The seven step series

[Bruno Marchal]

On 13 Jul 2009, at 09:38, Kim Jones wrote:

>
>
> On 11/07/2009, at 6:24 AM, Bruno Marchal wrote:
>
>>
>> I am also a bit anxious about Kim, who is the one who suggested me  
>> the
>> initial explanations, but who seems to have disappear right now.
>
>
>
> OK - I'm back.

I am quite glad you are back, and alive!



> Since May 27 to two days ago I have been without
> Internet access.

Well, that is form of death nowadays. I wish you an happy resurrection.

>
>
> I made the mistake of upgrading my Broadband plan to add Internet
> phone. It took two telcos a month to complete this ridiculously basic
> operation with mistakes made and attendant extra waiting times.
>
> Then, just as the connection was restored at the beginning of July,
> the plumbing in this block of apartments fell apart and a major
> excavation work went ahead and this time the plumbers cut the phone
> cable and didn't realise it which meant I wasted another week trying
> to get the problem diagnosed.

Matter kicks back, even with comp!

>
>
> So now finally everything is back to normal. I have just started
> reading this thread and can see that the class is a very exclusive
> one! I will try my best to follow through on the exercises and the
> comments, corrections. I feel I have access to the correct
> mathematical symbols on my Mac now but *time* is the thing that I
> don't have much of anymore, so I feel a bit depressed about the level
> of effort I can devote to it. If only we didn't have to work for a
> living things would be vastly easier.

You are right. Take your time and ask any question. I will try to sum  
up through the next posts.


>
>
> The notion of sets is indeed a tricky one. I am just now going over
> the initial exercises again. Do not wait for me. I am also trying to
> catch up on about 4,000 emails.

Gosh!


>
>
> Bruno - my sincerest apologies for this hiatus. You seem eager to get
> to the seventh and eighth steps. Why wouldn't you be.


No, we have all the time. I just answered to posts, and made  
elementary recalls of math. But I realize that some people have not  
follow any course in set theory. In Belgium it was taught as "modern  
math" during 10 years in high school, and then it has disappeared  
(alas 1000X). The discrepancy between the math level of different  
people is very big, but the shortcut I am pursuing now, should not be  
insurmountable. I will make the next posts as self-contained as  
possible, but please interrupt when you feel I am esoteric.

Have a good day,

Bruno



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




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[Bruno Marchal]

Hi Kim, Marty, Johnathan, John, Mirek, and all...

We were studying a bit of elementary set theory to prepare ourself to Cantor's theorem, and then Kleene's theorem, which are keys to a good understanding of the universal numbers, and to Church thesis, which are the keys of the seven steps. 

I intend to bring you to the comp enlightenment :)

But first some revision. Read the following with attention!

A set is a collection of things, which in general can themselves be anything. Its use consists in making a many into a one.

If something, say x,  belongs to a set S, it is usually called "element" of S. We abbreviate this by (x \belongs-to S).

Example:

A = {1, 2, 56}. A is a set with three elements which are the numbers 1, 2 and 56.

We write:

(1 \belongs-to {1, 2, 56}), or (1 \belongs-to A), or simply 1 \belongs-to A, when no confusions exist. The parentheses "(" and ")" are just delimiters for easing the reading. I write \belongs-to the relation "belongs to" to remind it is a mathematical symbol.

B = {Kim, Marty, Russell, Bruno, George, Jurgen} is  a set with 5 elements which are supposed to be humans.

C = {34, 54, Paul, {3, 4}}

For this one, you may be in need of spectacles. In case of doubt, you can expand it a little bit:

C = {    34,     54,    Paul,   {3, 4}     }

You see that C is a sort of hybrid set which has 4 elements:

   - the number 34

   - the number 54

   - the human person Paul

   - the set {3, 4}

Two key remarks:

1) the number 3 is NOT an element of C. Nor is the number 4 an element of C. 3 and 4 are elements of {3, 4}, which is an element of C. But, generally, elements of elements are not elements! It could happen that element of element are element, like in D = {3, 4, {3, 4}}, the number 3 is both an element of D and element of an element of D ({3, 4}), but this is a special circumstance due to the way D is defined.

2) How do I know that "Paul" is a human, and not a dog. How do I know that "Paul" does not refer just to the string "paul". Obvioulsy the expression "paul" is ambiguous, and will usually be understood only in some context. This will not been a problem because the context will be clear. Actually we will consider only set of numbers, or set of mathematical objects which have already been defined. Here I have use the person Paul just to remind that typically set can have as elements any object you can conceive.

What is the set of even prime number strictly bigger than 2. Well, to solve this just recall that ALL prime numbers are odd, except 2. So this set is empty. The empty set { } is the set which has no elements. It plays the role of 0 in the world of sets.

We have seen some operations defined on sets.

We have seen INTERSECTION, and UNION.

The intersection of the two sets S1 = {1, 2, 3} and S2 = {2, 3, 7, 8} will be written (S1 \inter S2), and is equal to the set of elements which belongs to both S1 and S2. We have

(S1 \inter S2) = {2, 3}

We can define (S1 \inter S2) = {x such-that ((x belongs-to S1) and (x belongs-to S2))}

2 belongs to (S1 \inter S2) because ((2 belongs-to S1) and (2 belongs-to S2))

8 does not belongs to (S1 \inter S2) because it is false that ((2 belongs-to S1) and (2 belongs-to S2)). Indeed 8 does not belong to S1.

