A question on a<b

[Anton A. Ermolenko]

Originally posted by Organic
Dear Anton A. Ermolenko,

By [null] do you mean that you accept my point of view
which is:

An analogy: [null] is like an empty sheet of paper with no written thing on it.


Organic

No, I don't.
My post is there only because of a technical problem.

 

[Organic]

In my opinion, the empty set is the building block of the idea of the set.

Let us call it a zero-order logic object.


A zero-order logic is based on one and only one sentence which is:

"There is x".

In the case of the idea of the set, x is the empty set.


Orgnic

 

[Anton A. Ermolenko]

Originally posted by Organic
In my opinion, the empty set is the building block of the idea of the set.

Let us call it a zero-order logic object.

Why do you need another axiom? Don't you think that we could do with existing ones?

Originally posted by Organic
A zero-order logic is based on one and only one sentence which is:

"There is x".

In the case of the idea of the set, x is the empty set.

In the case of the idea of the set, x is only an uncertain set, nothing else.

 

[Hurkyl]

I've been keeping tabs on this thread... but I can't seem to figure out what your point is, Anton, can you enlighten me?


Some comments on the empty set that haven't been mentioned:

Given the existence of any set S, the ZF axiom of subsets can be used to construct the empty set:

ExAy:(y in x) <=> (y in S and ~(y in S))


It's easy enough to prove that a set constructed this way is unique and doesn't depend on the identity of S. I'll call the empty set 0 because I don't know the code to give me the empty set symbol.

It is easy enough to show that 0, constructed this way, satisfies, for any set T, 0 U T = T, 0 x T = 0, and it is also easy enough to show that 0 is the unique set satisfying these two properties. In fact, in ZF, the four statements:

z = 0
Ay: ~(y in z)
AT: z U T = T
AT: z x T = z

are equivalent; if one holds for a set z, then they all hold.


Why include the axiom of the empty set if you can use the axiom of subsets to construct it? Minimalism. In order to get started, the existence of at least one set must be postulated. Because no matter what set gets postulated one can construct the empty set, the most minimal postulate you can state is that the empty set exists.


Oh, and anyone have an online reference to the axioms of NBG?

 

[Anton A. Ermolenko]

Originally posted by Hurkyl
I've been keeping tabs on this thread... but I can't seem to figure out what your point is, Anton, can you enlighten me?


Some comments on the empty set that haven't been mentioned:

Given the existence of any set S, the ZF axiom of subsets can be used to construct the empty set:

ExAy:(y in x) <=> (y in S and ~(y in S))


It's easy enough to prove that a set constructed this way is unique and doesn't depend on the identity of S. I'll call the empty set 0 because I don't know the code to give me the empty set symbol.

It is easy enough to show that 0, constructed this way, satisfies, for any set T, 0 U T = T, 0 x T = 0, and it is also easy enough to show that 0 is the unique set satisfying these two properties. In fact, in ZF, the four statements:
Why include the axiom of the empty set if you can use the axiom of subsets to construct it? Minimalism. In order to get started, the existence of at least one set must be postulated. Because no matter what set gets postulated one can construct the empty set, the most minimal postulate you can state is that the empty set exists.

Of course! Bingo! And I think so, too! Maybe I was not clear enough. My point is that ZF exclude an axiom of the empty set because the empty set concept is a theorem within the framework of ZF, not an axiom. But I asked Organic to give the empty set definition of his own, because the ZF empty set definition doesn't allow to make the deductions he has made.

Originally posted by Hurkyl z = 0
Ay: ~(y in z)
AT: z U T = T
AT: z x T = z

are equivalent; if one holds for a set z, then they all hold.[/B]

Could you specify what you mean? And where are the contradictions with my definitions?

 

[Hurkyl]

y point is that ZF exclude an axiom of the empty set because the empty set concept is a theorem within the framework of ZF, not an axiom.

That is not the case!

The theorem is that if there exists a set, there exists an empty set. IOW:
(Ex) => (EyAz:!(z is in y))


The catch is that the axiom of the empty set is the only ZFC axiom capable of proving Ex. (Note: you can't even state the axiom of infinity unless the empty set is known to exist)


Let me restate this to emphasize what this means:

Without the axiom of the empty set, ZFC cannot prove the existence of even a single set.

Could you specify what you mean? And where are the contradictions with my definitions?

I am asserting:

Az: [ (z = 0) <=> (Ay: ~(y in z)) <=> (AT: z U T = T) <=> (AT: z x T = z) ]

(where 0 is as I defined it in my previous post)

I'm not sure why you ask about constradictions with your definitions; I think your definition (if I interpret you correctly) is equivalent to the usual definition, but is redundant, complicated, and confusing.

 

[Anton A. Ermolenko]

Originally posted by Hurkyl
That is not the case!

The theorem is that if there exists a set, there exists an empty set. IOW:
(Ex) => (EyAz:!(z is in y))


The catch is that the axiom of the empty set is the only ZFC axiom capable of proving Ex. (Note: you can't even state the axiom of infinity unless the empty set is known to exist)


Let me restate this to emphasize what this means:

Without the axiom of the empty set, ZFC cannot prove the existence of even a single set.





