Consciousness is information?

[Bruno Marchal]

Excellent!

Kim, are you OK with Marty's answers?

Does someone have a (non philosophical) problem?

I will be busy right now (9h22 am). This afternoon I will send the next seven exercises.

Bruno

On 02 Jun 2009, at 21:57, m.a. wrote:

Bruno,

            I appreciate the simplicity of the examples. My answers follow the questions.    marty a.

----- Original Message -----

From: "Bruno Marchal" <marchal@...>

> 
> 
> ============================================= begin  
> ===============================
> 
> 1) SET
> 
> Informal definition: a set is a collection of object, called elements,  
> with the idea that it, the collection or set, can be considered itself  
> as an object. It is a many seen as a one, if you want. If the set is  
> not to big, we can describe it exhaustively by listing the elements,  
> if the set is bigger, we can describe it by some other way. Usually we  
> use accolades "{", followed by the elements, separated by commas, and  
> then "}", in the exhaustive description of a set.
> 
> Example/exercise:
> 
> 1) The set of odd natural numbers which are little than 10. This is a  
> well defined, and not to big set, so we can describe it exhaustively by
> {1, 3, 5, 7, 9}. In this case we say that 7 belongs to  {1, 3, 5, 7, 9}.
> Exercise 1: does the number 24 belongs to the set {1, 3, 5, 7, 9}?    NO
> 
> 2) the set of even natural number  which are little than 13. It is {0,  
> 2, 4, 6, 8, 10, 12}. OK? Some people can have a difficulty which is  
> not related to the notion of set, for example they can ask themselves  
> if zero (0) is really an even number. We will come back to this.          
> 
> 3) The set of odd natural numbers which are little than 100. This set  
> is already too big to describe exhaustively. We will freely describe  
> such a set by a quasi exhaustion like {1, 3, 5, 7, 9, 11, ... 95, 97,  
> 99}.
> Exercise 2: does the number 93 belongs to the set of odd natural  
> numbers which are little than 100, that is: does 93 belongs to {1, 3,  
> 5, 7, 9, 11, ... 95, 97, 99}?                                                               YES
> 
> 4) The set of all natural numbers. This set is hard to define, yet I  
> hope you agree we can describe it by the infinite quasi exhaustion by  
> {0, 1, 2, 3, ...}.
> Exercise 3: does the number 666 belongs to the set of natural numbers,  
> that is does 666 belongs to {0, 1, 2, 3, ...}.                                         YES
> Exercice 4: does the real number square-root(2) belongs to {0, 1, 2,  
> 3, ...}?                                                                                              NO (a guess)
> 
> 
> 5) When a set is too big or cumbersome, mathematician like to give  
> them a name. They will usually say: let S be the set {14, 345, 78}.  
> Then we can say that 14 belongs to S, for example.
> Exercise 5: does 345 belongs to S?                                                    YES
> 
> A set is entirely defined by its elements. Put in another way, we will  
> say that two sets are equal if they have the same elements.
> Exercise 6. Let S be the set {0, 1, 45} and let M be the set described  
> by {45, 0, 1}. Is it true or false that S is equal to M?                              YES
> Exercise 7. Let S be the set {666} and M be the set {6, 6, 6}. Is is  
> true or false that S is equal to M?                                                        NO                                                      
> 
> Seven exercises are enough. Are you ready to answer them. I hope you  
> don't find them too much easy, because I intend to proceed in a way  
> such that all exercise will be as easy, despite we will climb toward  
> very much deeper notion. Feel free to ask question, comments, etc. I  
> will try to adapt myself.                                                                       SO FAR SO GOOD
> 
> Next: we will see some operation on sets (union, intersection), and  
> the notion of subset. If all this work, I will build a latex document,  
> and make it the standard reference for the seventh step for the non  
> mathematician, or for the beginners in mathematics.
> 
> Bruno
> 
> 
> 
> http://iridia.ulb.ac.be/~marchal/
> 
> 
> 
> 
> 

http://iridia.ulb.ac.be/~marchal/

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[Torgny Tholerus]

Bruno Marchal skrev:

> On 02 Jun 2009, at 19:43, Torgny Tholerus wrote:
>
>  
>> Bruno Marchal skrev:
>>    
>>> 4) The set of all natural numbers. This set is hard to define, yet I
>>> hope you agree we can describe it by the infinite quasi exhaustion by
>>> {0, 1, 2, 3, ...}.
>>>
>>>      
>> Let N be the biggest number in the set {0, 1, 2, 3, ...}.
>>
>> Exercise: does the number N+1 belongs to the set of natural numbers,
>> that is does N+1 belongs to {0, 1, 2, 3, ...}?
>>    
>
>
> Yes. N+1 belongs to {0, 1, 2, 3, ...}.
> This follows from classical logic and the fact that the proposition "N  
> be the biggest number in the set {0, 1, 2, 3, ...}" is always false.  
> And false implies all propositions.
>  

No, you are wrong.  The answer is No.

Proof:

Define "biggest number" as:

a is the biggest number in the set S if and only if for every element e
in S you have e < a or e = a.

Now assume that N+1 belongs to the set of natural numbers.

Then you have N+1 < N or N+1 = N.

But this is a contradiction.  So the assumption must be false.  So we
have proved that N+1 does not belongs to the set of natural numbers.

--
Torgny Tholerus

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[Quentin Anciaux-2]

2009/6/3 Torgny Tholerus <torgny@...>:

>
> Bruno Marchal skrev:
>> On 02 Jun 2009, at 19:43, Torgny Tholerus wrote:
>>
>>
>>> Bruno Marchal skrev:
>>>
>>>> 4) The set of all natural numbers. This set is hard to define, yet I
>>>> hope you agree we can describe it by the infinite quasi exhaustion by
>>>> {0, 1, 2, 3, ...}.
>>>>
>>>>
>>> Let N be the biggest number in the set {0, 1, 2, 3, ...}.
>>>
>>> Exercise: does the number N+1 belongs to the set of natural numbers,
>>> that is does N+1 belongs to {0, 1, 2, 3, ...}?
>>>
>>
>>
>> Yes. N+1 belongs to {0, 1, 2, 3, ...}.
>> This follows from classical logic and the fact that the proposition "N
>> be the biggest number in the set {0, 1, 2, 3, ...}" is always false.
>> And false implies all propositions.
>>
>
> No, you are wrong.  The answer is No.
>
> Proof:
>
> Define "biggest number" as:
>
> a is the biggest number in the set S if and only if for every element e
> in S you have e < a or e = a.
>
> Now assume that N+1 belongs to the set of natural numbers.
>
> Then you have N+1 < N or N+1 = N.
>
> But this is a contradiction.  So the assumption must be false.  So we
> have proved that N+1 does not belongs to the set of natural numbers.

Hi,

No, what you've demonstrated is that there is no biggest number (you
falsified the hypothesis which is there exists a biggest number). You
did a "demonstration par l'absurde" (in french, don't know how it is
called in english). And you have shown a contradiction, which implies
that your assumption is wrong (there exists a biggest number), not
that this number is not in the set.

Regards,
Quentin


> --
> Torgny Tholerus
>
> >
>



--
All those moments will be lost in time, like tears in rain.