Of course some sets can be disjoint, that is, can have an empty intersection:

{1, 2, 3} \inter {4, 5, 6} = { }.

Similarly we can define (S1 \union S2) by the set of the elements belonging to S1 or belonging to S2:

(S1 \union S2) = {x such-that ((x belongs-to S1) or (x belongs-to S2))}

We have, with S1 and S2 the same as above (S1 = {1, 2, 3} and S2 = {2, 3, 7, 8}):

(S1 \union S2) = {1, 2, 3, 7, 8}.

OK. I suggest you reread the preceding post, and let me know in case you have a problem.

We have seen also a key relation defined on sets: the relation of inclusion.

We say that (A \included-in B) is true when all elements of A are also elements of B.

Example:

The set of ferocious dogs is included in the set of ferocious animals.

The set of even numbers is included in the set of natural numbers.

The set {2, 6, 8} is included in the set {2, 3, 4, 5, 6, 7, 8}

The set {2, 6, 8} is NOT included in the set  {2, 3, 4, 5, 7, 8}.

When a set A is included in a set B, A is called a subset of B.

We were interested in looking to all subsets of a some set.

What are the subsets of {a, b} ?

They are { }, {a}, {b}, {a, b}. Why?

{a, b} is included in {a, b}. This is obvious. All elements of {a,b} are elements of {a, b}.

{a} is included in {a, b}, because all elements of {a} are elements of {a, b}

The same for {b}.

You see that to verify that a set with n elements is a subset of some set, you have to make n verifications. 

So, to see that the empty set is a subset of some set, you have to verify 0 things. So the empty set is a subset of any set.

proposition: { } is included-in any set.

So the subsets of {a, b} are { }, {a}, {b}, {a, b}.

But set have been invented to make a ONE from a MANY, and it is natural to consider THE set of all subsets of a set. It is called the powerset of that set.

So the powerset of {a, b} is THE set {{ }, {a}, {b}, {a, b}}. OK?

Train yourself on the following exercises:

What is the powerset of { }

What is the powerset of {a}

What is the powerset of {a, b, c}

Any question?

This was a bit of revision, to let Kim catch up.

The sequel will appear asap. Be sure everything is OK, and please, ask question if it is not. 

You can also ask any question on the first sixth steps of UDA ('course).

Bruno

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

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[Miroslav Dobsicek]

Hi Bruno,

I am in a good mood and a bit picky :-) Do you know how many entries
google gave me upon entering
 Theaetetical -marchal -bruno

Mirek

> for some people I think. It is just unusual.


> theorem and the Theaetetical definitions of knowledge.





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[Bruno Marchal]

Hi  Mirek,

>
> Hi Bruno,
>
> I am in a good mood and a bit picky :-) Do you know how many entries
> google gave me upon entering
> Theaetetical -marchal -bruno


Well 144?

Good way to find my papers on that. The pages refer quickly to this  
list or the FOR list.

Tomorrow I will not be there.

I let those interested to meditate on two questions (N is {0, 1, 2, 3,  
4, ...}):

1) What is common between the set of all subsets of a set with n  
elements, and the set of all finite sequences of "0" and "1" of length  
n.
2) What is common between the set of all subsets of N, and the set of  
all infinite sequences of "0" and "1".

Just some (finite and infinite) bread for surviving the day :)

Bruno


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




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[John Mikes]

Bruno,

I appreciate your grade-school teaching. We (I for one) can use it.

I still find that whatever you explain is an 'extract' of what can be thought of a 'set' (a one representing a many).

Your 'powerset' is my example.

All those elements you put into { }s are the same as were the physical objects to Aristotle in his 'total' - the SUM of which was always MORE than the additives of those objects.

Relations!

The set is not an inordinate heap (correct me please, if I am off) of the elements, the elements are in SOME relation to each other and the "set"-idea of their ensemble, to form a SET.

You stop short at the naked elements together, as I see.

They wear cloths and hold hands. Mortar is among them.

Maybe your math-idea can tolerate any sequence and hiatus concerning to the 'set', and it still stays the same, as far as the "math-idea you need" goes,  but if I go further (and you indicated that ANYTHING can form a set) the relations of the set-partners comes into play. Not only those which WE choose for 'interesting' to such set, but ALL OF THEM influencing the character of that "ONE".

Just musing.

 

John

 

On Tue, Jul 14, 2009 at 4:40 AM, Bruno Marchal <marchal@...> wrote:

Hi Kim, Marty, Johnathan, John, Mirek, and all...

We were studying a bit of elementary set theory to prepare ourself to Cantor's theorem, and then Kleene's theorem, which are keys to a good understanding of the universal numbers, and to Church thesis, which are the keys of the seven steps. 

I intend to bring you to the comp enlightenment :)

But first some revision. Read the following with attention!

A set is a collection of things, which in general can themselves be anything. Its use consists in making a many into a one.

