The Grades of Initiation

[Daniel]

12 minutes ago, whocoulditbe? said:

I'm saying ∃ isn't enough. You need ∀ to cover A={} B={}. If you prove that, then I'm forced to pick between {} ⊆ {} and {} ∩ {} = {}. For now I maintain that both are true. It now occurs to me that proving ∀A: A ⊆ A => A ∩ A ≠ {} would also be enough to do the trick for what you're arguing.

 

Fair is fair, I answered your question, please answer my question:

 

If {} is disjoint from itself, {} ⊆ {} is true or false?  If it's true then it cannot be disjoint from itself.  If it's false it cannot be s subset of every set.

 

Pick one!  Is it true or false?

 

1 hour ago, whocoulditbe? said:

PS, Sorry if I'm being a bit aggressive in my replies. It's kind of fun to be able to talk about math on here.

 

Perfectly fine.  please answer my question:

[Daniel]

1 minute ago, whocoulditbe? said:

Thats what ∀ means...

 

I know!

[whocoulditbe?]

5 minutes ago, Daniel said:

{} ⊆ {}

is true, and {} ∩ {} = {}, i.e. it is disjoint from itself.

Edited September 14, 2023 by whocoulditbe?
i.e.

[Daniel]

2 hours ago, whocoulditbe? said:

I should have been more specific:
Please show that ∀A, B: A ⊆ B => A ∩ B ≠ {}.

 

Ahhhhh!  I missed this reply!

 

No Problem!

 

Per the truth table P => Q, it's only false if A ⊆ B is true AND A ∩ B ≠ {} is false!  So, my strategy first would be to prove that A ∩ B ≠ {} canot be false by contradiction, meaning it's always and forever true!  A ⊆ B, true/false is completely irrelevant.  Like I said => is the weakst stanadard for truth.

 

I can probably do that.

[Daniel]

5 minutes ago, whocoulditbe? said:

is true, and {} ∩ {} = {}, i.e. it is disjoint from itself.

 

Can't be, that contradicts the defintion of 'disjointed'.

[whocoulditbe?]

1 minute ago, Daniel said:

Ahhhhh!  I missed this reply!

All the funnest arguments are the result of a trivial miscommunication!

 

Quote

I can probably do that.

I'm interested to see the strategy even though I think it will turn out to be false

 

2 minutes ago, Daniel said:

Can't be, that contradicts the defintion of 'disjointed'.

A and B are disjoint if A ∩ B = {}. What definition are you using?

[Daniel]

6 minutes ago, whocoulditbe? said:

is true, and {} ∩ {} = {}, i.e. it is disjoint from itself.

 

I already posted this, did I miss your reply?

 

A ∩ A = A is not disjointed,

B ∩ B = B is not disjointed,

C ∩ C = C is not disjointed

 

{} ∩ {} = {} is not disjointed

 

etc....

 

unless ∩ has a different meaning when used with {}... just like ∩....  

[Daniel]

2 minutes ago, whocoulditbe? said:

All the funnest arguments are the result of a trivial miscommunication!

 

What I brought was true!

 

4 minutes ago, whocoulditbe? said:

A and B are disjoint if A ∩ B = {}. What definition are you using?

 

 

[whocoulditbe?]

5 minutes ago, Daniel said:

 

I already posted this, did I miss your reply?

 

A ∩ A = A is not disjointed,

B ∩ B = B is not disjointed,

C ∩ C = C is not disjointed

 

{} ∩ {} = {} is not disjointed

 

etc....

 

unless ∩ has a different meaning when used with {}... just like ∩....  

That's not a definition. The pattern S ∩ S = S only has one set, but disjointedness concerns two.

Edited September 14, 2023 by whocoulditbe?
typo

[Daniel]

sorry, emergency at the office, I have to run away now....

[whocoulditbe?]

See you. Hope it's nothing too bad.