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[Torgny Tholerus]

Quentin Anciaux skrev:

> 2009/6/3 Torgny Tholerus <torgny@...>:
>  
>> Bruno Marchal skrev:
>>    
>>> On 02 Jun 2009, at 19:43, Torgny Tholerus wrote:
>>>
>>>
>>>      
>>>> Bruno Marchal skrev:
>>>>
>>>>        
>>>>> 4) The set of all natural numbers. This set is hard to define, yet I
>>>>> hope you agree we can describe it by the infinite quasi exhaustion by
>>>>> {0, 1, 2, 3, ...}.
>>>>>
>>>>>
>>>>>          
>>>> Let N be the biggest number in the set {0, 1, 2, 3, ...}.
>>>>
>>>> Exercise: does the number N+1 belongs to the set of natural numbers,
>>>> that is does N+1 belongs to {0, 1, 2, 3, ...}?
>>>>
>>>>        
>>> Yes. N+1 belongs to {0, 1, 2, 3, ...}.
>>> This follows from classical logic and the fact that the proposition "N
>>> be the biggest number in the set {0, 1, 2, 3, ...}" is always false.
>>> And false implies all propositions.
>>>
>>>      
>> No, you are wrong.  The answer is No.
>>
>> Proof:
>>
>> Define "biggest number" as:
>>
>> a is the biggest number in the set S if and only if for every element e
>> in S you have e < a or e = a.
>>
>> Now assume that N+1 belongs to the set of natural numbers.
>>
>> Then you have N+1 < N or N+1 = N.
>>
>> But this is a contradiction.  So the assumption must be false.  So we
>> have proved that N+1 does not belongs to the set of natural numbers.
>>    
>
> Hi,
>
> No, what you've demonstrated is that there is no biggest number (you
> falsified the hypothesis which is there exists a biggest number). You
> did a "demonstration par l'absurde" (in french, don't know how it is
> called in english). And you have shown a contradiction, which implies
> that your assumption is wrong (there exists a biggest number), not
> that this number is not in the set.
>  

How do you know that there is no biggest number?  Have you examined all
the natural numbers?  How do you prove that there is no biggest number?

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[Quentin Anciaux-2]

2009/6/3 Torgny Tholerus <torgny@...>:

>
> Quentin Anciaux skrev:
>> 2009/6/3 Torgny Tholerus <torgny@...>:
>>
>>> Bruno Marchal skrev:
>>>
>>>> On 02 Jun 2009, at 19:43, Torgny Tholerus wrote:
>>>>
>>>>
>>>>
>>>>> Bruno Marchal skrev:
>>>>>
>>>>>
>>>>>> 4) The set of all natural numbers. This set is hard to define, yet I
>>>>>> hope you agree we can describe it by the infinite quasi exhaustion by
>>>>>> {0, 1, 2, 3, ...}.
>>>>>>
>>>>>>
>>>>>>
>>>>> Let N be the biggest number in the set {0, 1, 2, 3, ...}.
>>>>>
>>>>> Exercise: does the number N+1 belongs to the set of natural numbers,
>>>>> that is does N+1 belongs to {0, 1, 2, 3, ...}?
>>>>>
>>>>>
>>>> Yes. N+1 belongs to {0, 1, 2, 3, ...}.
>>>> This follows from classical logic and the fact that the proposition "N
>>>> be the biggest number in the set {0, 1, 2, 3, ...}" is always false.
>>>> And false implies all propositions.
>>>>
>>>>
>>> No, you are wrong.  The answer is No.
>>>
>>> Proof:
>>>
>>> Define "biggest number" as:
>>>
>>> a is the biggest number in the set S if and only if for every element e
>>> in S you have e < a or e = a.
>>>
>>> Now assume that N+1 belongs to the set of natural numbers.
>>>
>>> Then you have N+1 < N or N+1 = N.
>>>
>>> But this is a contradiction.  So the assumption must be false.  So we
>>> have proved that N+1 does not belongs to the set of natural numbers.
>>>
>>
>> Hi,
>>
>> No, what you've demonstrated is that there is no biggest number (you
>> falsified the hypothesis which is there exists a biggest number). You
>> did a "demonstration par l'absurde" (in french, don't know how it is
>> called in english). And you have shown a contradiction, which implies
>> that your assumption is wrong (there exists a biggest number), not
>> that this number is not in the set.
>>
>
> How do you know that there is no biggest number?

You just did.
You shown that by assuming there is one it entails a contradiction.

> Have you examined all
> the natural numbers?

No, that's what demonstration is all about.

> How do you prove that there is no biggest number?

You did it.

>
> >
>



--
All those moments will be lost in time, like tears in rain.

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[Juho Pennanen]

Quentin Anciaux kirjoitti:
 > 2009/6/3 Torgny Tholerus <torgny@...>:
...
 >> How do you know that there is no biggest number?
 >
 > You just did.
 > You shown that by assuming there is one it entails a contradiction.
 >
 >> Have you examined all
 >> the natural numbers?
 >
 > No, that's what demonstration is all about.
 >

Clearly you two disagree on what {0, 1, 2, 3, ...} means.

All definitions of natural numbers I have seen imply that N+1 is a
natural number whenever N is. Then there clearly is no biggest number.

But I can see someone could have philosophical objections to the
conventional definition. I've heard of ultrafinitists, e.g., but have
not checked how they define natural numbers (if they do).

jp


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[Jesse Mazer]

Some parts of this message have been removed. Learn more about Nabble's security policy.

> Date: Wed, 3 Jun 2009 13:14:16 +0200
> Subject: Re: The seven step-Mathematical preliminaries
> From: allcolor@...
> To: everything-list@...
>
>
> 2009/6/3 Torgny Tholerus <torgny@...>:
>>
>> Bruno Marchal skrev:
>>> On 02 Jun 2009, at 19:43, Torgny Tholerus wrote:
>>>
>>>
>>>> Bruno Marchal skrev:
>>>>
>>>>> 4) The set of all natural numbers. This set is hard to define, yet I
>>>>> hope you agree we can describe it by the infinite quasi exhaustion by
>>>>> {0, 1, 2, 3, ...}.
>>>>>
>>>>>
>>>> Let N be the biggest number in the set {0, 1, 2, 3, ...}.
>>>>
>>>> Exercise: does the number N+1 belongs to the set of natural numbers,
>>>> that is does N+1 belongs to {0, 1, 2, 3, ...}?
>>>>
>>>
>>>
>>> Yes. N+1 belongs to {0, 1, 2, 3, ...}.
>>> This follows from classical logic and the fact that the proposition "N
>>> be the biggest number in the set {0, 1, 2, 3, ...}" is always false.
>>> And false implies all propositions.
>>>
>>
>> No, you are wrong.  The answer is No.
>>
>> Proof:
>>
>> Define "biggest number" as:
>>
>> a is the biggest number in the set S if and only if for every element e
>> in S you have e < a or e = a.
>>
>> Now assume that N+1 belongs to the set of natural numbers.
>>
>> Then you have N+1 < N or N+1 = N.
>>
>> But this is a contradiction.  So the assumption must be false.  So we
>> have proved that N+1 does not belongs to the set of natural numbers.
>
> Hi,
>
> No, what you've demonstrated is that there is no biggest number (you
> falsified the hypothesis which is there exists a biggest number). You
> did a "demonstration par l'absurde" (in french, don't know how it is
> called in english). And you have shown a contradiction, which implies
> that your assumption is wrong (there exists a biggest number), not
> that this number is not in the set.

The English term for this is "proof by contradiction":

http://en.wikipedia.org/wiki/Proof_by_contradiction 

Of course, Torgny's conclusion is a little off--he did not show the assumption "N+1 belongs to the set of natural numbers" must be wrong as he suggested, rather he showed the assumption "N is the largest natural number" must have been wrong. Just by the usual definition of natural numbers, if N is a natural number then N+1 must be one too (the page at http://en.wikipedia.org/wiki/Recursion#Formal_definitions_of_recursion says that natural numbers are defined in a recursive way: 'the formal definition of natural numbers in set theory is: 1 is a natural number, and each natural number has a successor, which is also a natural number'). If Torgny doesn't agree, I think he needs to provide an alternate definition of "natural number" where it would not be true *by definition* that N+1 is a natural number if N is.

