# A question on a<b

Hurkyl
Staff Emeritus
Gold Member
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 existance 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.

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 existance 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!).

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Anton A. Ermolenko,

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

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Hurkyl
Staff Emeritus
Gold Member
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
Staff Emeritus
Gold Member
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.

Originally posted by Organic
Anton A. Ermolenko,

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.

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Hurkyl
Staff Emeritus
Gold Member
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 existance 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.

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

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