If something, say x,  belongs to a set S, it is usually called "element" of S. We abbreviate this by (x \belongs-to S).

Example:

A = {1, 2, 56}. A is a set with three elements which are the numbers 1, 2 and 56.

We write:

(1 \belongs-to {1, 2, 56}), or (1 \belongs-to A), or simply 1 \belongs-to A, when no confusions exist. The parentheses "(" and ")" are just delimiters for easing the reading. I write \belongs-to the relation "belongs to" to remind it is a mathematical symbol.

B = {Kim, Marty, Russell, Bruno, George, Jurgen} is  a set with 5 elements which are supposed to be humans.

C = {34, 54, Paul, {3, 4}}

For this one, you may be in need of spectacles. In case of doubt, you can expand it a little bit:

C = {    34,     54,    Paul,   {3, 4}     }

You see that C is a sort of hybrid set which has 4 elements:

   - the number 34

   - the number 54

   - the human person Paul

   - the set {3, 4}

Two key remarks:

1) the number 3 is NOT an element of C. Nor is the number 4 an element of C. 3 and 4 are elements of {3, 4}, which is an element of C. But, generally, elements of elements are not elements! It could happen that element of element are element, like in D = {3, 4, {3, 4}}, the number 3 is both an element of D and element of an element of D ({3, 4}), but this is a special circumstance due to the way D is defined.

2) How do I know that "Paul" is a human, and not a dog. How do I know that "Paul" does not refer just to the string "paul". Obvioulsy the expression "paul" is ambiguous, and will usually be understood only in some context. This will not been a problem because the context will be clear. Actually we will consider only set of numbers, or set of mathematical objects which have already been defined. Here I have use the person Paul just to remind that typically set can have as elements any object you can conceive.

What is the set of even prime number strictly bigger than 2. Well, to solve this just recall that ALL prime numbers are odd, except 2. So this set is empty. The empty set { } is the set which has no elements. It plays the role of 0 in the world of sets.

We have seen some operations defined on sets.

We have seen INTERSECTION, and UNION.

The intersection of the two sets S1 = {1, 2, 3} and S2 = {2, 3, 7, 8} will be written (S1 \inter S2), and is equal to the set of elements which belongs to both S1 and S2. We have

(S1 \inter S2) = {2, 3}

We can define (S1 \inter S2) = {x such-that ((x belongs-to S1) and (x belongs-to S2))}

2 belongs to (S1 \inter S2) because ((2 belongs-to S1) and (2 belongs-to S2))

8 does not belongs to (S1 \inter S2) because it is false that ((2 belongs-to S1) and (2 belongs-to S2)). Indeed 8 does not belong to S1.

Of course some sets can be disjoint, that is, can have an empty intersection:

{1, 2, 3} \inter {4, 5, 6} = { }.

Similarly we can define (S1 \union S2) by the set of the elements belonging to S1 or belonging to S2:

(S1 \union S2) = {x such-that ((x belongs-to S1) or (x belongs-to S2))}

We have, with S1 and S2 the same as above (S1 = {1, 2, 3} and S2 = {2, 3, 7, 8}):

(S1 \union S2) = {1, 2, 3, 7, 8}.

OK. I suggest you reread the preceding post, and let me know in case you have a problem.

We have seen also a key relation defined on sets: the relation of inclusion.

We say that (A \included-in B) is true when all elements of A are also elements of B.

Example:

The set of ferocious dogs is included in the set of ferocious animals.

The set of even numbers is included in the set of natural numbers.

The set {2, 6, 8} is included in the set {2, 3, 4, 5, 6, 7, 8}

The set {2, 6, 8} is NOT included in the set  {2, 3, 4, 5, 7, 8}.

When a set A is included in a set B, A is called a subset of B.

We were interested in looking to all subsets of a some set.

What are the subsets of {a, b} ?

They are { }, {a}, {b}, {a, b}. Why?

{a, b} is included in {a, b}. This is obvious. All elements of {a,b} are elements of {a, b}.

{a} is included in {a, b}, because all elements of {a} are elements of {a, b}

The same for {b}.

You see that to verify that a set with n elements is a subset of some set, you have to make n verifications. 

So, to see that the empty set is a subset of some set, you have to verify 0 things. So the empty set is a subset of any set.

proposition: { } is included-in any set.

So the subsets of {a, b} are { }, {a}, {b}, {a, b}.

But set have been invented to make a ONE from a MANY, and it is natural to consider THE set of all subsets of a set. It is called the powerset of that set.

So the powerset of {a, b} is THE set {{ }, {a}, {b}, {a, b}}. OK?

Train yourself on the following exercises:

What is the powerset of { }

What is the powerset of {a}

What is the powerset of {a, b, c}

Any question?

This was a bit of revision, to let Kim catch up.

The sequel will appear asap. Be sure everything is OK, and please, ask question if it is not. 

You can also ask any question on the first sixth steps of UDA ('course).

Bruno

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

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[m.a.-2]

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

 

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

From: marchal@...

To: everything-list@...

Sent: Tuesday, July 14, 2009 4:40 AM

Subject: Re: The seven step series

Hi Kim, Marty, Johnathan, John, Mirek, and all...

