Is this a paradox? what is 'equal'?

[wij]

https://en.wikipedia.org/wiki/Set_(mathematics)
A set is the mathematical model for a collection of different[1]
things;...

https://en.wikipedia.org/wiki/Axiom_of_extensionality
"two sets A and B are equal if and only if A and B have the same
members."...
Extensionality: ∀A∀B(∀x(x∈A <=> x∈B)) <=> A=B

Let S={ x| x is a set. ∀x,y∈S, x=y } // '=' defined by Axiom of
extensionality.

Let A={1,3,5..}, B={2,4,6,..}. A and B satisfies the property of the
axiom of
extensionality.
So, both A and B are in S by the definition of S. But the definition
of set
requires the elements be different, where 'different' is the negation
of the
axiom of extensionality (definition?). Therefore, A and B cannot both
be in S.

So, the question: Are A,B in S?

[immibis]

On 1/23/24 19:46, wij wrote:
> "two sets A and B are equal if and only if A and B have the same
> members."...
> Extensionality: ∀A∀B(∀x(x∈A <=> x∈B)) <=> A=B
>

> Let A={1,3,5..}, B={2,4,6,..}. A and B satisfies the property of the
> axiom of extensionality.

No they don't. 1 is an x which is in A but not in B.

> So, both A and B are in S by the definition of S. But the definition
> of set
> requires the elements be different, where 'different' is the negation
> of the
> axiom of extensionality (definition?). Therefore, A and B cannot both
> be in S.

A and B satisfy the negation of the axiom of extensionality so they are
different.

However, even if they were the same, then if one was in S they would
both be in S.

It is like saying:

Fred = 1
George = 1
Fred is in the set {1} and George is also in the set {1}, even though
{1} only has one element.

[wij]

On Tue, 2024-01-23 at 20:04 +0100, immibis wrote:
> On 1/23/24 19:46, wij wrote:
> > "two sets A and B are equal if and only if A and B have the same
> > members."...
> > Extensionality: ∀A∀B(∀x(x∈A <=> x∈B)) <=> A=B
> >
> > Let A={1,3,5..}, B={2,4,6,..}. A and B satisfies the property of
> > the
> > axiom of extensionality.
>
> No they don't. 1 is an x which is in A but not in B.

Idiot. Read it carefully "Let ...."

[immibis]

On 1/23/24 20:09, wij wrote:
> On Tue, 2024-01-23 at 20:04 +0100, immibis wrote:
>> On 1/23/24 19:46, wij wrote:
>>> "two sets A and B are equal if and only if A and B have the same
>>> members."...
>>> Extensionality: ∀A∀B(∀x(x∈A <=> x∈B)) <=> A=B
>>>
>>> Let A={1,3,5..}, B={2,4,6,..}. A and B satisfies the property of
>>> the
>>> axiom of extensionality.
>>
>> No they don't. 1 is an x which is in A but not in B.
>
> Idiot. Read it carefully "Let ...."

A is the set of odd natural numbers and B is the set of even natural
numbers (except for 0), right?

So 1 is the x which disproves extensionality for A and B.

[wij]

On Wed, 2024-01-24 at 03:09 +0800, wij wrote:
> On Tue, 2024-01-23 at 20:04 +0100, immibis wrote:
> > On 1/23/24 19:46, wij wrote:
> > > "two sets A and B are equal if and only if A and B have the same
> > > members."...
> > > Extensionality: ∀A∀B(∀x(x∈A <=> x∈B)) <=> A=B
> > >
> > > Let A={1,3,5..}, B={2,4,6,..}. A and B satisfies the property of
> > > the
> > > axiom of extensionality.
> >
> > No they don't. 1 is an x which is in A but not in B.
>
> Idiot. Read it carefully "Let ...."

Sorry! Looks like I am the idiot!!