Jesse

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[Brian Tenneson]

I don't know if Bruno is about to answer this in messages I haven't
checked yet but one can visualize the square root of 2.  If you draw a
square one meter by one meter, then the length of the diagonal is the
square root of 2 meters.  It is approximately 1.4.  What's relevant to
Bruno's question is that the square root of two is greater than one but
less than two, according to the geometry of the diagonal: the diagonal
is more than one and less than two (a picture really helps drive this
point home).
Now since the square root of two is more than one and less than two, it
does -not- belong to the set {0,1, 2, 3, 4, ...}.
In other words, the square root of two is not a natural number.

kimjones@... wrote:

>
> On Wed Jun  3  0:39 , Bruno Marchal <marchal@...> sent:
>
>  
>> Hi Kim, Hi Marty and others,
>>
>> So it is perhaps time to do some math.
>>    
>
>
> It is
>
>
>  
>> Obviously this is a not a course in math, but it is an explanation  
>>    
> >from scratch of the seven step of the universal dovetailer argument.  
>  
>> It is a shortcut, and most probably we will make some digression from  
>> time to time, but let us try not to digress too much.
>>
>> Kim, you are OK with this?
>>    
>
>
> I am
>
>
>
>
> I have to take into account the problem you  
>  
>> did have with math, and which makes this lesson a bit challenging for  
>> me, and I guess for you too.
>>    
>
>
> Hopefully my innocence will allow me to bypass the pedantry and orthodoxies of the field and allow a
> shortcut to a high level of understanding of the UDA. Only a complete neophyte would have the gall to
> say something like that!
>
>
>
>  
>> I begin with the very useful and elementary notion of set, as  
>> explained in what is called "naive set theory", and which is the base  
>> of almost all part of math.
>>
>> ============================================= begin  
>> ===============================
>>
>> 1) SET
>>
>> Informal definition: a set is a collection of object, called elements,  
>> with the idea that it, the collection or set, can be considered itself  
>> as an object. It is a many seen as a one, if you want. If the set is  
>> not to big, we can describe it exhaustively by listing the elements,  
>> if the set is bigger, we can describe it by some other way. Usually we  
>> use accolades "{", followed by the elements, separated by commas, and  
>> then "}", in the exhaustive description of a set.
>>
>> Example/exercise:
>>
>> 1) The set of odd natural numbers which are little than 10. This is a  
>> well defined, and not to big set, so we can describe it exhaustively by
>> {1, 3, 5, 7, 9}. In this case we say that 7 belongs to  {1, 3, 5, 7, 9}.
>> Exercise 1: does the number 24 belongs to the set {1, 3, 5, 7, 9}?
>>    
>
>
> No
>
>
>  
>> 2) the set of even natural number  which are little than 13. It is {0,  
>> 2, 4, 6, 8, 10, 12}. OK? Some people can have a difficulty which is  
>> not related to the notion of set, for example they can ask themselves  
>> if zero (0) is really an even number. We will come back to this.
>>
>> 3) The set of odd natural numbers which are little than 100. This set  
>> is already too big to describe exhaustively. We will freely describe  
>> such a set by a quasi exhaustion like {1, 3, 5, 7, 9, 11, ... 95, 97,  
>> 99}.
>> Exercise 2: does the number 93 belongs to the set of odd natural  
>> numbers which are little than 100, that is: does 93 belongs to {1, 3,  
>> 5, 7, 9, 11, ... 95, 97, 99}?
>>    
>
>
> Yes
>
>
>
>  
>> 4) The set of all natural numbers. This set is hard to define, yet I  
>> hope you agree we can describe it by the infinite quasi exhaustion by  
>> {0, 1, 2, 3, ...}.
>> Exercise 3: does the number 666 belongs to the set of natural numbers,  
>> that is does 666 belongs to {0, 1, 2, 3, ...}.
>>    
>
>
> Yes
>
>
>  
>> Exercice 4: does the real number square-root(2) belongs to {0, 1, 2,  
>> 3, ...}?
>>    
>
>
> No idea what square-root(2) means. When I said I was innumerate I wasn't kidding! I could of course look
> it up or ask my mathematics teacher friends but I just know your explanation will make theirs seem trite.
>
>
>  
>> 5) When a set is too big or cumbersome, mathematician like to give  
>> them a name. They will usually say: let S be the set {14, 345, 78}.  
>> Then we can say that 14 belongs to S, for example.
>> Exercise 5: does 345 belongs to S?
>>    
>
>
> Clearly, yes
>
>
>
>  
>> A set is entirely defined by its elements. Put in another way, we will  
>> say that two sets are equal if they have the same elements.
>> Exercise 6. Let S be the set {0, 1, 45} and let M be the set described  
>> by {45, 0, 1}. Is it true or false that S is equal to M?
>>    
>
>
> True - unless integer position within a given sequence in a set plays a role. I will guess that it does not
>
>
>
>  
>> Exercise 7. Let S be the set {666} and M be the set {6, 6, 6}. Is is  
>> true or false that S is equal to M?
>>    
>
>
> False - the commas separate each natural number
>
>
>  
>> Seven exercises are enough. Are you ready to answer them.
>>    
>
>
> Done - apart from the square root question
>
>
>  I hope you  
>  
>> don't find them too much easy, because I intend to proceed in a way  
>> such that all exercise will be as easy, despite we will climb toward  
>> very much deeper notion. Feel free to ask question, comments, etc. I  
>> will try to adapt myself.
>>    
>
>
> Very excited about doing this. If you can make it all as approachable as this I am over the moon!
>
>
>
>  
>> Next: we will see some operation on sets (union, intersection), and  
>> the notion of subset. If all this work, I will build a latex document,  
>> and make it the standard reference for the seventh step for the non  
>> mathematician, or for the beginners in mathematics.
>>    
>
>
> What a wonderful idea!
>
> Kim
>
>
>  
>> Bruno
>>
>>
>>
>> http://iridia.ulb.ac.be/~marchal/
>>
>>    
>
>
> >
>
>  

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[Brian Tenneson]

> How do you know that there is no biggest number?  Have you examined all
> the natural numbers?  How do you prove that there is no biggest number?
>
>  
In my opinion those are excellent questions.  I will attempt to answer
them.  The intended audience of my answer is everyone, so please forgive
me if I say something you already know.

Firstly, no one has or can examine all the natural numbers.  By that I
mean no human.  Maybe there is an omniscient machine (or a "maximally
knowledgeable" in some paraconsistent way)  who can examine all numbers
but that is definitely putting the cart before the horse.

Secondly, the question boils down to a difference in philosophy:
mathematicians would say that it is not necessary to examine all natural
numbers.  The mathematician would argue that it suffices to examine all
essential properties of natural numbers, rather than all natural numbers.

There are a variety of equivalent ways to define a natural number but
the essential features of natural numbers are that
(a) there is an ordering on the set of natural numbers, a well
ordering.  To say a set is well ordered entails that every =nonempty=
subset of it has a least element.
(b) the set of natural numbers has a least element (note that it is
customary to either say 0 is this least element or say 1 is this least
element--in some sense it does not matter what the starting point is)
(c) every natural number has a natural number successor.  By successor
of a natural number, I mean anything for which the well ordering always
places the successor as larger than the predecessor.

Then the set of natural numbers, N, is the set containing the least
element (0 or 1) and every successor of the least element, and only
successors of the least element.

There is nothing wrong with a proof by contradiction but I think a
"forward" proof might just be more convincing.

Consider the following statement:
Whenever S is a subset of N, S has a largest element if, and only if,
the complement of S has a least element.

By complement of S, I mean the set of all elements of N that are not
elements of S.

Before I give a longer argument, would you agree that statement is
true?  One can actually be arbitrarily explicit: M is the largest
element of S if, and only if, the successor of M is the least element of
the compliment of S.

If so, then that statement proves that there is no largest element of N:
Letting S be N in particular, note that N is a subset of N (albeit not a
"proper" subset).
Then the statement reads as the following for this particular choice S:
N has a largest element if, and only if, the complement of N has a least
element.

The compliment of N is the empty set.  To elaborate: the compliment of N
is the set of all elements of N that are not elements of N.  No elements
can both be and not be elements of N, so this set is empty.

The empty set does not have a least element.  In fact, it has no
elements at all.

Therefore, N does not have a largest element.