Bruno: May I advise you about an instance of English usage? The word "supposed" in the next sentence is often used as sarcasm to imply serious doubt about the statement. In this context it can be interpreted as a slight. I think you meant to say "assumed" which implies an evident fact. Please don't apologize, we are most grateful for your efforts in using English and are happy to make allowances for minor slips.

B = {Kim, Marty, Russell, Bruno, George, Jurgen} is  a set with 5 elements which are supposed to be humans.

 

 

 

 

I also have a question: see below:

 

We have seen INTERSECTION, and UNION.

The intersection of the two sets S1 = {1, 2, 3} and S2 = {2, 3, 7, 8} will be written (S1 \inter S2), and is equal to the set of elements which belongs to both S1 and S2. We have

(S1 \inter S2) = {2, 3}

We can define (S1 \inter S2) = {x such-that ((x belongs-to S1) and (x belongs-to S2))}

2 belongs to (S1 \inter S2) because ((2 belongs-to S1) and (2 belongs-to S2))

8 does not belongs to (S1 \inter S2) because it is false that ((2 belongs-to S1) and (2 belongs-to S2)). Indeed 8 does not belong to S1.

Doesn't the statement in bold (above) contradict the statement immediately preceding (also in bold)?

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[Kim Jones-2]

On 14/07/2009, at 6:40 PM, Bruno Marchal wrote:

> The intersection of the two sets S1 = {1, 2, 3} and S2 = {2, 3, 7,  
> 8} will be written (S1 \inter S2), and is equal to the set of  
> elements which belongs to both S1 and S2. We have
>
> (S1 \inter S2) = {2, 3}
>
> We can define (S1 \inter S2) = {x such-that ((x belongs-to S1) and  
> (x belongs-to S2))}
>
> 2 belongs to (S1 \inter S2) because ((2 belongs-to S1) and (2  
> belongs-to S2))
> 8 does not belongs to (S1 \inter S2) because it is false that ((2  
> belongs-to S1) and (2 belongs-to S2)). Indeed 8 does not belong to S1.
>

Quick (silly) questions:

1.

why do you have to write "\inter"  ? Why not just write "inter"  ?

Typing "\" causes me to make use of a key on my keyboard I have never  
used before which is scary ;-)


2.

"such-that" is surely "such that" but the hyphen might just mean  
something
(this is mathematics; there are dots and dashes and slashes all over  
the place so you have to know what they all mean)

likewise

"belongs-to" would still mean the same thing if we wrote "belongs to"  
would it not?


best

K






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[Bruno Marchal]

On 15 Jul 2009, at 00:50, John Mikes wrote:

Bruno,

I appreciate your grade-school teaching. We (I for one) can use it.

I still find that whatever you explain is an 'extract' of what can be thought of a 'set' (a one representing a many).

Your 'powerset' is my example.

All those elements you put into { }s are the same as were the physical objects to Aristotle in his 'total' - the SUM of which was always MORE than the additives of those objects.

Relations!

The set is not an inordinate heap (correct me please, if I am off) of the elements, the elements are in SOME relation to each other and the "set"-idea of their ensemble, to form a SET.

You stop short at the naked elements together, as I see.

You get the idea. 

We can add structure to sets, by explicitly endowing them with operations and relations.

They wear cloths and hold hands. Mortar is among them.

Maybe your math-idea can tolerate any sequence and hiatus concerning to the 'set', and it still stays the same, as far as the "math-idea you need" goes,  

Yes, it is the methodology.

but if I go further (and you indicated that ANYTHING can form a set)

More precisily, we can form a set of multiple thing we can conceive or defined.

the relations of the set-partners comes into play. Not only those which WE choose for 'interesting' to such set, but ALL OF THEM influencing the character of that "ONE".

Just musing.

It is OK. The idea consists in simplifying the things as much as possible, and then to realize that despite such simplification we are quickly driven to the unprovable, unnameable, un-reductible, far sooner than we could have imagine. 

Bruno

On Tue, Jul 14, 2009 at 4:40 AM, Bruno Marchal <marchal@...> wrote:

Hi Kim, Marty, Johnathan, John, Mirek, and all...

We were studying a bit of elementary set theory to prepare ourself to Cantor's theorem, and then Kleene's theorem, which are keys to a good understanding of the universal numbers, and to Church thesis, which are the keys of the seven steps. 

I intend to bring you to the comp enlightenment :)

But first some revision. Read the following with attention!

A set is a collection of things, which in general can themselves be anything. Its use consists in making a many into a one.

If something, say x,  belongs to a set S, it is usually called "element" of S. We abbreviate this by (x \belongs-to S).

Example:

A = {1, 2, 56}. A is a set with three elements which are the numbers 1, 2 and 56.

We write:

(1 \belongs-to {1, 2, 56}), or (1 \belongs-to A), or simply 1 \belongs-to A, when no confusions exist. The parentheses "(" and ")" are just delimiters for easing the reading. I write \belongs-to the relation "belongs to" to remind it is a mathematical symbol.

B = {Kim, Marty, Russell, Bruno, George, Jurgen} is  a set with 5 elements which are supposed to be humans.

