Deeper than primes

[HatRack]

At the top of the half post, there is no comparison between the members of sets.

At the bottom of the half post, there is a comparison between the members of sets.

So, in Doronetics, when you are not comparing the members of sets, 0 = {} is true. But, when you are comparing the members of sets, 0 = {} is false.

I've got to hand it to you Doron, your latest attempt to evade the existence of the empty set is probably the most foolish thing I've ever seen in writing.

 

[HatRack]

It is also fun to compare and contrast the past with the present. Consider this not so recent example:

1={{}}

2={{},{{}}}

3={{},{{}},{{},{{}}}}

...

in comparison to this:

{} ≠ 0

{{}} ≠ {0} ≠ 1

{{},{{}}} ≠ {0,1} ≠ 2

{{},{{}},{{},{{}}}} ≠ {0,1,2} ≠ 3

So Doron, how do you reconcile these two completely contradictory posts? You're going to have to decipher your own gibberish from all the way back in July of 09. Good luck!

 

[doronshadmi]

It is also fun to compare and contrast the past with the present. Consider this not so recent example:

1={{}}

2={{},{{}}}

3={{},{{}},{{},{{}}}}

...

EDIT:

So Doron, how do you reconcile these two completely contradictory posts?

I made a mistake by writing this, the right one is:


1=|{{}}|

2=|{{},{{}}}|

3=|{{},{{}},{{},{{}}}}|

...

 

[doronshadmi]

At the following part there is no comparison between the members of sets (there is only a comparison between cardinality's value):

|{}| = 0

|{{}}| = |{0}| = 1

|{{},{{}}}| = |{0,1}| = 2

|{{},{{}},{{},{{}}}}| = |{0,1,2}| = 3

...

At the following part there is a comparison between the members of sets and also between sets and cardinalities (which are not sets):

{} ≠ 0

{{}} ≠ {0} ≠ 1

{{},{{}}} ≠ {0,1} ≠ 2

{{},{{}},{{},{{}}}} ≠ {0,1,2} ≠ 3

...

In other words HatRack, you have no case.

 

[HatRack]

EDIT:


I made a mistake by writing this, the right one is:


1=|{{}}|

2=|{{},{{}}}|

3=|{{},{{}},{{},{{}}}}|

...

Now wait a minute, I thought the empty set didn't exist in Doronetics. Is it one of those things that exists in some sentences but not in others?

 

[doronshadmi]

Now wait a minute, I thought the empty set didn't exist in Doronetics.

I simply don't believe it, after all the posts between us you still do not get the difference between the existence of {} and the "content" of the existing {}.

EDIT: No wonder that you can't get posts like (for example) http://www.internationalskeptics.com/forums/showpost.php?p=6706832&postcount=13578 .

 

[doronshadmi]

You said sets share members with their power set.

This is what I have said:

... the set and the power set share members that are based on the same rule and there are always members of the power set that are not in the range of the set (whether X is finite or not), then no set is complete exactly because every set has a power set and every power set has also power set etc... ad infinitum ...

So you are wrong.

This is also an untrue statement. But you have no concept of what you said, do you Doron? Maybe it is not just a reading comprehension issue you continually manifest, but a more general cognition failure.

You statement quite clearly states that the power set of, for example, the set {A,B,C} is 2^|{A,B,C}| = 8. That is clearly wrong.

I swear, Doron, you have the talent to mess up a pig's sty.

The cardinality of the power set of {A,B,C} is 2^|{A,B,C}| = 8, whether you like is or not.

 

[HatRack]

I simply don't believe it, after all the posts between us you still do not get the difference between the existence of {} and the "content" of the existing {}.

EDIT: No wonder that you can't get posts like (for example) http://www.internationalskeptics.com/forums/showpost.php?p=6706832&postcount=13578 .

So {} doesn't exist, but its contents do. How about enlightening us as to exactly what the "content" of {} is.

 

[HatRack]

The cardinality of the power set of {A,B,C} is 2^|{A,B,C}| = 8, whether you like is or not.

That's not what you said here:

Actually the power set of some set is 2^(the number of the distinct objects of that set)

In this quote, you mixed power set and power set cardinality up, which is what jsfisher was referring to.

 

[jsfisher]

You said sets share members with their power set.

This is what I have said:

... the set and the power set share members that are based on the same rule and there are always members of the power set that are not in the range of the set (whether X is finite or not), then no set is complete exactly because every set has a power set and every power set has also power set etc... ad infinitum ...

