understand why what you wrote is actually useful. A vector space is something that satisfies the vector space axioms. It's as simple as that. In today's mathematics a set is something that satisfies ZF. What more do you want?

 

[micromass]

Thank God!

but,

A is STILL a set⇔∃z(z∈A∨A=∅)

This is valid whether we're talking ZF, NBG, New Foundations or the Bhagavad Gita ...

This is not true in NBG! Consider the class of all sets, this satisfies the right-hand side of your equivalence, but it's not a set.

It is true in ZF however, but iit's trivial since everything is a set there. Why don't you just accept the fact that mathematicians don't do things your way? I'm not saying it's wrong what you say, it's just not what we do.

 

[xxxx0xxxx]

This is not true in NBG! Consider the class of all sets, this satisfies the right-hand side of your equivalence, but it's not a set.

It is true in ZF however, but it's trivial since everything is a set there. Why don't you just accept the fact that mathematicians don't do things your way? I'm not saying it's wrong what you say, it's just not what we do.

Ok rename set to class and the definition is valid... now

\mbox{A is a class} \Leftrightarrow \exists z( z \in A \vee A=\emptyset)

for NBG, then derive set from class, BTW now there is no class of classes, but get set "a" as a representation of class "A". All it is is renaming the model elements, only we use the word "class" for the word set. Now NBG is a model in ZF.

And... we can turn it around and make ZF a model in NBG.

 

[xxxx0xxxx]

Why is "still" in all caps?
I personally do not understand why what you wrote is actually useful. A vector space is something that satisfies the vector space axioms. It's as simple as that. In today's mathematics a set is something that satisfies ZF. What more do you want?

Because, ZF is not the end all, be all of set theories, there are infinities of set theories.

ZF just happens to be the one that most people use; all the other set theories can be modeled in ZF and vice versa.

 

[micromass]

Ok rename set to class and the definition is valid... now

\mbox{A is a class} \Leftrightarrow \exists z( z \in A \vee A=\emptyset)

for NBG, then derive set from class, BTW now there is no class of classes, but get set "a" as a representation of class "A". All it is is renaming the model elements, only we use the word "class" for the word set. Now NBG is a model in ZF.

And... we can turn it around and make ZF a model in NBG.

A bit of a silly definition isn't it? Once your definition defines a set, and once a class. I don't like it. And I don't think many mathematicians do...

Because, ZF is not the end all, be all of set theories, there are infinities of set theories.

ZF just happens to be the one that most people use; all the other set theories can be modeled in ZF and vice versa.

Of course not all set theory can be modeled in ZF. You really need to open a set theory book...

 

[Hurkyl]

\mbox{A is a class} \Leftrightarrow \exists z( z \in A \vee A=\emptyset)

As an aside, in NBG, both sides of the equivalence in NBG are tautological predicates of one variable (A).

 

[xxxx0xxxx]

As an aside, in NBG, both sides of the equivalence in NBG are tautological predicates of one variable (A).

Yes, the convention is usually to drop the universal quantifier...

\forall A(\mbox{A is a class} \Leftrightarrow \exists z(z \in A \vee A = \emptyset))

All definitions are tautologies.

 

[Hurkyl]

All definitions are tautologies.

No, I mean, for example,

A is a class​

is a tautology.

 

[xxxx0xxxx]

No, I mean, for example,

A is a class​

is a tautology.

Well, that may be because you are not making a distinction between object language and metalanguage.

 

[micromass]

There is no metalanguage in mathematics...

 

[disregardthat]

There is no metalanguage in mathematics...

What he is probably referring to an axiomatic formalization of the logical inferences within set theory, known as a metatheory, in which one could e.g. formally say "\phi is unprovable" for a statement \phi in set theory, whereas set theory can only say "\phi" (or \neg \phi, or any combinations of conjunctions, negations, etc.. of set theoretical statements). A metalanguage consists of statements about statements in some axiomatic setting, like set theory. Such a metalanguage is necessary to be able to prove that certain things are unprovable in set theory, such as the continuum hypothesis, so called metatheorems. It is also used to prove that the axiom of choice is logically independent from ZF.

 

[Hurkyl]

Well, that may be because you are not making a distinction between object language and metalanguage.

What you wrote was a statement in the language of NBG, and you even stated it as "for NBG". If you mean something other than what you wrote, it's up to you to make that explicitly clear. It is not up to us to divine your intent, nor even to assume your being meaningful at all.

 

[xxxx0xxxx]

What you wrote was a statement in the language of NBG, and you even stated it as "for NBG". If you mean something other than what you wrote, it's up to you to make that explicitly clear. It is not up to us to divine your intent, nor even to assume your being meaningful at all.

Well, you need not divine much,

"A" is in the object language

"is a class" is a statement in the metalanguage.

