Deeper than primes

[HatRack]

The axiom of the empty set uses the definition of being a set with no members.

Correct.

Yet one of the members must be the empty set, otherwise the case {{}} is not covered by this axiom, and the empty set does not exist (as the axiom claims).

Okay, I finally see the mistake that you are making, and it's even dumber than I initially thought, which is why I failed to decipher it in the first place. Read this statement carefully: the axiom of the empty set asserts nothing about {{}}. You need another ZFC axiom, in addition to axiom of the empty set, to deduce the existence of {{}}. I bet you can't figure out which one.

Furthermore a set can't be a factor of its own terms of existence, and in this case the term is that {} is not a member of {}, which is a circular reasoning.

I already explained to you why this particular line of reasoning is asinine. Hint: read the reply where I used the word asinine.

 

[zooterkin]

No, all this time I meat closed minds like you, that no matter how many times they observe it, that can't can the non-local property of a stick, which is not less than at AND not at a given circle drown on the floor.

There are no parts here.

Um, what?

 

[doronshadmi]

the axiom of the empty set asserts nothing about {{}}

It asserts about sets that are not its members. If one of the sets that are not the members of the empty set is the empty set, then a circular reasoning is used, because we cannot determine something about the empty set by using the empty set.

In order to avoid this circular reasoning the case where the empty set is not the member of the empty set, is omitted, but then the axiom does not cover {{}} case and the empty set does not exist.

 

[doronshadmi]

Um, what?

The stick is at AND not at the given circle exactly because it is a non-local object that has no parts.

 

[HatRack]

It asserts about sets that are not its members.

No. It does not say anything about any other set but the empty set. The assertion that the empty set has no elements is a property of the empty set alone.

Definition: The empty set is the set that has no elements.

You're not understanding the meaning of the implied universal quantifier here. This definition is really saying: "for all sets X, X is not an element of the empty set". Using the universal quantifier to say "for all sets X" is equivalent to saying "if a set X exists". Worded differently:

Definition: The empty set is the set such that if a set X exists then X is not an element of the empty set.

This is just a definition. There's nothing circular about it. Now for the axiom:

Axiom: The empty set exists.

Worded differently, with the definition of empty set substituted, and letting {} denote the empty set:

Axiom: The set {}, such that if a set X exists then X is not an element of {}, exists.

Now, the axiom tells us that {} exists: the set that does not contain a set X if X exists.

At this point, we only know that one set, {}, exists. And by definition of {}, {} is not an element of {} (let X = {} to obtain this statement).

Because we only know the existence of one set ({}) at this point, it is invalid logic to deduce a proposition involving any other set like you keep doing. The only useful proposition you can possibly deduce at this point is the one I just did about {} not being an element of itself.

There is no contradiction here for the reason that the definition of the empty set only asserts something about sets that are already known to exist. It does not say anything about sets that are not known to exist in the axiomatic system at this point, including the set {{}}.

I cannot put it more plainly Doron. Either you get it, or you don't. I'm not the one who came up with this stuff, centuries of ingenious mathematicians did. Either every mathematician in the world is wrong, or you simply can't understand something as basic as the Axiom of the Empty Set.

 

[doronshadmi]

Axiom: The set {}, such that if a set X exists then X is not an element of {}, exists.

Now, the axiom tells us that {} exists: the set that does not contain a set X if X exists.

At this point, we only know that one set, {}, exists. And by definition of {}, {} is not an element of {} (let X = {} to obtain this statement).

But if {} exists, then it can't be used as one of the sets that are not its members, because then you are using {} in order to determine its emptiness by avoiding {} to be a member of {}. This self avoidance is the guarantee that {} is empty, and it is defiantly a circular reasoning, because we can't use {} as the guarantee that {} is empty (where the guarantee is: "{} is not a member of {}").

Things ars not changed even if {} is one of the rest of the sets that are not the members of {}.

 

[zooterkin]

But if {} exists, then it can't be used as one of the sets that are not its members, because then you are using {} in order to determine its emptiness by avoiding {} to be a member of {}. This self avoidance is the guarantee that {} is empty, and it is defiantly a circular reasoning, because we can't use {} as the guarantee that {} is empty.

Things ars not changed even if {} is one of the rest of the sets that are not the members of {}.

No. I see circular reasoning is another thing you don't understand.

 

[HatRack]

But if {} exists, then it can't be used as one of the sets that are not its members, because then you are using {} in order to determine its emptiness by avoiding {} to be a member of {}. This self avoidance is the guarantee that {} is empty, and it is defiantly a circular reasoning, because we can't use {} as the guarantee that {} is empty.

No Doron. The existence of {} implies that {} is not a member of {}, not the other way around. You have it ass backwards. Here is the argument in an even more basic form, with (1) the definition, (2) the axiom, and (3) the conclusion:

(1) {} is the set such that if the set X exists then X is not its member.
(2) {} exists.
(3) Therefore, {} is not a member of {}.

(1) is a definition. There's no arguing with definitions, it is simply a short representation of something that you commonly say.

(2) is an axiom. An axiom can say whatever it wants, it needs no justification. Of course, an axiom gets judged on other grounds: whether it is consistent with other axioms and whether it leads to anything useful. Since this is a ZFC axiom, it is not known to be inconsistent and it is known to, when combined with other axioms, lead to many useful things.

Perhaps you need to master more basic logical deductions before you can understand the Axiom of the Empty Set. Try this one:

(1) All men are mortal.
(2) Socrates is a man.
(3) Therefore, Socrates is mortal.