[Daniel]

2 hours ago, whocoulditbe? said:

See you. Hope it's nothing too bad.

 

Nope.  Not too bad.  Boss calls and says "I only have a few seconds, we can't take in any payments. ~click~"  I heard "patients".  Not taking in any "payments" is much much less of a problem.  However, once I was there, it was a lot of "... oh!  while you're here..."

 

"∀A, B: A ⊆ B => A ∩ B ≠ {}."

 

OK!  I had time to ponder.  And this ^^ doesn't accomplish anything in regards to the question I asked.  I asked:

 

If {} is disjoint from itself, {} ⊆ {} is true or false?  If it's true then it cannot be disjoint from itself.  If it's false it cannot be s subset of every set.

 

---------------------------------------------------------------------------------------------------------------------------------------------------------------------

 

Case 1:  A and B are disjointed.

 

The most important question, due to the truth table for entailment is,  can A ∩ B ≠ {} ever be false?  Answer:  of course!!!  Not sure why I didn't realize that immediately, but, anyways.  It always false if A and B are disjointed.  So... now I need to go back and consider A ⊆ B.  If A, and B are disjointed, then, A ⊆ B, is always false.

 

Returning to P and Q from before:

 

Premise 'P':  A ⊆ B

Premise "Q": A ∩ B ≠ {}

 

Truth Table for P "=>" Q

 

P | Q | True/False

T | T | T

T | F | F

F | T | T

F | F | T

 

If A & B are disjointed then, P is false and Q is false.

 

P => Q | True/False

F   | F   | True

 

Case 1: A & B are disjointed:  A ⊆ B => A ∩ B ≠ {} is True.

 

---------------------------------------------------------------------------------------------------------------------------------------------------------------------

 

Case 2:  A and B are not disjointed.

 

If A & B are not disjointed then A ∩ B ≠ {} is always true AND A ⊆ B is always true.

 

Premise 'P':  A ⊆ B

Premise "Q": A ∩ B ≠ {}

 

Truth Table for P "=>" Q

 

P | Q | True/False

T | T | T

T | F | F

F | T | T

F | F | T

 

If A & B are disjointed then, P is true and Q is true.

 

P => Q | True/False

T   | T   | True

 

Case 2: A & B are NOT disjointed:  A ⊆ B => A ∩ B ≠ {} is True.

 

---------------------------------------------------------------------------------------------------------------------------------------------------------------------

 

Therefore "∀A, B: A ⊆ B => A ∩ B ≠ {}." is always and forever true.

 

A and B can be disjoint or not.

 

---------------------------------------------------------------------------------------------------------------------------------------------------------------------

 

I asked:  

 

If {} is disjoint from itself, {} ⊆ {} is true or false?  If it's true then it cannot be disjoint from itself.  If it's false it cannot be s subset of every set.

 

You answered:

 

Please show:  Please show that ∀A, B: A ⊆ B => A ∩ B ≠ {}.

 

Your answer is irrelevant to evaluating {} ⊆ {}.  Would you please answer my question now? I've been doing all the work here.  It's fun, but, throw me a bone here, will 'ya?

 

3 hours ago, whocoulditbe? said:

A and B are disjoint if A ∩ B = {}. What definition are you using?

 

I do believe it's the standard defintion.  Set A and B are disjoint if ( would prefer Iff, but, I don't make the rules ):

 

A ∉ B and B ∉ A.

 

{} ∉ {} means {} is disjoint from itself.  

 

 

[Daniel]

4 hours ago, whocoulditbe? said:

That's not a definition. The pattern S ∩ S = S only has one set, but disjointedness concerns two.

 

Important note!  I typo'd.

 

"unless ∩ has a different meaning when used with {}... just like ∩.... " I meant to copy paste subset, not intersection.

 

"unless ∩ has a different meaning when used with {}... just like ⊆..."

 

very important.

 

Remember how I said if {} is interpretted properly everything fits and plays nicely? 