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[Brent Meeker-2]

Quentin Anciaux wrote:

> 2009/6/3 Torgny Tholerus <torgny@...>:
>  
>> Bruno Marchal skrev:
>>    
>>> On 02 Jun 2009, at 19:43, Torgny Tholerus wrote:
>>>
>>>
>>>      
>>>> Bruno Marchal skrev:
>>>>
>>>>        
>>>>> 4) The set of all natural numbers. This set is hard to define, yet I
>>>>> hope you agree we can describe it by the infinite quasi exhaustion by
>>>>> {0, 1, 2, 3, ...}.
>>>>>
>>>>>
>>>>>          
>>>> Let N be the biggest number in the set {0, 1, 2, 3, ...}.
>>>>
>>>> Exercise: does the number N+1 belongs to the set of natural numbers,
>>>> that is does N+1 belongs to {0, 1, 2, 3, ...}?
>>>>
>>>>        
>>> Yes. N+1 belongs to {0, 1, 2, 3, ...}.
>>> This follows from classical logic and the fact that the proposition "N
>>> be the biggest number in the set {0, 1, 2, 3, ...}" is always false.
>>> And false implies all propositions.
>>>
>>>      
>> No, you are wrong.  The answer is No.
>>
>> Proof:
>>
>> Define "biggest number" as:
>>
>> a is the biggest number in the set S if and only if for every element e
>> in S you have e < a or e = a.
>>
>> Now assume that N+1 belongs to the set of natural numbers.
>>
>> Then you have N+1 < N or N+1 = N.
>>
>> But this is a contradiction.  So the assumption must be false.  So we
>> have proved that N+1 does not belongs to the set of natural numbers.
>>    
>
> Hi,
>
> No, what you've demonstrated is that there is no biggest number (you
> falsified the hypothesis which is there exists a biggest number). You
> did a "demonstration par l'absurde" (in french, don't know how it is
> called in english). And you have shown a contradiction, which implies
> that your assumption is wrong (there exists a biggest number), not
> that this number is not in the set.
>
> Regards,
> Quentin

When you arrive at a contradiction it doesn't tell you which assumption
is wrong.

Brent

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[Quentin Anciaux-2]

2009/6/3 Brent Meeker <meekerdb@...>:

>
> Quentin Anciaux wrote:
>> 2009/6/3 Torgny Tholerus <torgny@...>:
>>
>>> Bruno Marchal skrev:
>>>
>>>> On 02 Jun 2009, at 19:43, Torgny Tholerus wrote:
>>>>
>>>>
>>>>
>>>>> Bruno Marchal skrev:
>>>>>
>>>>>
>>>>>> 4) The set of all natural numbers. This set is hard to define, yet I
>>>>>> hope you agree we can describe it by the infinite quasi exhaustion by
>>>>>> {0, 1, 2, 3, ...}.
>>>>>>
>>>>>>
>>>>>>
>>>>> Let N be the biggest number in the set {0, 1, 2, 3, ...}.
>>>>>
>>>>> Exercise: does the number N+1 belongs to the set of natural numbers,
>>>>> that is does N+1 belongs to {0, 1, 2, 3, ...}?
>>>>>
>>>>>
>>>> Yes. N+1 belongs to {0, 1, 2, 3, ...}.
>>>> This follows from classical logic and the fact that the proposition "N
>>>> be the biggest number in the set {0, 1, 2, 3, ...}" is always false.
>>>> And false implies all propositions.
>>>>
>>>>
>>> No, you are wrong.  The answer is No.
>>>
>>> Proof:
>>>
>>> Define "biggest number" as:
>>>
>>> a is the biggest number in the set S if and only if for every element e
>>> in S you have e < a or e = a.
>>>
>>> Now assume that N+1 belongs to the set of natural numbers.
>>>
>>> Then you have N+1 < N or N+1 = N.
>>>
>>> But this is a contradiction.  So the assumption must be false.  So we
>>> have proved that N+1 does not belongs to the set of natural numbers.
>>>
>>
>> Hi,
>>
>> No, what you've demonstrated is that there is no biggest number (you
>> falsified the hypothesis which is there exists a biggest number). You
>> did a "demonstration par l'absurde" (in french, don't know how it is
>> called in english). And you have shown a contradiction, which implies
>> that your assumption is wrong (there exists a biggest number), not
>> that this number is not in the set.
>>
>> Regards,
>> Quentin
> When you arrive at a contradiction it doesn't tell you which assumption
> is wrong.
>
> Brent

Well I agree, but the second assumption depends on the first which is
N exists and well defined. If it was, the second assumption is
trivially false.

Quentin


>
> >
>



--
All those moments will be lost in time, like tears in rain.

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[Bruno Marchal]

Very good answer, Kim, 

Just a few comments. and then the sequel.

Exercice 4: does the real number square-root(2) belongs to {0, 1, 2,  

3, ...}?

No idea what square-root(2) means. When I said I was innumerate I wasn't kidding! I could of course look
it up or ask my mathematics teacher friends but I just know your explanation will make theirs seem trite.

Well thanks. 

The square root of 2 is a number x, such that x*x (x times x, x multiplied by itself) gives 2.

For example, the square root of 4 is 2, because 2*2 is 4. The square root of 9 is 3, because 3*3 is 9. Her by "square root" I mean the positive square root, because we will see (more later that soon) that numbers can have negative square root, but please forget this.

At this stage, with this definition, you can guess that the square root of 2 cannot be a natural number. 1*1 = 1, and 2*2 = 4, and it would be astonishing that x could be bigger than 2. So if there is number x such that x*x is 2, we can guess that such a x cannot be a natural number, that is an element of {0, 1, 2, 3 ...}, and the answer of exercise 4 is "no". The square root of two will reappear recurrently, but more in examples, than in the sequence of notions which are strictly needed for UDA-7.

A set is entirely defined by its elements. Put in another way, we will  

say that two sets are equal if they have the same elements.

Exercise 6. Let S be the set {0, 1, 45} and let M be the set described  

by {45, 0, 1}. Is it true or false that S is equal to M?

True - unless integer position within a given sequence in a set plays a role. I will guess that it does not.

You are right.

Exercise 7. Let S be the set {666} and M be the set {6, 6, 6}. Is is  

true or false that S is equal to M?

False - the commas separate each natural number

You are right. Also note that there is only one element in the set {6, 6, 6}. It is just a redundant description of the set {6}.

Very excited about doing this. If you can make it all as approachable as this I am over the moon!

I will try, and it is very kind to play such a candid role. I appreciate that you have the ability to say "I don't know <something>". It is very helpful for me to remain approachable, and eventually it will help everybody. 

So let us continue.

=============== Intension and extension ====================

Before defining "intersection, union and the notion of subset, I would like to come back on the ways we can define some specific sets.

In the case of finite and "little" set we have seen that we can define them by exhaustion. This means we can give an explicit complete description of all element of the set. 

Example. A = {0, 1, 2, 77, 98, 5}

When the set is still finite and too big, or if we are lazy, we can sometimes define the set by quasi exhaustion. This means we describe enough elements of the set in a manner which, by requiring some good will and some imagination, we can estimate having define the set.

Example. B = {3, 6, 9, 12, ... 99}. We can understand in this case that we meant the set of multiple of the number three, below 100.

A fortiori, when a set in not finite, that is, when the set is infinite, we have to use either quasi-exhaustion, or we have to use some sentence or phrase or proposition describing the elements of the set.

Definition.

I will say that a set is defined IN EXTENSIO, or simply, in extension, when it is defined in exhaustion or quasi-exhaustion.

I will say that a set is defined IN INTENSIO, or simply in intension, with a "s", when it is defined by a sentence explaining the typical attribute of the elements.

Example: Let A be the set {2, 4, 6, 8, 10, ... 100}. We can easily define A in intension:  A = the set of numbers which are even and more little than 100. mathematician will condense this by the following:

A = {x such that x is even and little than 100}  = {x ⎮ x is even & x < 100}. "⎮" is a special character, abbreviating "such that", and I hope it goes through the mail. If not I will use "such that", or s.t., or things like that.

The expression {x ⎮ x is even} is literally read as:  the set of object x, (or number x if we are in a context where we talk about number) such that x is even.

Exercise 1: Could you define in intension the following infinite set C = {101, 103, 105, ...}

C = ?

Exercise 2: I will say that a natural number is a multiple of 4 if it can be written as 4*y, for some y. For example 0 is a multiple of 4, (0 = 4*0), but also 28, 400, 404, ...  Could you define in extension the following set D = {x ⎮ x < 10  &  x is a multiple of 4}. 