C = {34, 54, Paul, {3, 4}}

For this one, you may be in need of spectacles. In case of doubt, you can expand it a little bit:

C = {    34,     54,    Paul,   {3, 4}     }

You see that C is a sort of hybrid set which has 4 elements:

   - the number 34

   - the number 54

   - the human person Paul

   - the set {3, 4}

Two key remarks:

1) the number 3 is NOT an element of C. Nor is the number 4 an element of C. 3 and 4 are elements of {3, 4}, which is an element of C. But, generally, elements of elements are not elements! It could happen that element of element are element, like in D = {3, 4, {3, 4}}, the number 3 is both an element of D and element of an element of D ({3, 4}), but this is a special circumstance due to the way D is defined.

2) How do I know that "Paul" is a human, and not a dog. How do I know that "Paul" does not refer just to the string "paul". Obvioulsy the expression "paul" is ambiguous, and will usually be understood only in some context. This will not been a problem because the context will be clear. Actually we will consider only set of numbers, or set of mathematical objects which have already been defined. Here I have use the person Paul just to remind that typically set can have as elements any object you can conceive.

What is the set of even prime number strictly bigger than 2. Well, to solve this just recall that ALL prime numbers are odd, except 2. So this set is empty. The empty set { } is the set which has no elements. It plays the role of 0 in the world of sets.

We have seen some operations defined on sets.

We have seen INTERSECTION, and UNION.

The intersection of the two sets S1 = {1, 2, 3} and S2 = {2, 3, 7, 8} will be written (S1 \inter S2), and is equal to the set of elements which belongs to both S1 and S2. We have

(S1 \inter S2) = {2, 3}

We can define (S1 \inter S2) = {x such-that ((x belongs-to S1) and (x belongs-to S2))}

2 belongs to (S1 \inter S2) because ((2 belongs-to S1) and (2 belongs-to S2))

8 does not belongs to (S1 \inter S2) because it is false that ((2 belongs-to S1) and (2 belongs-to S2)). Indeed 8 does not belong to S1.

Of course some sets can be disjoint, that is, can have an empty intersection:

{1, 2, 3} \inter {4, 5, 6} = { }.

Similarly we can define (S1 \union S2) by the set of the elements belonging to S1 or belonging to S2:

(S1 \union S2) = {x such-that ((x belongs-to S1) or (x belongs-to S2))}

We have, with S1 and S2 the same as above (S1 = {1, 2, 3} and S2 = {2, 3, 7, 8}):

(S1 \union S2) = {1, 2, 3, 7, 8}.

OK. I suggest you reread the preceding post, and let me know in case you have a problem.

We have seen also a key relation defined on sets: the relation of inclusion.

We say that (A \included-in B) is true when all elements of A are also elements of B.

Example:

The set of ferocious dogs is included in the set of ferocious animals.

The set of even numbers is included in the set of natural numbers.

The set {2, 6, 8} is included in the set {2, 3, 4, 5, 6, 7, 8}

The set {2, 6, 8} is NOT included in the set  {2, 3, 4, 5, 7, 8}.

When a set A is included in a set B, A is called a subset of B.

We were interested in looking to all subsets of a some set.

What are the subsets of {a, b} ?

They are { }, {a}, {b}, {a, b}. Why?

{a, b} is included in {a, b}. This is obvious. All elements of {a,b} are elements of {a, b}.

{a} is included in {a, b}, because all elements of {a} are elements of {a, b}

The same for {b}.

You see that to verify that a set with n elements is a subset of some set, you have to make n verifications. 

So, to see that the empty set is a subset of some set, you have to verify 0 things. So the empty set is a subset of any set.

proposition: { } is included-in any set.

So the subsets of {a, b} are { }, {a}, {b}, {a, b}.

But set have been invented to make a ONE from a MANY, and it is natural to consider THE set of all subsets of a set. It is called the powerset of that set.

So the powerset of {a, b} is THE set {{ }, {a}, {b}, {a, b}}. OK?

Train yourself on the following exercises:

What is the powerset of { }

What is the powerset of {a}

What is the powerset of {a, b, c}

Any question?

This was a bit of revision, to let Kim catch up.

The sequel will appear asap. Be sure everything is OK, and please, ask question if it is not. 

You can also ask any question on the first sixth steps of UDA ('course).

Bruno

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

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

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[Bruno Marchal]

On 15 Jul 2009, at 04:15, m.a. wrote:

 

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

From: marchal@...

To: everything-list@...

Sent: Tuesday, July 14, 2009 4:40 AM

Subject: Re: The seven step series

Hi Kim, Marty, Johnathan, John, Mirek, and all...

Bruno: May I advise you about an instance of English usage? The word "supposed" in the next sentence is often used as sarcasm to imply serious doubt about the statement. In this context it can be interpreted as a slight. I think you meant to say "assumed" which implies an evident fact. Please don't apologize, we are most grateful for your efforts in using English and are happy to make allowances for minor slips.

B = {Kim, Marty, Russell, Bruno, George, Jurgen} is  a set with 5 elements which are supposed to be humans.