 

[doronshadmi]

Now, the axiom tells us that {} exists: the set that does not contain a set X if X exists.

You still do not get that the existence of {} as empty set can't be determined by {} (such that {} is not a member of {}), because in this case the empty set is empty because the empty set is not the member of the empty set, which is a circular reasoning.

You can't use X in order to determine the property of its own existence, as a part of the axiom that determine its existence.

 

[doronshadmi]

The existence of {} implies that {} is not a member of {},

I disagree with you, the existence of X (where its property is essential to its existence) can't be defined by X (by the same property which is essential to its existence), as a part of the axiom that determine the existence of X, otherwise we are using X in order to determine the existence of X, which is a circular reasoning.

 

[doronshadmi]

(1) {} is the set such that if the set X exists then X is not its member.

Set {} exists according to some properties, which can't be used by the set that has the same properties.

If you do that you get this: {} exists if {} exists.

 

[doronshadmi]

(1) All men are mortal.
(2) Socrates is a man.
(3) Therefore, Socrates is mortal.

(1) All ZF members are sets.
(2) X is a ZF member.
(3) Therefore X is a set.
(4) If X is a set according to an axiom, then the properties of that set

 

[zooterkin]

You still do not get that the existence of {} as empty set can't be determined by {} (such that {} is not a member of {}), because in this case the empty set is empty because the empty set is not the member of the empty set, which is a circular reasoning.

No, it isn't, it's the definition.

You can't use X in order to determine the property of its own existence, as a part of the axiom that determine its existence.

Just as well that's not what is being done, then.

{} is the empty set. {{}} is not an empty set, it is a set with the empty set as a member. Where is the contradiction?

 

[HatRack]

You still do not get that the existence of {} as empty set can't be determined by {} (such that {} is not a member of {}), because in this case the empty set is empty because the empty set is not the member of the empty set, which is a circular reasoning.

The only one using circular reasoning here is you. And ironically, you don't even understand what circular reasoning is.

You can't use X in order to determine the property of its own existence, as a part of the axiom that determine its existence.

I can do whatever I want in an axiom, as long as a contradiction cannot be derived.

I disagree with you, the existence of X (where its property is essential to its existence) can't be defined by X (by the same property which is essential to its existence), as a part of the axiom that determine the existence of X, otherwise we are using X in order to determine the existence of X, which is a circular reasoning.

There's no disagreement in math: either you're right or you're wrong. If you can't understand why you're wrong, then that's your problem.

Set {} exists according to some properties, which can't be used by the set that has the same properties.

If you do that you get this: {} exists if {} exists.

Complete nonsense, as I've proven with a simple syllogism, which you can't even understand.

(1) All ZF members are sets.
(2) X is a ZF member.
(3) Therefore X is a set.
(4) If X is a set according to an axiom, then the properties of that set

And what is this supposed to prove?

 

[doronshadmi]

I can do whatever I want in an axiom, as long as a contradiction cannot be derived.

Or circular reasoning is not used.

And what is this supposed to prove?

(1) All ZF members are sets.
(2) X is a ZF member.
(3) Therefore X is a set.
(4) If X is determined an axiom, then the properties of X can't be determined by itself.

For example: Axiom A determines the existence of a set that has no members. No members is an essential property of this existence. If one uses this kind of existence in order to determine this kind of existence by the set, then one uses a circular reasoning as a part of the determination of axiom A.

 

[HatRack]

Or circular reasoning is not used.

There's no such thing as circular reasoning in an axiom. An axiom is an assumption, you don't need any type of reasoning to deduce it. Is it really that hard to understand?

(1) All ZF members are sets.
(2) X is a ZF member.
(3) Therefore X is a set.
(4) If X is determined an axiom, then the properties of X can't be determined by itself.

For example: Axiom A determines the existence of a set that has no members. No members is an essential property of this existence. If one uses this kind of existence in order to determine this kind of existence by axiom A, the one uses a circular reasoning as a part of the determination of axiom A.

Here you go again, trying to smuggle your false conclusion in as an axiom.

 

[doronshadmi]

There's no such thing as circular reasoning in an axiom. An axiom is an assumption, you don't need any type of reasoning to deduce it. Is it really that hard to understand?

No, an axiom is a determination of things, and it is not allowed to use the determined things to determine themselves under a given axiom, otherwise we are using a circular reasoning.

Here you go again, trying to smuggle your false conclusion in as an axiom.

I made a mistake in this part, here is the correct one:

(4) If X is determined by an axiom, then the properties of X can't be determined by itself.

For example: Axiom A determines the existence of a set that has no members. No members is an essential property of its existence. If one uses this kind of existence in order to determine this kind of existence by that set (this set is not a member of itself), then one uses a circular reasoning as a part of the determination of axiom A.

 

[jsfisher]

See what I mean about the Axiom of the Empty set? I told you so.

Keep in mind, Doronetics is founded on the mathematical and social skills of a 5-year-old, including the playground taunts.

 

[doronshadmi]

See what I mean about the Axiom of the Empty set? I told you so.

See what? that a mathematician thinks that he/she avoids circular reasoning just because he/she ignores it?

 

[jsfisher]

By the way:

Emptiness is that has no predecessor. [sic]​

In other words:

Emptiness is a thing that has no thing less than it.​

Must not emptiness be one of the things that isn't less than emptiness? Is this circular reasoning? OMG!!!!