So you are wrong.

See the highlighted part? You said sets and their power sets share members. You were wrong then, and you are wrong now trying to deny it.

This is also an untrue statement. But you have no concept of what you said, do you Doron? Maybe it is not just a reading comprehension issue you continually manifest, but a more general cognition failure.

You statement quite clearly states that the power set of, for example, the set {A,B,C} is 2^|{A,B,C}| = 8. That is clearly wrong.

I swear, Doron, you have the talent to mess up a pig's sty.

The cardinality of the power set of {A,B,C} is 2^|{A,B,C}| = 8, whether you like is or not.

It is not a question of what I like or dislike, merely one of what you said. You said the power set of a set, not the cardinality of it, was 2^cardinality of the set. You were wrong.

Your statements are sloppy. Your logic is sloppy. Your concepts are sloppy. It is no wonder you make so many contradictory, irrelevant, and nonsensical posts.

 

[jsfisher]

EDIT:

I made a mistake by writing this, the right one is:

1=|{{}}|

2=|{{},{{}}}|

3=|{{},{{}},{{},{{}}}}|

...

You make many mistakes:

For example:

N={ {{}}, {{},{{}}}}, {{},{{}},{{},{{}}}}, ... }

Now:

The set of natural numbers is N={ {{}}, {{},{{}}}, {{},{{}},{{},{{}}}}, ... }

The set of even numbers is E={ {{},{{}}}, {{},{{}},{{},{{}}},{{},{{}},{{},{{}}}}} ... }

Now:

The set of natural numbers is N={ {{}}, {{},{{}}}, {{},{{}},{{},{{}}}}, ... }

The set of even numbers is E={ {{},{{}}}, {{},{{}},{{},{{}}},{{},{{}},{{},{{}}}}} ... }

So, according to the new case:

The set of natural numbers is N={ {{}}, {{},{{}}}, {{},{{}},{{},{{}}}}, ... }

The set of even numbers is E={ {{},{{}}}, {{},{{}},{{},{{}}},{{},{{}},{{},{{}}}}} ... }

 

[zooterkin]

You are wrong, look:

You didn't understand what I said at all, did you?

 

[doronshadmi]

So {} doesn't exist, but its contents do.

Exactly the opposite, HatRack, exactly the opposite.

 

[epix]

It is also fun to compare and contrast the past with the present. Consider this not so recent example:

Originally Posted by doronshadmi
1={{}}

2={{},{{}}}

3={{},{{}},{{},{{}}}}

in comparison to this:

Originally Posted by doronshadmi
{} ≠ 0

{{}} ≠ {0} ≠ 1

{{},{{}}} ≠ {0,1} ≠ 2

{{},{{}},{{},{{}}}} ≠ {0,1,2} ≠ 3

= and ≠ are different tools
but explain that to the King of Fools

 

[epix]

EDIT:


I made a mistake by writing this, the right one is:


1=|{{}}|

2=|{{},{{}}}|

3=|{{},{{}},{{},{{}}}}|

...

No, you didn't, coz || is an elementary character error, but humans use the symbol to express cardinality and absolute value. Since "error" has 5 letters, and 5 decimal equals V Roman, you take the V and fit it into ||, so that

||ISTAKE becomes repaired MISTAKE.

Isn't it so, mate? Fix this, Haldegard:
||en at Work.

 

[doronshadmi]

See the highlighted part? You said sets and their power sets share members. You were wrong then, and you are wrong now trying to deny it.

You used only a part of what I wrote, so?

Here is this part without your cutting:

…the set and the power set share members that are based on the same rule and there are always members of the power set that are not in the range of the set…

If you really read it (something that you did not do) you find that:

a) The set and the power set members share the same rule of construction.

b) There are always members of the power set that are not members of the set because they are not in the range of the set.

It is not a question of what I like or dislike, merely one of what you said. You said the power set of a set, not the cardinality of it, was 2^cardinality of the set. You were wrong.

jsfisher, I show that the power set of a set is not necessarily a collection of subsets, and the construction of the members of set X and its power set 2^X are based on the same rule, even if the members of the power set are not in the range of the set.

This construction does not follow ZF because by ZF agreement also the members must be sets.

Actually according to the same rule (which does not follow the agreement of ZF), also the members of N are actually not sets as follows:
N={ |{{}}|, |{{},{{}}}|, |{{},{{}},{{},{{}}}}|, ... }, (by ZF The set of natural numbers is N={ {{}}, {{},{{}}}, {{},{{}},{{},{{}}}}, ... }).