Thanks for letting me know. In french "I assume" is the same as "I suppose". I'm afraid it will take time for me not doing that error again. But don't hesitate to remind me of the "false friend" behavior. Sorry for the unintended sarcasm.

To be sure it is not really an assumption, and a "supposition" means more like an "obvious implicit fact we should take into account without mentioning", as opposed to an "assumption" which is more akin to "a key hypothesis". Here I was referring to conventions only, but then, as yopu point out, an non intended sarcasm could be see. Difficult. That is why I prefer to stick on less ambiguous, purely mathematical examples of sets.

I also have a question: see below:

 

We have seen INTERSECTION, and UNION.

The intersection of the two sets S1 = {1, 2, 3} and S2 = {2, 3, 7, 8} will be written (S1 \inter S2), and is equal to the set of elements which belongs to both S1 and S2. We have

(S1 \inter S2) = {2, 3}

We can define (S1 \inter S2) = {x such-that ((x belongs-to S1) and (x belongs-to S2))}

2 belongs to (S1 \inter S2) because ((2 belongs-to S1) and (2 belongs-to S2))

8 does not belongs to (S1 \inter S2) because it is false that ((2 belongs-to S1) and (2 belongs-to S2)). Indeed 8 does not belong to S1.

Doesn't the statement in bold (above) contradict the statement immediately preceding (also in bold)?

You are completely right. I just did a copy and past, and forget to substitute the "2" by the "8".

So the statements are:

2 belongs to (S1 \inter S2) because ((2 belongs-to S1) and (2 belongs-to S2))

8 does not belongs to (S1 \inter S2) because it is false that ((8 belongs-to S1) and (8 belongs-to S2)).

Bruno

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

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[Bruno Marchal]

On 15 Jul 2009, at 09:09, Kim Jones wrote:

On 14/07/2009, at 6:40 PM, Bruno Marchal wrote:

The intersection of the two sets S1 = {1, 2, 3} and S2 = {2, 3, 7,  

8} will be written (S1 \inter S2), and is equal to the set of  

elements which belongs to both S1 and S2. We have

(S1 \inter S2) = {2, 3}

We can define (S1 \inter S2) = {x such-that ((x belongs-to S1) and  

(x belongs-to S2))}

2 belongs to (S1 \inter S2) because ((2 belongs-to S1) and (2  

belongs-to S2))

8 does not belongs to (S1 \inter S2) because it is false that ((2  

belongs-to S1) and (2 belongs-to S2)). Indeed 8 does not belong to S1.

Quick (silly) questions:

1.

why do you have to write "\inter"  ? Why not just write "inter"  ?

Typing "\" causes me to make use of a key on my keyboard I have never  
used before which is scary ;-)

For the intersection of two sets S1 and S1,  I have used  

1)  S1 ∩ S2

But the math symbol "∩"  did not go through all emailing system, so, I have used

2) S1 INTERSECTION S2

But then I recall that in mails, capital letters seems aggressive, loudly ..., so I have used

3) S1 intersection S2

But then the difference between what is supposed to be represent a mathematical symbols, and a word in english, disappears, so I have used

4) S1 \intersection S2

But this, on the last post, seems to me to be a little too long, and so I am using now

5) S1 \inter S2

Only God knows what I will use tomorrow. You should learn that there is no standard of mathematical notations, and no two mathematicians use the same symbols, and not one mathematician use the same symbols in two different texts. What is nice, is that, usually, mathematicians quickly redefine what they mean by any symbols at the beginning of their books and papers.

Of course doing math on mails aggravates apparently this search for the symbols which could satisfy everyone.

Sorry to scare you,

2.

"such-that" is surely "such that" but the hyphen might just mean  
something
(this is mathematics; there are dots and dashes and slashes all over  
the place so you have to know what they all mean)

likewise

"belongs-to" would still mean the same thing if we wrote "belongs to"  
would it not?

Same remark. I should have use \such-that, and \belongs-to.

Note that at this stage it is really not important to be aware of the difference between a symbol and what the symbols refer too, but in logic such differences acquire some importance at some point, and I just try to prepare you for such nuances, having the sequel of this introduction in my mind.

You question are not silly and makes sense, as you see,

Bruno

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

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[John Mikes]

Please read between your lines included in bold letters

John

On Thu, Jul 16, 2009 at 4:13 AM, Bruno Marchal <marchal@...> wrote:

On 15 Jul 2009, at 00:50, John Mikes wrote:

Bruno,

I appreciate your grade-school teaching. We (I for one) can use it.

I still find that whatever you explain is an 'extract' of what can be thought of a 'set' (a one representing a many).

Your 'powerset' is my example.

All those elements you put into { }s are the same as were the physical objects to Aristotle in his 'total' - the SUM of which was always MORE than the additives of those objects.

Relations!

The set is not an inordinate heap (correct me please, if I am off) of the elements, the elements are in SOME relation to each other and the "set"-idea of their ensemble, to form a SET.

You stop short at the naked elements together, as I see.

You get the idea. 

We can add structure to sets, by explicitly endowing them with operations and relations.

Furhter below you also expose the contrary (to simplify) - I am afraid your "operations and relations" are restricted to the numbers-based (math?) domain, which is not what I mean by 'totality'.