A last notational, but important symbol. Sets have elements. For example the set A = {1, 2, 3} has three elements 1, 2 and 3. For saying that 3 is an element of A in an a short way, we usually write 3 ∈ A.  this is read as "3 belongs to A", or "3 is in A". Now 4 does not belong to A. To write this in a short way, we will write 4 ∉ A, or we will write ¬ (4 ∈ A) or sometimes just NOT(4 ∈ A). It is read: 4 does not belong to A, or: it is not the case that 4 belongs to A.

Having those notions and notations at our disposition we can speed up on the notion of union and intersection.

The intersection of the sets A and B is the (new) set of those elements which belongs to both A and B. Put in another way: 

The intersection of the sets A with the set B is the set of those elements which belongs to A and which belongs to B. 

This new set, obtained from A and B is written A ∩ B, or A inter. B (in case the special character doesn't go through).

With our notations we can write or define the intersection A ∩ B directly

A ∩ B = {x ⎮ x ∈ A and x ∈ B}.

Example {3, 4, 5, 6, 8} ∩ {5, 6, 7, 9} = {5, 6}

Similarly, we can directly define the union of two sets A and B, written A ∪ B in the following way:

A ∪ B = {x ⎮ x ∈ A or x ∈ B}.    Here we use the usual logical "or". p or q is suppose to be true if p is true or q is true (or both are true). It is not the exclusive "or".

Example {3, 4, 5, 6, 8} ∪ {5, 6, 7, 9} = {3, 4, 5, 6, 7, 8}.

Exercice 3. 

Let N = {0, 1, 2, 3, ...}

Let A = {x ⎮ x < 10}

Let B = {x ⎮ x is even}

Describe in extension (that is: exhaustion or quasi-exhaustion) the following sets:

N ∪ A =

N ∪ B =

A ∪ B =

B ∪ A =

N ∩ A =

B ∩ A =

N ∩ B =

A ∩ B =

Exercice 4

Is it true that A ∩ B = B ∩ A, whatever A and B are? 

Is it true that A ∪ B = B ∪ A, whatever A and B are?

Now, I could give you exercise so that you would be lead to discoveries, but I prefer to be as simple and approachable as possible, and my goal is not even to give you the taste for doing research, so I will do the discovery by myself here and now. Indeed a natural question occurs. What will happen if we try to find the intersection of two sets which have no elements in common? For example, what is the intersection of A = {x ⎮ x is even} with B = {x ⎮ x is odd} ? At first sight we could say that there is no intersection, given that A and B have no elements in common. But a set is just a bit more than its elements. And if there is no elements in the intersection, it means simply that the set A ∩ B has no elements. So we are very inspired if we let that bizarre set to exist, so we give it a name, and call it the empty set, and we can describe it easily in exhaustion by { }, although many describe it as ∅. So, if A and B have no elements in common, A ∩ B is still well defined and is equal to ∅. having a new toy, we can play with it:

Exercise 5, with A and B the same as in exercise 3.

∅ ∪ A =

∅ ∪ B =

A ∪ ∅ =

B ∪ ∅ =

N ∩ ∅ =

B ∩ ∅ =

∅ ∩ B =

∅ ∩ ∅ =

∅ ∪ ∅ =

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

SUBSET

We will say that A is a subset of B (A and B being sets) if, whatever object x represents, each time x belongs to A, it belongs to B. Put in another way it means that IF x belongs to A, THEN x belongs to B. It means that all the elements of A are also elements of B. We can write, with 

x ∈ A -> x ∈ B.

             

And this we abbreviate as A ⊆ B, and we read it: A is included in B.

Example:

1) Let us look if the set A = {1, 2} is included in the set B = {1, 2, 3}.  Here A has two elements. To see if A is included in B, we have to look at each element in the set A, and we have to see if they belongs to B. Now A has two elements, 1, and 2, so we have two tasks to accomplish, or two questions to answer:

does 1 belongs also to B. The answer is yes.

does 2 belongs also to B. The answer is yes.

We have thus verify that all elements of A are also elements of B, and thus we can conclude that A is indeed included in B.

2) Let us look if the set A = {1} is included in B = {1, 2, 3}.  Now, A has only one element. So we are lucky, we have only one task to accomplish! Is 1 an element of B? The answer is yes. Thus we have {1} is included in {1, 2, 3}.

3) Let us look if the set A=  { }, the empty set ∅,  is included in B = {1, 2, 3}. Now A has no element. So we are even more lucky, we have no task to accomplish at all. The condition is trivially satisfied. So the empty set is included in {1, 2, 3}. And this shows that the empty set is included in any set. In particular we have that ∅ ⊆ ∅.

Note that all set is a subset of itself. Trivially, all elements of A is an element of A.

Exercise 6 

We will say that a set A is a subset of a set B, if A is included in B.

Could you give all the subsets of the set  {1, 2}.

Could you give all the subsets of the set  {1}

Could you give all the subsets of the set  { }.

The post is long enough, so I spare you the seventh exercise. Also I have to go, I hope there are not to many typo errors and spelling mistakes, and well, I pray for the special symbols going trough. It is possible that they go through for most mailing systems, but not all. Let me know.

Bon courage,

Bruno

http://iridia.ulb.ac.be/~marchal/

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[Brent Meeker-2]

Brian Tenneson wrote:

>  
>> How do you know that there is no biggest number?  Have you examined all
>> the natural numbers?  How do you prove that there is no biggest number?
>>
>>  
>>    
> In my opinion those are excellent questions.  I will attempt to answer
> them.  The intended audience of my answer is everyone, so please forgive
> me if I say something you already know.
>
> Firstly, no one has or can examine all the natural numbers.  By that I
> mean no human.  Maybe there is an omniscient machine (or a "maximally
> knowledgeable" in some paraconsistent way)  who can examine all numbers
> but that is definitely putting the cart before the horse.
>
> Secondly, the question boils down to a difference in philosophy:
> mathematicians would say that it is not necessary to examine all natural
> numbers.  The mathematician would argue that it suffices to examine all
> essential properties of natural numbers, rather than all natural numbers.
>
> There are a variety of equivalent ways to define a natural number but
> the essential features of natural numbers are that
> (a) there is an ordering on the set of natural numbers, a well
> ordering.  To say a set is well ordered entails that every =nonempty=
> subset of it has a least element.
> (b) the set of natural numbers has a least element (note that it is
> customary to either say 0 is this least element or say 1 is this least
> element--in some sense it does not matter what the starting point is)
> (c) every natural number has a natural number successor.  By successor
> of a natural number, I mean anything for which the well ordering always
> places the successor as larger than the predecessor.
>
> Then the set of natural numbers, N, is the set containing the least
> element (0 or 1) and every successor of the least element, and only
> successors of the least element.
>
> There is nothing wrong with a proof by contradiction but I think a
> "forward" proof might just be more convincing.
>
> Consider the following statement:
> Whenever S is a subset of N, S has a largest element if, and only if,
> the complement of S has a least element.
>  

Let S={even numbers}  the complement of S, ~S={odd numbers}  ~S has a
least element, 1.  Therefore there is a largest even number.

Brent


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[Jason Resch-2]

On Fri, May 22, 2009 at 4:37 PM, Bruno Marchal <marchal@...> wrote:

>> Do you believe if we create a computer in this physical
>> universe that it could be made conscious,
>
> But a computer is never conscious, nor is a brain. Only a person is
> conscious, and a computer or a brain can only make it possible for a
> person to be conscious relatively to another computer. So your
> question is ambiguous.
> It is not my brain which is conscious, it is me who is conscious. My
> brain appears to make it possible for my consciousness to manifest
> itself relatively to you. Remember that we are supposed to no more
> count on the physical supervenience thesis.
> It remains locally correct to attribute a consciousness through a
> brain or a body to a person we judged succesfully implemented locally
> in some piece of matter (like when we say yes to a doctor).  But the
> piece of matter is not the subject of the consciousness. It is only
> the "abstract person" or "program" who is the subject of consciousness.
> To say a brain is conscious consists in doing Searle's'mistake when he
> confused levels of computations in the Chinese room, as well seen
> already by Hofstadter and Dennett in Mind's I.
>
>

Thanks for your response, if I understand you correctly, you are
saying that if we run a simulation of a mind, we are not creating
consciousness, only adding an additional instantiation to a mind which
already has an infinity of indeterminable instantiations.  Is that
right?