Your statements are sloppy. Your logic is sloppy. Your concepts are sloppy. It is no wonder you make so many contradictory, irrelevant, and nonsensical posts.

Since you are closed under ZF agreement you can't see anything beyond it.

 

[doronshadmi]

You didn't understand what I said at all, did you?

You do not get the following:

For example, the set {1} is defined as the set such that 1 is its member and every other set X (including {1} itself) is not a member.

I said that 1 (if we are not following ZF) is not itself a set, so in this case there is a difference between 1 (which is not a set) and {1} (which is a set), and HatRack's argument is not equivalent to my argument about set {} that is not a member of set {} (which is a circular reasoning because set {} is a factor that deremines set {}).

 

[jsfisher]

See the highlighted part? You said sets and their power sets share members. You were wrong then, and you are wrong now trying to deny it.

You used only a part of what I wrote, so?

Here is this part without your cutting:

…the set and the power set share members that are based on the same rule and there are always members of the power set that are not in the range of the set…

If you really read it (something that you did not do) you find that:

a) The set and the power set members share the same rule of construction.

No, you clearly said "share members". That part is flatly wrong, and it makes no difference whatsoever what rule might exist by which they share members. If any such rule did exist, it would be, as is usually the case with things you post, irrelevant.

On the other hand, your attempt to rewrite history by swapping two words creates a new bit of wrongness. The rule for set membership can be arbitrary whereas the rule for membership in its power set is fixed. So, even your attempt to dodge and cover your original mistake only compounds it with an additional two.

b) There are always members of the power set that are not members of the set because they are not in the range of the set.

I never said otherwise. You, however, did.

It is not a question of what I like or dislike, merely one of what you said. You said the power set of a set, not the cardinality of it, was 2^cardinality of the set. You were wrong.

jsfisher, I show that the power set of a set is not necessarily a collection of subsets, and the construction of the members of set X and its power set 2^X are based on the same rule, even if the members of the power set are not in the range of the set.

This is again irrelevant to the point. You made a bogus statement. It was wrong. Trying to change the subject, as you are now trying to do, doesn't change the fact you made a sloppy error by omitting the phrase "cardinality of".

That said, you did not show that the power set of a set is not necessarily a collection of subsets. That is the bloody definition of a power set. Try as you may (and you do try to do this frequently), definitions are not something you can disprove. Moreover, as already stated, there is no "same rule".

 

[doronshadmi]

Every set uses itself as a factor in its definition. For example, the set {1} is defined as the set such that 1 is its member and every other set X (including {1} itself) is not a member. This type of "self-reference" is inescapable in defining sets.

OK, let us follow ZF where any member must be also a set.

1) This rule does not hold in the case of the empty set, because the "member" of {} is not a set.

2) Let us say that 0={} and 1={{}}, so in this case {1}={{{}}}.

3) Any other set (accept {{}}) is not a member of {{{}}}, so also in this case a given set is used as a factor in order to determine itself as a part of its own derermination.

So thank you HatRack, according to (1) and (3) we see that ZF is based on an invalid reasoning.

 

[doronshadmi]

No, you clearly said "share members". That part is flatly wrong,

Because you ignore the rest of what I wrote, you are really js.

On the other hand, your attempt to rewrite history by swapping two words creates a new bit of wrongness. The rule for set membership can be arbitrary whereas the rule for membership in its power set is fixed. So, even your attempt to dodge and cover your original mistake only compounds it with an additional two.

Nonsense.

The rule for set membership is understood by you only in terms of ZF where a member of a set must be also a set, so sell your "arbitrary membership" to your community, they will buy it.

This is again irrelevant to the point.

Yes I know jsfisher the relevant thing is your limited ZF agreement, which according to it, a member of a set must be also a set. Enjoy your also in 2011 with your community.

That said, you did not show that the power set of a set is not necessarily a collection of subsets.

{000, 001, 010, …} is not necessarily a collection of subsets, unless you force, for example, 000, it to be a subset by some agreement, as it was done in the case of ZF.

That is the bloody definition of a power set. Try as you may (and you do try to do this frequently), definitions are not something you can disprove. Moreover, as already stated, there is no "same rule".

Definitions are nothing but agreements, generalize the notion of some agreement and your definition is fundamentally changed.