 

Consider:

 

1 + ( +1 ) = 1 + 1

1 + ( - 1 ) = 1 - 1. 

1 - ( - 1 ) = 1 + 1.

 

What's happening here?

 

+ is intuitive

- is the inversion

 

1 + ( +1 ) is intuitive, it is adding.

1 + ( - 1 ) is an inversion, it is subracting.

1 - ( - 1 ) is an inversion of the inversion, it is adding.

 

A set is a collection of elements.  It is **inclusive**.  That is it's defintion.

What's {}????  It is NOT a collection of elements.  It can never ever under any circumstances possess an element.  It is the inversion of a set.  It is **exclusive**, excluding ( verb ).

 

A set includes, {} excludes.

 

OK, ok, now.... what if the same idea for " 1 + ( -1 ) " is applied to set operators with {}?  It's beautiful.  Everything fits.

 

{} ∉ {}?  What is ∉?  Is it inclusive or exclusive?  Exclusive, right?  It is saying there are no elements in {}.  {} is the inverse of a set.  it is exclusive.  Exclusive agrees and is consistent with ∉.  Therefore it is interpretted intuitively.  {} IS disjoint from {}.  

 

{} ⊆ A?  What is ⊆?  Is it inclusive or exclusive?  Inclusive, right?  It is saying there is at least 1 matching element in A.  {} is the inverse of a set.  It is exclusive.  The exclusive {} DISAGREES with the inclusive ⊆, just like the + and - disagree in " 1 + ( -1 ) ".

 

If {} ⊆ A is true, then, what if I invert its interpretation?  It's weird, but it works.  {} ⊆ A means, in english, "there are **no elements** in A other than A" because {} has no elements.  If so, can't this be said of each and every set?  

 

If I consider { 1 ,2 , 3 }, it's like I have a glass, and it is just the right size to fit { 1 , 2 , 3 } and nothing more.  When I say "nothing more", that IS {}.  "nothing more" = {}.  Can it be said that {} is a subset of every set?  Kind of. It's a subset, but not like any other subset.

 

When I say "nothing more" in english, what does that mean?

 

Doesn't it mean what I said perviously?

 

If I have a set A:{ 1 , 2 , 3 } and nothing more, then, A: { 1 , 2 , 3 , {} }. ( ignore the cardinality ).

 

A: { 1 , 2 , 3 , { ... not-1, not-2, not-3, not-4, not-5, not-6, not-7, not-8, not-9 .... } } and it goes on forever? { ... not-shirt, not-shoes, not-hat, not-gloves, not-aardvark, not-dog, not-sitting, not-flying, not-dreams, not-dreaming, not-milk, not-non-dairy-milk-eventhough-it-doesn't-exsist, ... }

 

So, is {} even a set?  Once it is considered as a subset, yeah. That's because it was put in a box. { {} }.  Before then, is it a set?

 

{ A, B , C } ∩ { 1 , 2 , 3 } = ??? <------ that's not a set.  It's not a collection.  It's a class, an action.  A never ending action.  A super simple algorithm which produces a realization of 無.

 

------------------------------------------------------------------------------------------------------------------------

 

Now the question can be answered

 

If {} is disjoint from itself, {} ⊆ {} is true or false?  If it's true then it cannot be disjoint from itself.  If it's false it cannot be s subset of every set.

 

{} ∉ {} is true!  This can be interpretted naturally.  {} is consistent with ∉; both are exclusive.  {} IS disjoint from itself.

 

{} ⊆ {} is true, BUT, it must be interpretted in the inverse.  {} is inconsistent with ⊆.  {} is exclusive, ⊆ is inclusive.  {} is NOT a subset of itself or any other set.  But, {} ⊆ {} is still true, if the quality of nullification, "nothing more, always nothing more..." is understood about it.  This means that any set, always has a subset included, it's partner, a never-ending-negation.  Including this faux-subset means "that's it, nothing more is being included".  And every set includes this as a consequence of being a collection of elements.