"A is a class" is well-formed-formula by virtue of the definiens:

\exists z (z \in A \vee A = \emptyset)

which is composed of nothing but elements of the object language.

 

[xxxx0xxxx]

What he is probably referring to an axiomatic formalization of the logical inferences within set theory, known as a metatheory, in which one could e.g. formally say "\phi is unprovable" for a statement \phi in set theory, whereas set theory can only say "\phi" (or \neg \phi, or any combinations of conjunctions, negations, etc.. of set theoretical statements). A metalanguage consists of statements about statements in some axiomatic setting, like set theory. Such a metalanguage is necessary to be able to prove that certain things are unprovable in set theory, such as the continuum hypothesis, so called metatheorems. It is also used to prove that the axiom of choice is logically independent from ZF.

Seems pretty straightforward to me.

ZF can serve as a metalanguage for NBG, or any other model.

 

[micromass]

Seems pretty straightforward to me.

ZF can serve as a metalanguage for NBG, or any other model.

I'd love to see some kind of proof for that...

 

[Hurkyl]

"is a class" is a statement in the metalanguage.

"is a class" is a predicate in the language of NBG. (the trivially true one, to be precise)

I have no idea how you manage to read it as a "statement" "in the metalanguage" or what you could possibly mean by such a claim.

(maybe... you just aren't familiar with forms of first-order logic that offer a richer syntax for doing routine logical tasks?)




That's not quite true -- I was wondering if you were thinking of an interpretation of NBG, and you were using "class" to refer not to the type of that name in NBG (actually, I think it's usually shortened to Cls) but some metatype you're interpreting it into. But, of course, if you were doing that "A is a class" would make no sense if A was a variable in the language of NBG.

 

[xxxx0xxxx]

I'd love to see some kind of proof for that...

"is a class" is a predicate in the language of NBG. (the trivially true one, to be precise)

I have no idea how you manage to read it as a "statement" "in the metalanguage" or what you could possibly mean by such a claim.

(maybe... you just aren't familiar with forms of first-order logic that offer a richer syntax for doing routine logical tasks?)




That's not quite true -- I was wondering if you were thinking of an interpretation of NBG, and you were using "class" to refer not to the type of that name in NBG (actually, I think it's usually shortened to Cls) but some metatype you're interpreting it into. But, of course, if you were doing that "A is a class" would make no sense if A was a variable in the language of NBG.

See the attached...

Moderator's note -- removed copyrighted material[/color]

 

[micromass]

See the attached...

I don't see how that answers my question...
Firstly, your attachment assumes the existence of a inaccessible cardinal, which is a huge assumption. In fact it cannot be proven that ZFC+inaccessible cardinal is relatively consistent with ZFC. Thus it is an assumption I don't like to see.
Furthermore, a statement in ZFC is provable iff it is provable in NBG. Thus the assumption of a large cardinal seems to be a bit too much.

Also, you said that "any set theory can be phrased in terms of ZF". I have yet to see a proof for this. How would you phrase NF or MK in terms of ZF?

 

[xxxx0xxxx]

I'd love to see some kind of proof for that...

See the attached...


http://matwbn.icm.edu.pl/ksiazki/fm/fm64/fm64126.pdf

Moderator's note: duplication of copyrighted material replaced with link to source[/color]
(article title: "Novak's result by Henkin's method")

 

[micromass]

That doesn't answer the question...

 

[xxxx0xxxx]

That doesn't answer the question...

Well,

having failed to answer your question in which you replaced my word "model" with your word "rephrased," and shown why mathematics is very much like arguing "how many angels can fit on the head of pin?", I return to my original statement to the OP:

A \mbox{ is a set} \Leftrightarrow \exists z (z \in A \vee A = \emptyset)

without any of the encumbrances peculiar to a particular theory of sets.

Best regards to all...

except those who absolutely have to have the last word...

 

[micromass]

Well,

having failed to answer your question in which you replaced my word "model" with your word "rephrased,"

OK, sorry. Then provide a proof why ZF can be seen as model of other set theories? Or whatever it is you meant.

and shown why mathematics is very much like arguing "how many angels can fit on the head of pin?"

Let me remind you how mathematics works: we work with clear definitions and we prove things. Therefore, if we make a statement, then the statement should be precise and provable. If you actually think that mathematics is like arguing "how many angels can fit on the head of a pin", then I'm sorry but you haven't understood mathematics at all.

I return to my original statement to the OP:

A \mbox{ is a set} \Leftrightarrow \exists z (z \in A \vee A = \emptyset)

Doesn't work in NBG.

except those who absolutely have to have the last word...

It's not about having the last word, it's about correct obviously false statements.

 

[Hurkyl]

Ah, we've moved onto the "try to drown them in a deluge of material without any explanation or attempting to connect it to the issue at hand" stage of crackpottery. I think now's a good time to close it.