They wear cloths and hold hands. Mortar is among them.

Maybe your math-idea can tolerate any sequence and hiatus concerning to the 'set', and it still stays the same, as far as the "math-idea you need" goes,  

Yes, it is the methodology.

but if I go further (and you indicated that ANYTHING can form a set)

More precisily, we can form a set of multiple thing we can conceive or defined.

 

I would not restrict 'a set' to what WE can conceive, or define now. (Not even within the 'math'-related domain).

the relations of the set-partners comes into play. Not only those which WE choose for 'interesting' to such set, but ALL OF THEM influencing the character of that "ONE".

Just musing.

It is OK. The idea consists in simplifying the things as much as possible, and then to realize that despite such simplification we are quickly driven to the unprovable, unnameable, un-reductible, far sooner than we could have imagine. 

 

I may suggest (or: assume?) that instead of "despite" it would make more sense to write: "AS A CONSEQUENCE"

- think about it.

 

Bruno

John

On Tue, Jul 14, 2009 at 4:40 AM, Bruno Marchal <marchal@...> wrote:

Hi Kim, Marty, Johnathan, John, Mirek, and all...

We were studying a bit of elementary set theory to prepare ourself to Cantor's theorem, and then Kleene's theorem, which are keys to a good understanding of the universal numbers, and to Church thesis, which are the keys of the seven steps. 

I intend to bring you to the comp enlightenment :)

But first some revision. Read the following with attention!

A set is a collection of things, which in general can themselves be anything. Its use consists in making a many into a one.

If something, say x,  belongs to a set S, it is usually called "element" of S. We abbreviate this by (x \belongs-to S).

Example:

A = {1, 2, 56}. A is a set with three elements which are the numbers 1, 2 and 56.

We write:

(1 \belongs-to {1, 2, 56}), or (1 \belongs-to A), or simply 1 \belongs-to A, when no confusions exist. The parentheses "(" and ")" are just delimiters for easing the reading. I write \belongs-to the relation "belongs to" to remind it is a mathematical symbol.

B = {Kim, Marty, Russell, Bruno, George, Jurgen} is  a set with 5 elements which are supposed to be humans.

C = {34, 54, Paul, {3, 4}}

For this one, you may be in need of spectacles. In case of doubt, you can expand it a little bit:

C = {    34,     54,    Paul,   {3, 4}     }

You see that C is a sort of hybrid set which has 4 elements:

   - the number 34

   - the number 54

   - the human person Paul

   - the set {3, 4}

Two key remarks:

1) the number 3 is NOT an element of C. Nor is the number 4 an element of C. 3 and 4 are elements of {3, 4}, which is an element of C. But, generally, elements of elements are not elements! It could happen that element of element are element, like in D = {3, 4, {3, 4}}, the number 3 is both an element of D and element of an element of D ({3, 4}), but this is a special circumstance due to the way D is defined.

2) How do I know that "Paul" is a human, and not a dog. How do I know that "Paul" does not refer just to the string "paul". Obvioulsy the expression "paul" is ambiguous, and will usually be understood only in some context. This will not been a problem because the context will be clear. Actually we will consider only set of numbers, or set of mathematical objects which have already been defined. Here I have use the person Paul just to remind that typically set can have as elements any object you can conceive.

What is the set of even prime number strictly bigger than 2. Well, to solve this just recall that ALL prime numbers are odd, except 2. So this set is empty. The empty set { } is the set which has no elements. It plays the role of 0 in the world of sets.

We have seen some operations defined on sets.

We have seen INTERSECTION, and UNION.

The intersection of the two sets S1 = {1, 2, 3} and S2 = {2, 3, 7, 8} will be written (S1 \inter S2), and is equal to the set of elements which belongs to both S1 and S2. We have

(S1 \inter S2) = {2, 3}

We can define (S1 \inter S2) = {x such-that ((x belongs-to S1) and (x belongs-to S2))}

2 belongs to (S1 \inter S2) because ((2 belongs-to S1) and (2 belongs-to S2))

8 does not belongs to (S1 \inter S2) because it is false that ((2 belongs-to S1) and (2 belongs-to S2)). Indeed 8 does not belong to S1.

Of course some sets can be disjoint, that is, can have an empty intersection:

{1, 2, 3} \inter {4, 5, 6} = { }.

Similarly we can define (S1 \union S2) by the set of the elements belonging to S1 or belonging to S2:

(S1 \union S2) = {x such-that ((x belongs-to S1) or (x belongs-to S2))}

We have, with S1 and S2 the same as above (S1 = {1, 2, 3} and S2 = {2, 3, 7, 8}):

(S1 \union S2) = {1, 2, 3, 7, 8}.

OK. I suggest you reread the preceding post, and let me know in case you have a problem.

We have seen also a key relation defined on sets: the relation of inclusion.

We say that (A \included-in B) is true when all elements of A are also elements of B.

Example:

The set of ferocious dogs is included in the set of ferocious animals.

The set of even numbers is included in the set of natural numbers.

The set {2, 6, 8} is included in the set {2, 3, 4, 5, 6, 7, 8}

The set {2, 6, 8} is NOT included in the set  {2, 3, 4, 5, 7, 8}.