 

Now we can look at the examples of intersections.

 

A ∩ B = {} is disjointed

{} ∩ A = {} is ?????

 

{} is inconsistent with ∩.  {} is exclusive, ∩ is inclusive.  {} ∩ A is true, but, it means that there is NOT an element in {} which is also in A.  

 

{} ∩ A = {} is disjointed.

 

{} ∩ {} = {} is also disjointed.

 

Here's a fun one.

 

{} = {} ???? is it true or false?

Edited September 15, 2023 by Daniel

[whocoulditbe?]

13 hours ago, Daniel said:

If A, and B are disjointed, then, A ⊆ B, is always false.

I'm denying this. It's not in any axiom and you haven't shown it to be true. You've put an exact equivalent of your conclusion inside the premise. (A => B means the same thing as ~B => ~A by contraposition)

[whocoulditbe?]

13 hours ago, Daniel said:

If A & B are not disjointed then A ∩ B ≠ {} is always true AND A ⊆ B is always true.

This is plainly false. try A = {1, 2} and B = {2, 3}.

Edited September 15, 2023 by whocoulditbe?

[whocoulditbe?]

12 hours ago, Daniel said:

A set is a collection of elements.  It is **inclusive**.  That is it's defintion.

What's {}????  It is NOT a collection of elements.  It can never ever under any circumstances possess an element.  It is the inversion of a set.  It is **exclusive**, excluding ( verb ).

 

A set includes, {} excludes.

A set is something that can include. It also turns out that all sets do exclude, because there is no universal set. There is nothing especially strange about an empty box, an empty set, a silent room, or a muon no heya. You are projecting some kind of horror vacui onto set theory, yet it works perfectly well as it is.

[Daniel]

1 hour ago, whocoulditbe? said:

This is plainly false. try A = {1, 2} and B = {2, 3}.

 

Ah yes, I missed that case.  Oops, sorry, and thank you.  It still doesn't matter.  In that case, : A ⊆ B => A ∩ B ≠ {} is still true.

 

If A ⊆ B is false, then "=>" is always true in the above proposal.  But, I'll update what I wrote in a bit, to add that case.  It seems like we have 3 of the four possibilities covered from the truth table, right?  All that's missing is the elusive, P is true and Q is false.

 

P => Q | True/False

T  |  T   |  True  ( example:  A:{ 1 }  B:{ 1 , 2 } ) 

T  |  F   |  False ?????

F  |  T   |  True  ( example:  A:{ 1 ,2 }  B:{ 2 , 3 } )

F  |  F   |  True  ( example:  A:{ 1 }  B: { 2 } )

 

And this is honestly an important point.  why choose => instead of boolean AND for defintions?  It's the weakest of the weak standard for truth?

 

1 hour ago, whocoulditbe? said:

A set is something that can include. It also turns out that all sets do exclude, because there is no universal set. There is nothing especially strange about an empty box, an empty set, a silent room, or a muon no heya. You are projecting some kind of horror vacui onto set theory, yet it works perfectly well as it is.

 

That is just ZFC.  The set of all sets exists in other set theories.  I know that ZFC works.  I never said it didn't.  I brought multiple sources.  5 I think.  All are showing that {} ... drumroll  .... doesn't exist.

 

Senior PHD Physicist from Duke University explained it.  It's NOT an empty box.  That's a false analogy. I understand it as a box lacking a bottom, or a bag with a hole in the bottom, or a black hole.

 

It's interesting to me that you chose the words "horror story" to describe this.  As if this is something to be afraid of.  Again, it's like a witch trial.  I'm speaking blasphemy.  And you're crying out "burn him, burn him" like the The Holy Grail.  ( You're aware of the famous scene, right?  The witch trial is all about "logic" )

 

But I've said "everything fits and plays nicely if {} is interpretted properly".  So, what's the "horror story"?  A change in perspective?  Is that really so scary?  Really???