[Mike Terry]

On 23/01/2024 18:46, wij wrote:
> https://en.wikipedia.org/wiki/Set_(mathematics)
> A set is the mathematical model for a collection of different[1]
> things;...
>
> https://en.wikipedia.org/wiki/Axiom_of_extensionality
> "two sets A and B are equal if and only if A and B have the same
> members."...
> Extensionality: ∀A∀B(∀x(x∈A <=> x∈B)) <=> A=B
>
> Let S={ x| x is a set. ∀x,y∈S, x=y } // '=' defined by Axiom of
> extensionality.

That's not a proper definition for a set. To start with, The condition ∀x,y∈S, x=y which is
defining S contains the symbol S, so there is circularity. Also, if you are using ZFC then ZFC does
not allow sets to be constructed using unrestricted comprehension.

Also this is OT for comp.theory, but wouldn't be out of place in sci.math.

>
> Let A={1,3,5..}, B={2,4,6,..}. A and B satisfies the property of the
> axiom of
> extensionality.

What do you mean by that? You understand A and B are not equal, right?

> So, both A and B are in S by the definition of S. But the definition
> of set
> requires the elements be different, where 'different' is the negation
> of the
> axiom of extensionality (definition?). Therefore, A and B cannot both
> be in S.

S in not properly defined.
A and B are not equal, and so could both be (different) members of some set that was properly
defined, E.g. A and B are both in {A,B}.
You are badly misunderstanding something, but I can't see what...

>
> So, the question: Are A,B in S?
>

Regards,
Mike.

[wij]

On Tue, 2024-01-23 at 20:34 +0000, Mike Terry wrote:
> On 23/01/2024 18:46, wij wrote:
> >   https://en.wikipedia.org/wiki/Set_(mathematics)
> >   A set is the mathematical model for a collection of different[1]
> > things;...
> >
> >   https://en.wikipedia.org/wiki/Axiom_of_extensionality
> >   "two sets A and B are equal if and only if A and B have the same
> > members."...
> >   Extensionality: ∀A∀B(∀x(x∈A <=> x∈B)) <=> A=B
> >
> >   Let S={ x| x is a set. ∀x,y∈S, x=y }  // '=' defined by Axiom of
> > extensionality.
>
> That's not a proper definition for a set.  To start with, The
> condition ∀x,y∈S, x=y which is
> defining S contains the symbol S, so there is circularity.  Also, if
> you are using ZFC then ZFC does
> not allow sets to be constructed using unrestricted comprehension.

Let's modify it to: S={ all the set x,y that x=y }

I don't know how to express the idea 'formally'. I feel it is like
Russell's paradox. But the expression (x∈x) in Russell's paradox is
apparently problematic and WILL cause lots of problems. But it was
accepted. As it is your? logic, I wonder how you understand it?

> Also this is OT for comp.theory, but wouldn't be out of place in
> sci.math.
>
> >
> >   Let A={1,3,5..}, B={2,4,6,..}. A and B satisfies the property of
> > the
> > axiom of
> >   extensionality.
>
> What do you mean by that?  You understand A and B are not equal,
> right?
>

Let A=B (A and B are defined equal under the meaning of extensionality)

[immibis]

On 1/23/24 22:03, wij wrote:
> On Tue, 2024-01-23 at 20:34 +0000, Mike Terry wrote:
>> On 23/01/2024 18:46, wij wrote:
>>>   https://en.wikipedia.org/wiki/Set_(mathematics)
>>>   A set is the mathematical model for a collection of different[1]
>>> things;...
>>>
>>>   https://en.wikipedia.org/wiki/Axiom_of_extensionality
>>>   "two sets A and B are equal if and only if A and B have the same
>>> members."...
>>>   Extensionality: ∀A∀B(∀x(x∈A <=> x∈B)) <=> A=B
>>>
>>>   Let S={ x| x is a set. ∀x,y∈S, x=y }  // '=' defined by Axiom of
>>> extensionality.
>>
>> That's not a proper definition for a set.  To start with, The
>> condition ∀x,y∈S, x=y which is
>> defining S contains the symbol S, so there is circularity.  Also, if
>> you are using ZFC then ZFC does
>> not allow sets to be constructed using unrestricted comprehension.
>
> Let's modify it to: S={ all the set x,y that x=y }
>
> I don't know how to express the idea 'formally'. I feel it is like
> Russell's paradox. But the expression (x∈x) in Russell's paradox is
> apparently problematic and WILL cause lots of problems. But it was
> accepted. As it is your? logic, I wonder how you understand it?
>