I am asserting:

Az: [ (z = 0) <=> (Ay: ~(y in z)) <=> (AT: z U T = T) <=> (AT: z x T = z) ]

(where 0 is as I defined it in my previous post)

I'm not sure why you ask about constradictions with your definitions; I think your definition (if I interpret you correctly) is equivalent to the usual definition, but is redundant,
complicated, and confusing.

Once more, which of my definitions (formulas) exactly redundant and complicated? The emty set is only a formula, not an axiom (prove me that my definition of FORMULA of empty set is an axiom; the formulas !x=Æ and z=Æ (in Z7 and Z8) can be substituted with full definition of the empty set... am I worng? Just prove it!).
http://physics.nad.ru/img/Sets.gif

 

[Organic]

Anton A. Ermolenko,

Please answer to a simple question.

Who needs who to exist ?

Does the formula need the empty set, or the empty set needs the formula ?

I say: There is a set (a collector) with no content.

I don't need any formula to say that, because it is minimal and clear.

From this point any statement or formula that you give, will be more complicated then what I said, therefore cannot be in the same minimal level.

Therefore, you have no argument.



Organic

 

[Hurkyl]

Once more, which of my definitions (formulas) exactly redundant and complicated?

Specifically:
z = 0 is defined as
Au(u U z = u & u x z = z & Ev(v = z <=> (v x z = z & v U z = z) & AyAx (y * x = z <=> At(t in z <=> (t not in x & t not in y))))

(where * means set intersection)

is redundant and complicated.

Incidentally, I think it's complicated enough that you didn't even write down your intent correctly; "Ev" should have been "Av".


This formula is equivalent to each of the following statements:

(z = 0) := Au(u U z)
(z = 0) := Au(u x z = z)
(z = 0) := Au(~(u in z))

Your definition is redundant because you only need to have one of these three terms, and then the rest of your definition could be proven as a theorem.


And, as per the observation I made earlier, without the axiom of the empty set, 0 cannot be a formula. You cannot prove 0 exists, so you cannot use it as a constant. The basic formula involving 0 is the relation z = 0.

BTW, your page doesn't specify what Inf(x) is.

 

[Hurkyl]

I say: There is a set (a collector) with no content.

That is (directly translatable into) a formula:

EzAy: ~(y is in z)

There exists a z such that for all y, y is not in z.

 

[Anton A. Ermolenko]

Originally posted by Organic
Anton A. Ermolenko,

Please answer to a simple question.

Who needs who to exist ?

Does the formula need the empty set, or the empty set needs the formula ?

I say: There is a set (a collector) with no content.

I don't need any formula to say that, because it is minimal and clear.

From this point any statement or formula that you give, will be more complicated then what I said, therefore cannot be in the same minimal level.

Therefore, you have no argument.

Originally posted by Hurkyl

Specifically:
z = 0 is defined as
Au(u U z = u & u x z = z & Ev(v = z <=> (v x z = z & v U z = z) & AyAx (y * x = z <=> At(t in z <=> (t not in x & t not in y))))

(where * means set intersection)

is redundant and complicated.

Incidentally, I think it's complicated enough that you didn't even write down your intent correctly; "Ev" should have been "Av".


This formula is equivalent to each of the following statements:

(z = 0) := Au(u U z)
(z = 0) := Au(u x z = z)
(z = 0) := Au(~(u in z))

Your definition is redundant because you only need to have one of these three terms, and then the rest of your definition could be proven as a theorem.


And, as per the observation I made earlier, without the axiom of the empty set, 0 cannot be a formula. You cannot prove 0 exists, so you cannot use it as a constant. The basic formula involving 0 is the relation z = 0.

BTW, your page doesn't specify what Inf(x) is.

Well... one and the last time. The set theory operates sets by using operations such as x - direct product of sets, Å - direct addition of sets, U - join of sets, Ï - intersection of sets. The main requirement is that the results of all the above operations with sets are sets as well. Only this requrement makes the set theory use the empty set concept, therefore the empty set can only be defined by non-empty sets and operations. NOTHING ELSE! BECAUSE THE EQUALITY OR UNEQUALITY TO THE EMPTY SET (=Æ OR !=Æ) ARE ONLY FORMULAS. THE ICON Æ WITHOUT "=", "!=", "Î" or "Ï" MAKES NO SENSE. And this is really for the last time.

 

[Hurkyl]

Or...

0 is a constant in language of formal set theory that represents the empty set.


(incidentally, none of your math symbols appear as math symbols on my computer)


And I will repeat that without an axiom explicity specifying that some set exists, no set can be proven to exist... the existence of at least one set is a crutial part of the proof that (most of) the rest of mathematics is consistent relative to ZFC, so it really serve a practical purpose.

 

[Organic]

Hi Hurkyl,

That is (directly translatable into) a formula:

There exists a z such that for all y, y is not in z.

By using at least two variables (in this case z and y) we need some formula to describe the relations between them.

No set can be separated from the property of its content, therefore
we have an interesting situation here.

On one hand a collector can exist with no content, but on the other hand its property is depended on the property of its content.

But we also know that the content concept can't exist without a collector.

To define the exact definition of an existing thing z(a collector), is not in the same level of to define the existence of y(a content).

So z can exist with no clear property, but y can't exist at all without z.

Please tell me how Math language deals with these distinguished two levels.

If we say "There is a collector" , do you think that we can come to the conclusion that it has no content, as its minimal existence?



Thank you.



Organic