 

------------------------------------------------------------------------------------------------------------

 

Ask yourself this.   And maybe let's skip it for now.  If {} is a subset of itself, doesn't it suffer from the same paradox as the set of all sets?  You've repeatedly asserted "the universal set does not exist".  OK. 

 

{} ⊆ {} ?

 

Is it {}?

or is it { {} }?

or is it { { {} } }?

or is it { { { {} } } }?

or is it { { { { {} } } } }?

etc.

 

so how can {} = {} ?  it doesn't even exist?

 

If {} exists, and {} = {}, then the set of all sets exists too and russel's paradox is fallacious.

 

Unless.  Like I said before.. ⊆ is interpretted in the inverse for {} ⊆ .

 

1 hour ago, whocoulditbe? said:

I'm denying this. It's not in any axiom and you haven't shown it to be true. You've put an exact equivalent of your conclusion inside the premise. (A => B means the same thing as ~B => ~A by contraposition)

 

This is the defintion I brought for disjoint:  ( edit: here's a link to that post:  LINK )

 

A & B are disjoint if A ∉ B and B ∉ A.  

 

And I don't see why the diagrams I brought are not sufficient.  

 

Edited September 15, 2023 by Daniel

[Nungali]

I love the way Daniel previously sushhhed  people so as not to interrupt HD's thread  / stream of thought         

 

 

Sush Nungali !      ......   You are interrupting Daniel's stream of thought ! 

 

Oh ... okay then ..... maybe I am trying to make something out of  nothing  ? 

[whocoulditbe?]

16 minutes ago, Nungali said:

I love the way Daniel previously sushhhed  people so as not to interrupt HD's thread  / stream of thought         

Well, I'm also to blame after idealising slower posting here. The discussions drift so much on this forum, but the mods don't seem to be in the habit of spliting threads.

Edited September 15, 2023 by whocoulditbe?

[whocoulditbe?]

8 hours ago, Daniel said:

 

Ask yourself this.   And maybe let's skip it for now.  If {} is a subset of itself, doesn't it suffer from the same paradox as the set of all sets?  You've repeatedly asserted "the universal set does not exist".  OK. 

 

{} ⊆ {} ?

 

Is it {}?

or is it { {} }?

or is it { { {} } }?

or is it { { { {} } } }?

or is it { { { { {} } } } }?

What do you think a subset is?

Edited September 15, 2023 by whocoulditbe?
typo

[Nungali]

In a desperate attempt to get back on track .....  while answering the question ;

 

The I  to IV degrees listed in HD's  post previously . 

 

( Ie. they are a subset  .... of the greater 'triad' of three grades  )

[Daniel]

10 minutes ago, whocoulditbe? said:

What do you think a subset is?

 

it's listed somewhere in the thread.  relevance?  all that matters is, the set of all sets suffers from a paradox of self-reference.  {} suffers from the same paradox.

 

edit:  unless.... like I've been saying.  the inclusive ⊆ actually is inverted and means ... something else.  I'll leave it at that.

 

Edited September 15, 2023 by Daniel

[Daniel]

11 minutes ago, whocoulditbe? said:

What do you think a subset is?

 

{}  ⊆ {} violates the axiom of formation/regularity?

[Daniel]

This one I did double check:

 

"The axiom of regularity together with the axiom of pairing implies that no set is an element of itself"

 

https://en.m.wikipedia.org/wiki/Axiom_of_regularity

 

So, it's both.  

 

{} ⊆ {} violates the axiom of regularity together with the axiom of pairing.

 

Edited September 15, 2023 by Daniel

[Nungali]

Well, I hope that settled it .

 

a )  It's both .

b ) It's neither .

c ) It's  both and neither .

d ) It's neither both nor neither

e ) It's none of the above .

f ) It's all of the above .

g ) It's all and none of the above .

h ) It's neither all or none  of the above .