The idea doesn't even seem to make sense informally. What does "all the
set x,y that x=y" even mean?

>> Also this is OT for comp.theory, but wouldn't be out of place in
>> sci.math.
>>
>>>
>>>   Let A={1,3,5..}, B={2,4,6,..}. A and B satisfies the property of
>>> the
>>> axiom of
>>>   extensionality.
>>
>> What do you mean by that?  You understand A and B are not equal,
>> right?
>>
>
> Let A=B (A and B are defined equal under the meaning of extensionality)

But {1,3,5,...} and {2,4,6,...} are unequal under the meaning of
extensionality.

The axiom of extensionality defines sets A and B as equal if every
element in A is also in B, and every element not in A is also not in B.
But 1 is in A and not in B.

[wij]

Ok, let's modify it again: S= { x,y| x=y } // x,y are equal in the
meaning of extensionality

I.e. Whenever I find a,b that satisfy the 'extensionality' principle, x
and y are in S.

> > > Also this is OT for comp.theory, but wouldn't be out of place in
> > > sci.math.
> > >
> > > >
> > > >    Let A={1,3,5..}, B={2,4,6,..}. A and B satisfies the
> > > > property of
> > > > the
> > > > axiom of
> > > >    extensionality.
> > >
> > > What do you mean by that?  You understand A and B are not equal,
> > > right?
> > >
> >
> > Let A=B (A and B are defined equal under the meaning of
> > extensionality)
>
> But {1,3,5,...} and {2,4,6,...} are unequal under the meaning of
> extensionality.
>
> The axiom of extensionality defines sets A and B as equal if every
> element in A is also in B, and every element not in A is also not in
> B.
> But 1 is in A and not in B.
>

A and B are ELEMENT of S not the element of A or B (1,2,3,4...)

[immibis]

So S would be a set containing every set (including S). That isn't
allowed in ZFC, but it's allowed in some versions of set theory. We can
say it's allowed for now.

>>>> Also this is OT for comp.theory, but wouldn't be out of place in
>>>> sci.math.
>>>>
>>>>>
>>>>>    Let A={1,3,5..}, B={2,4,6,..}. A and B satisfies the
>>>>> property of
>>>>> the
>>>>> axiom of
>>>>>    extensionality.
>>>>
>>>> What do you mean by that?  You understand A and B are not equal,
>>>> right?
>>>>
>>>
>>> Let A=B (A and B are defined equal under the meaning of
>>> extensionality)
>>
>> But {1,3,5,...} and {2,4,6,...} are unequal under the meaning of
>> extensionality.
>>
>> The axiom of extensionality defines sets A and B as equal if every
>> element in A is also in B, and every element not in A is also not in
>> B.
>> But 1 is in A and not in B.
>>
>
> A and B are ELEMENT of S not the element of A or B (1,2,3,4...)
>

And A is not extensional with B.

A is extensional with A, and B is extensional with B, so they are both in S.

[wij]

I saw it, I made a mistake. Thanks for your insight.

[Mikko]

If A can be in a set then there is the set {A}.
If B can be in a set then there is the set {B}.
If A = B then {A} = {B}.
If {A} = {B} then A = B.
If A can be in a set and B can be in a set then there is the set {A, B},
which is a singlet set (i.e., {A, B} = {A} = {B} if A = B and
a doublet set otherwise.
If A = B then either both are in S or neither is.

Mikko