Does this imply that it is impossible to create a simulation of a mind
that finds it lives in an environment without uncertainty?  If so is
it because even if the physical laws in one instantiation may be
certain, where some of the infinite number of computations that all
instantiate that mind may diverge and in particular which one that
mind will find itself in is not knowable?

The consequence being that all observers everywhere live in QM-like
environments?

Thanks, I look forward to your reply.

Jason

>
>> or do you count all
>> appearance of matter to be only a description of a computation and not
>> capable of "true" computation?
>
> "appearance of matter" is a qualia. It does not describe anything but
> is a subjective experience, which may correspond to something stable
> and reflecting the existence of a computation (in Platonia) capable to
> manifest itself relatively to you.
>
>
>> Do you believe that the only real
>> computation exists platonically and this is the only source of
>> conscious experience?
>
> Computations and their relative implementations exist only in
> platonia, yes. But even in Platonia, they exist in multiple relative
> version, all defined eventually through many multiple relations
> between numbers.
>
>
>>  If so I find this confusing, as could there not
>> be multiple levels?
>
> But they are multiple levels of computations in Platonia or
> Arithmetic. Even a huge number of them. That is why we have to take
> into account the first person indeterminacies.
>
>
>
>
>> For example would a platonic turing machine
>> simulating another turing machine, simulating a mind be consicous?
>
>
> A 3-machine is never conscious. A 3-entity is never conscious. Only a
> person is. First person can only be associated with the infinities of
> computations computing them in Platonia.
>
>
>
>
>>  If
>> so, how does that differ from a platonic turing machine simulating a
>> physical reality with matter, simulating a mind?
>
>
> You will have to introduce a magical (assuming comp) selection
> principle for attaching, in a persistent way, a mind to that "physical
> reality" simulation. The mind can only be attached to an infinity of
> such relative simulations, and this is why if that mind look at itself
> below its substitution level he will find a trace of those
> computations. Comp says you have to make the statistic on all the
> computations. So the Physical has to be a sum on all those computations.
> That such computations statistically interfere is not so difficult to
> show. That the comp interference gives the apparent quantum one is not
> yet discarded.
>
> I think you are not taking sufficiently into account the first person
> (hopefully plural) indeterminacy in front of the universal dovetailer,
> (or arithmetic) which defined the space of all computations.
>
> Does this help a bit?
>
>
> Bruno
>
>
> http://iridia.ulb.ac.be/~marchal/
>
>
>
>
> >
>

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[Brian Tenneson]

Thank you very much.  I realized I made some false statements as well.

It seems likely that reliance on (not P ->  Q and not Q) -> P being a
tautology is the easiest proof of there being no largest natural number.

Brent Meeker wrote:

> Brian Tenneson wrote:
>  
>>  
>>    
>>> How do you know that there is no biggest number?  Have you examined all
>>> the natural numbers?  How do you prove that there is no biggest number?
>>>
>>>  
>>>    
>>>      
>> In my opinion those are excellent questions.  I will attempt to answer
>> them.  The intended audience of my answer is everyone, so please forgive
>> me if I say something you already know.
>>
>> Firstly, no one has or can examine all the natural numbers.  By that I
>> mean no human.  Maybe there is an omniscient machine (or a "maximally
>> knowledgeable" in some paraconsistent way)  who can examine all numbers
>> but that is definitely putting the cart before the horse.
>>
>> Secondly, the question boils down to a difference in philosophy:
>> mathematicians would say that it is not necessary to examine all natural
>> numbers.  The mathematician would argue that it suffices to examine all
>> essential properties of natural numbers, rather than all natural numbers.
>>
>> There are a variety of equivalent ways to define a natural number but
>> the essential features of natural numbers are that
>> (a) there is an ordering on the set of natural numbers, a well
>> ordering.  To say a set is well ordered entails that every =nonempty=
>> subset of it has a least element.
>> (b) the set of natural numbers has a least element (note that it is
>> customary to either say 0 is this least element or say 1 is this least
>> element--in some sense it does not matter what the starting point is)
>> (c) every natural number has a natural number successor.  By successor
>> of a natural number, I mean anything for which the well ordering always
>> places the successor as larger than the predecessor.
>>
>> Then the set of natural numbers, N, is the set containing the least
>> element (0 or 1) and every successor of the least element, and only
>> successors of the least element.
>>
>> There is nothing wrong with a proof by contradiction but I think a
>> "forward" proof might just be more convincing.
>>
>> Consider the following statement:
>> Whenever S is a subset of N, S has a largest element if, and only if,
>> the complement of S has a least element.
>>  
>>    
>
> Let S={even numbers}  the complement of S, ~S={odd numbers}  ~S has a
> least element, 1.  Therefore there is a largest even number.
>
> Brent
>
>
> >
>
>  

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[m.a.-2]

Some parts of this message have been removed. Learn more about Nabble's security policy.

Bruno,

           I stopped half-way through because I'm not at all sure of my answers and would like to have them confirmed or corrected, if necessary, rather than go on giving wrong answers.   marty a.

----- Original Message -----

From: marchal@...

To: everything-list@...

Sent: Wednesday, June 03, 2009 1:15 PM

Subject: Re: The seven step-Mathematical preliminaries 2

 

=============== Intension and extension ====================

Before defining "intersection, union and the notion of subset, I would like to come back on the ways we can define some specific sets.

In the case of finite and "little" set we have seen that we can define them by exhaustion. This means we can give an explicit complete description of all element of the set. 

Example. A = {0, 1, 2, 77, 98, 5}

When the set is still finite and too big, or if we are lazy, we can sometimes define the set by quasi exhaustion. This means we describe enough elements of the set in a manner which, by requiring some good will and some imagination, we can estimate having define the set.

Example. B = {3, 6, 9, 12, ... 99}. We can understand in this case that we meant the set of multiple of the number three, below 100.

A fortiori, when a set in not finite, that is, when the set is infinite, we have to use either quasi-exhaustion, or we have to use some sentence or phrase or proposition describing the elements of the set.

Definition.

I will say that a set is defined IN EXTENSIO, or simply, in extension, when it is defined in exhaustion or quasi-exhaustion.

I will say that a set is defined IN INTENSIO, or simply in intension, with a "s", when it is defined by a sentence explaining the typical attribute of the elements.

Example: Let A be the set {2, 4, 6, 8, 10, ... 100}. We can easily define A in intension:  A = the set of numbers which are even and more little than 100. mathematician will condense this by the following:

A = {x such that x is even and little than 100}  = {x ⎮ x is even & x < 100}. "⎮" is a special character, abbreviating "such that", and I hope it goes through the mail. If not I will use "such that", or s.t., or things like that.

The expression {x ⎮ x is even} is literally read as:  the set of object x, (or number x if we are in a context where we talk about number) such that x is even.

Exercise 1: Could you define in intension the following infinite set C = {101, 103, 105, ...}

C = ?                          C={x such that x is odd & x <101}

Exercise 2: I will say that a natural number is a multiple of 4 if it can be written as 4*y, for some y. For example 0 is a multiple of 4, (0 = 4*0), but also 28, 400, 404, ...  Could you define in extension the following set D = {x ⎮ x < 10  &  x is a multiple of 4}.    D=4*x  where x = 0 (but also 1,2,3...10)

A last notational, but important symbol. Sets have elements. For example the set A = {1, 2, 3} has three elements 1, 2 and 3. For saying that 3 is an element of A in an a short way, we usually write 3 ∈ A.  this is read as "3 belongs to A", or "3 is in A". Now 4 does not belong to A. To write this in a short way, we will write 4 ∉ A, or we will write ¬ (4 ∈ A) or sometimes just NOT(4 ∈ A). It is read: 4 does not belong to A, or: it is not the case that 4 belongs to A.

Having those notions and notations at our disposition we can speed up on the notion of union and intersection.

The intersection of the sets A and B is the (new) set of those elements which belongs to both A and B. Put in another way: 

The intersection of the sets A with the set B is the set of those elements which belongs to A and which belongs to B. 

This new set, obtained from A and B is written A ∩ B, or A inter. B (in case the special character doesn't go through).

With our notations we can write or define the intersection A ∩ B directly

A ∩ B = {x ⎮ x ∈ A and x ∈ B}.

Example {3, 4, 5, 6, 8} ∩ {5, 6, 7, 9} = {5, 6}

Similarly, we can directly define the union of two sets A and B, written A ∪ B in the following way:

A ∪ B = {x ⎮ x ∈ A or x ∈ B}.    Here we use the usual logical "or". p or q is suppose to be true if p is true or q is true (or both are true). It is not the exclusive "or".

Example {3, 4, 5, 6, 8} ∪ {5, 6, 7, 9} = {3, 4, 5, 6, 7, 8}.   Question: In the example above, 5,6 were the intersection because they were the (only) two numbers BOTH groups had in common. But in this example, 7 is only in the second group yet it is included in the answer. Please explain.

Exercice 3. 

Let N = {0, 1, 2, 3, ...}

Let A = {x ⎮ x < 10}

Let B = {x ⎮ x is even}

Describe in extension (that is: exhaustion or quasi-exhaustion) the following sets:

N ∪ A = {0,1,2,3...} inter {x inter x<10}= {0,1,2,3...9}

N ∪ B = {0,1,2,3....} inter {x inter x is even}= {0,2,4,6...}

A ∪ B = {x inter x <10} inter {x inter x is even}= {0,2,4,6,8}

B ∪ A = {x inter x is even} inter {x inter x < 10}= {0,2,4,6,8}

 

N ∩ A = {0,1,2,3...} inter {x inter x<10}= {0,1,2,3...9}

B ∩ A =  {x inter x is even} inter {x inter x < 10}= {0,2,4,6,8}

N ∩ B =  {0,1,2,3....} inter {x inter x is even}= {0,2,4,6...}

A ∩ B =   {x inter x <10} inter {x inter x is even}= {0,2,4,6,8}

Exercice 4

Is it true that A ∩ B = B ∩ A, whatever A and B are?       yes

Is it true that A ∪ B = B ∪ A, whatever A and B are?      yes

Now, I could give you exercise so that you would be lead to discoveries, but I prefer to be as simple and approachable as possible, and my goal is not even to give you the taste for doing research, so I will do the discovery by myself here and now. Indeed a natural question occurs. What will happen if we try to find the intersection of two sets which have no elements in common? For example, what is the intersection of A = {x ⎮ x is even} with B = {x ⎮ x is odd} ? At first sight we could say that there is no intersection, given that A and B have no elements in common. But a set is just a bit more than its elements. And if there is no elements in the intersection, it means simply that the set A ∩ B has no elements. So we are very inspired if we let that bizarre set to exist, so we give it a name, and call it the empty set, and we can describe it easily in exhaustion by { }, although many describe it as ∅. So, if A and B have no elements in common, A ∩ B is still well defined and is equal to ∅. having a new toy, we can play with it:

Exercise 5, with A and B the same as in exercise 3.

∅ ∪ A =

∅ ∪ B =

A ∪ ∅ =

B ∪ ∅ =

N ∩ ∅ =

B ∩ ∅ =

∅ ∩ B =

∅ ∩ ∅ =

∅ ∪ ∅ =

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

SUBSET

We will say that A is a subset of B (A and B being sets) if, whatever object x represents, each time x belongs to A, it belongs to B. Put in another way it means that IF x belongs to A, THEN x belongs to B. It means that all the elements of A are also elements of B. We can write, with 

x ∈ A -> x ∈ B.

             

And this we abbreviate as A ⊆ B, and we read it: A is included in B.

Example:

1) Let us look if the set A = {1, 2} is included in the set B = {1, 2, 3}.  Here A has two elements. To see if A is included in B, we have to look at each element in the set A, and we have to see if they belongs to B. Now A has two elements, 1, and 2, so we have two tasks to accomplish, or two questions to answer:

does 1 belongs also to B. The answer is yes.

does 2 belongs also to B. The answer is yes.

We have thus verify that all elements of A are also elements of B, and thus we can conclude that A is indeed included in B.

2) Let us look if the set A = {1} is included in B = {1, 2, 3}.  Now, A has only one element. So we are lucky, we have only one task to accomplish! Is 1 an element of B? The answer is yes. Thus we have {1} is included in {1, 2, 3}.

3) Let us look if the set A=  { }, the empty set ∅,  is included in B = {1, 2, 3}. Now A has no element. So we are even more lucky, we have no task to accomplish at all. The condition is trivially satisfied. So the empty set is included in {1, 2, 3}. And this shows that the empty set is included in any set. In particular we have that ∅ ⊆ ∅.

Note that all set is a subset of itself. Trivially, all elements of A is an element of A.

Exercise 6 

We will say that a set A is a subset of a set B, if A is included in B.

Could you give all the subsets of the set  {1, 2}.

Could you give all the subsets of the set  {1}

Could you give all the subsets of the set  { }.

The post is long enough, so I spare you the seventh exercise. Also I have to go, I hope there are not to many typo errors and spelling mistakes, and well, I pray for the special symbols going trough. It is possible that they go through for most mailing systems, but not all. Let me know.

Bon courage,

Bruno

http://iridia.ulb.ac.be/~marchal/

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[russell standish-2]

On Wed, Jun 03, 2009 at 10:11:41AM -0400, Jesse Mazer wrote:
>
> The English term for this is "proof by contradiction":
> http://en.wikipedia.org/wiki/Proof_by_contradiction 
>

Funnily enough, we were taught to call this by the latin phrase
"reductio ad absurdum". I think my maths prof came from Cambridge :).

Cheers

--

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[Torgny Tholerus]

Brian Tenneson skrev:

>  
>> How do you know that there is no biggest number?  Have you examined all
>> the natural numbers?  How do you prove that there is no biggest number?
>>
>>  
>>    
> In my opinion those are excellent questions.  I will attempt to answer
> them.  The intended audience of my answer is everyone, so please forgive
> me if I say something you already know.
>
> Firstly, no one has or can examine all the natural numbers.  By that I
> mean no human.  Maybe there is an omniscient machine (or a "maximally
> knowledgeable" in some paraconsistent way)  who can examine all numbers
> but that is definitely putting the cart before the horse.
>
> Secondly, the question boils down to a difference in philosophy:
> mathematicians would say that it is not necessary to examine all natural
> numbers.  The mathematician would argue that it suffices to examine all
> essential properties of natural numbers, rather than all natural numbers.
>
> There are a variety of equivalent ways to define a natural number but
> the essential features of natural numbers are that
> (a) there is an ordering on the set of natural numbers, a well
> ordering.  To say a set is well ordered entails that every =nonempty=
> subset of it has a least element.
> (b) the set of natural numbers has a least element (note that it is
> customary to either say 0 is this least element or say 1 is this least
> element--in some sense it does not matter what the starting point is)
> (c) every natural number has a natural number successor.  By successor
> of a natural number, I mean anything for which the well ordering always
> places the successor as larger than the predecessor.
>
> Then the set of natural numbers, N, is the set containing the least
> element (0 or 1) and every successor of the least element, and only
> successors of the least element.
>
> There is nothing wrong with a proof by contradiction but I think a
> "forward" proof might just be more convincing.
>
> Consider the following statement:
> Whenever S is a subset of N, S has a largest element if, and only if,
> the complement of S has a least element.
>
> By complement of S, I mean the set of all elements of N that are not
> elements of S.
>
> Before I give a longer argument, would you agree that statement is
> true?  One can actually be arbitrarily explicit: M is the largest
> element of S if, and only if, the successor of M is the least element of
> the compliment of S.
>  

I do not agree that statement is true.  Because if you call the Biggest
natural number B, then you can describe N as = {1, 2, 3, ..., B}.  If
you take the complement of N you will get the empty set.  This set have
no least element, but still N has a biggest element.

In your statement you are presupposing that N has no biggest element,
and from that axiom you can trivially deduce that there is no biggest
element.

--
Torgny Tholerus

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[Quentin Anciaux-2]

If you are ultrafinitist then by definition the set N does not
exist... (nor any infinite set countably or not).

If you pose the assumption of a biggest number for N, you come to a
contradiction because you use the successor operation which cannot
admit a biggest number.(because N is well ordered any successor is
strictly bigger and the successor operation is always valid *by
definition of the operation*)

So either the set N does not exists in which case it makes no sense to
talk about the biggest number in N, or the set N does indeed exists
and it makes no sense to talk about the biggest number in N (while it
makes sense to talk about a number which is strictly bigger than any
natural number).

To come back to the proof by contradiction you gave, the assumption
(2) which is that BIGGEST+1 is in N, is completely defined by the mere
existence of BIGGEST. If BIGGEST exists and well defined it entails
that BIGGEST+1 is not in N (but this invalidate the successor
operation and hence the mere existence of N). If BIGGEST in contrary
does not exist (as such, means it is not the biggest) then BIGGEST+1
is in N by definition of N.

Regards,
Quentin

2009/6/4 Torgny Tholerus <torgny@...>:

>
> Brian Tenneson skrev:
>>
>>> How do you know that there is no biggest number?  Have you examined all
>>> the natural numbers?  How do you prove that there is no biggest number?
>>>
>>>
>>>
>> In my opinion those are excellent questions.  I will attempt to answer
>> them.  The intended audience of my answer is everyone, so please forgive
>> me if I say something you already know.
>>
>> Firstly, no one has or can examine all the natural numbers.  By that I
>> mean no human.  Maybe there is an omniscient machine (or a "maximally
>> knowledgeable" in some paraconsistent way)  who can examine all numbers
>> but that is definitely putting the cart before the horse.
>>
>> Secondly, the question boils down to a difference in philosophy:
>> mathematicians would say that it is not necessary to examine all natural
>> numbers.  The mathematician would argue that it suffices to examine all
>> essential properties of natural numbers, rather than all natural numbers.
>>
>> There are a variety of equivalent ways to define a natural number but
>> the essential features of natural numbers are that
>> (a) there is an ordering on the set of natural numbers, a well
>> ordering.  To say a set is well ordered entails that every =nonempty=
>> subset of it has a least element.
>> (b) the set of natural numbers has a least element (note that it is
>> customary to either say 0 is this least element or say 1 is this least
>> element--in some sense it does not matter what the starting point is)
>> (c) every natural number has a natural number successor.  By successor
>> of a natural number, I mean anything for which the well ordering always
>> places the successor as larger than the predecessor.
>>
>> Then the set of natural numbers, N, is the set containing the least
>> element (0 or 1) and every successor of the least element, and only
>> successors of the least element.
>>
>> There is nothing wrong with a proof by contradiction but I think a
>> "forward" proof might just be more convincing.
>>
>> Consider the following statement:
>> Whenever S is a subset of N, S has a largest element if, and only if,
>> the complement of S has a least element.
>>
>> By complement of S, I mean the set of all elements of N that are not
>> elements of S.
>>
>> Before I give a longer argument, would you agree that statement is
>> true?  One can actually be arbitrarily explicit: M is the largest
>> element of S if, and only if, the successor of M is the least element of
>> the compliment of S.
>>
>
> I do not agree that statement is true.  Because if you call the Biggest
> natural number B, then you can describe N as = {1, 2, 3, ..., B}.  If
> you take the complement of N you will get the empty set.  This set have
> no least element, but still N has a biggest element.
>
> In your statement you are presupposing that N has no biggest element,
> and from that axiom you can trivially deduce that there is no biggest
> element.
>
> --
> Torgny Tholerus
>
> >
>

--
All those moments will be lost in time, like tears in rain.

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[Kim Jones-2]

On Thu Jun  4  1:15 , Bruno Marchal  sent:

>Very good answer, Kim, 
>Just a few comments. and then the sequel.
>Exercice 4: does the real number square-root(2) belongs to {0, 1, 2,  
>3, ...}?
>
>
>No idea what square-root(2) means. When I said I was innumerate I wasn't kidding! I
could of course look
>it up or ask my mathematics teacher friends but I just know your explanation will make
theirs seem trite.
>
>Well thanks. The square root of 2 is a number x, such that x*x (x times x, x multiplied by
itself) gives 2.For example, the square root of 4 is 2, because 2*2 is 4. The square root of
9 is 3, because 3*3 is 9. Her by "square root" I mean the positive square root, because we
will see (more later that soon) that numbers can have negative square root, but please
forget this. At this stage, with this definition, you can guess that the square root of 2
cannot be a natural number. 1*1 = 1, and 2*2 = 4, and it would be astonishing that x
could be bigger than 2. So if there is number x such that x*x is 2, we can guess that such
a x cannot be a natural number, that is an element of {0, 1, 2, 3 ...}, and the answer of
exercise 4 is "no". The square root of two will reappear recurrently, but more in examples,
than in the sequence of notions which are strictly needed for UDA-7.


OK - I find this quite mind-blowing; probably because I now understand it for the first
time in my life. So how did it get this rather ridiculous name of "square root"? What's it
called in French?

(snip)

>=============== Intension and extension ====================
>
>Before defining "intersection, union and the notion of subset, I would like to come back
on the ways we can define some specific sets.
>In the case of finite and "little" set we have seen that we can define them by exhaustion.
This means we can give an explicit complete description of all element of the
set. Example. A = {0, 1, 2, 77, 98, 5}
>When the set is still finite and too big, or if we are lazy, we can sometimes define the set
by quasi exhaustion. This means we describe enough elements of the set in a manner
which, by requiring some good will and some imagination, we can estimate having define
the set.
>Example. B = {3, 6, 9, 12, ... 99}. We can understand in this case that we meant the set of
multiple of the number three, below 100.

>A fortiori, when a set in not finite, that is, when the set is infinite, we have to use either
quasi-exhaustion, or we have to use some sentence or phrase or proposition describing
the elements of the set.

>Definition. I will say that a set is defined IN EXTENSIO, or simply, in extension, when it is
defined in exhaustion or quasi-exhaustion. I will say that a set is defined IN INTENSIO, or
simply in intension, with an "s", when it is defined by a sentence explaining the typical
attribute of the elements.

>Example: Let A be the set {2, 4, 6, 8, 10, ... 100}. We can easily define A in intension:  A
= the set of numbers which are even and smaller than 100. Mathematicians will condense
this by the following:
>A = {x such that x is even and smaller than 100}  = {x ⎮ x is even & x
special character, abbreviating "such that", and I hope it goes through the mail.




Just an upright line? It comes through as that. I can't seem to get this symbol happening so I will
use "such that"




 If not I will use "such that", or s.t., or things like that.The expression {x ⎮ x is even} is
literally read as:  the set of objects x, (or number x if we are in a context where we talk
about numbers) such that x is even.

>Exercise 1: Could you define in intension the following infinite set C = {101, 103, 105,
...}C = ?


C = {x such that x is odd and x > 101}


>Exercise 2: I will say that a natural number is a multiple of 4 if it can be written as 4*y,
for some y. For example 0 is a multiple of 4, (0 = 4*0), but also 28, 400, 404, ...  Could
you define in extension the following set D = {x ⎮ x < 10 and x is a multiple of 4}?

D = 4*x where x = 0 but also { 1, 2, 3, 4, 8 }


I now realise I am doomed for the next set of exercises because I cannot get to the special
symbols required (yet). As I am adding Internet Phone to my system, I am currently using an
ancient Mac without the correct symbol pallette while somebody spends a few days to flip a single
switch...as soon as I can get back to my regular machine I will complete the rest.

In the meantime I am enjoying the N+1 disagreement - how refreshing it is to see that classical
mathematics remains somewhat controversial!


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