When a set A is included in a set B, A is called a subset of B.

We were interested in looking to all subsets of a some set.

What are the subsets of {a, b} ?

They are { }, {a}, {b}, {a, b}. Why?

{a, b} is included in {a, b}. This is obvious. All elements of {a,b} are elements of {a, b}.

{a} is included in {a, b}, because all elements of {a} are elements of {a, b}

The same for {b}.

You see that to verify that a set with n elements is a subset of some set, you have to make n verifications. 

So, to see that the empty set is a subset of some set, you have to verify 0 things. So the empty set is a subset of any set.

proposition: { } is included-in any set.

So the subsets of {a, b} are { }, {a}, {b}, {a, b}.

But set have been invented to make a ONE from a MANY, and it is natural to consider THE set of all subsets of a set. It is called the powerset of that set.

So the powerset of {a, b} is THE set {{ }, {a}, {b}, {a, b}}. OK?

Train yourself on the following exercises:

What is the powerset of { }

What is the powerset of {a}

What is the powerset of {a, b, c}

Any question?

This was a bit of revision, to let Kim catch up.

The sequel will appear asap. Be sure everything is OK, and please, ask question if it is not. 

You can also ask any question on the first sixth steps of UDA ('course).

Bruno

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

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

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[Bruno Marchal]

On 14 Jul 2009, at 10:40, Bruno Marchal wrote:

<snip>

> So the subsets of {a, b} are { }, {a}, {b}, {a, b}.
>
> But set have been invented to make a ONE from a MANY, and it is  
> natural to consider THE set of all subsets of a set. It is called  
> the powerset of that set.
>
> So the powerset of {a, b} is THE set {{ }, {a}, {b}, {a, b}}. OK?
>
> Train yourself on the following exercises:
>
> What is the powerset of { }
> What is the powerset of {a}
> What is the powerset of {a, b, c}

I give the answer, and I continue slowly.

1) What is the powerset of {a, b, c}?

By definition, the powerset of {a, b, c} is the set of all subsets of  
{a, b, c}.
I go slowly.

Is the set {d, e, f} a subset of {a, b, d}? No. None of the elements  
of {d, e, f} are elements of {a, b, c}. The question was ridiculous.

Is the set {a, b, d} a subset of {a, b, c}? No. One element of {a, b,  
d}, indeed, d, does not belong to {a, b, c}, so {a, b, d} cannot be a  
subset of {a, b, c}. The question was ridiculous again, but less  
obviously so.

Is the set {a, b, c} a subset of {a, b, c}. Yes. All elements of {a,  
b, c} are elements of {a, b, c}. {a, b, c} is included in {a, b, c}.

Can we conclude from this that the powerset of {a, b, c} is {{a, b,  
c}}. No. We can conclude only that {{a, b, c}} is included in the  
powerset. It is very plausible that there are other subsets!

Indeed,

Is {a, b} included in {a, b, c}? Yes, all elements of {a, b} are  
elements of {a, b, c}. This take two verifications: we have to verify  
that a belongs to {a, b, c}. And that b belongs to {a, b, c}.

Can we conclude from this that the powerset of {a, b, c} is {{a, b, c}  
{a, b}}. No, we could still miss other subsets.

I accelerate a little bit.
Is {a, c} a subset of {a, b, c}? Yes, by again two easy verification.

Is there another doubleton (set with two elements) having elements in  
{a, b, c}? Yes. {c, b}. It is easy to miss them, so you have to be  
careful. All two elements of {c, b} are elements of {a, b, c}, as can  
be verified by two easy verification.

  Is {b, c} a subset of {a, b, c}. Yes, but we have already consider  
it. Indeed the set {b, c} is the same set as {c, b}.

Is there another doubleton? No. Why? I search and don't find it.

is there yet some subset to find?

Yes, the set with one element, notably. They are called singleton.

Here it is easy to guess that there will be as many singletons  
included in (a, b, c} that there is elements in {a, b, c}. So the  
singletons are {a}, {b}, and {c}. This can be verified by one  
verification for each.

Are there still subset? Yes. We have seen that the empty set { } is  
included in any set. This can be (re)verify by 0 verifications, given  
that there is 0 element in { }..

Conclusion:
There are 8 subsets in {a, b, c}, which are { }, {a}, {b}, {c}.{a, b},  
{a, c}, {c, b} and {a, b, c}. And thus,

The powerset of {a, b, c}  is the set { { }, {a}, {b}, {c}.{a, b}, {a,  
c}, {c, b} {a, b, c}}.


2) What is the powerset of {a}?

Answer {{ } {a}}. It has two elements.

3) What is the powerset of { }
We could think at first sight that there are no subsets, given that  
{ } is empty. But we have seen that { } is included in any set. So { }  
is included in { }. Again you can verify this by zero verification!  
But then the powerset of { }, which is the set of sets included in { }  
is not empty:  It has one element, the empty set. It is {{ }}. Think  
that {{ }} is a box containing that empty box.


Attempt toward a more general conclusion.

The powerset of a set with 0 element has been shown having 1 elements,  
and no more.
The powerset of a set with 1 element has been shown having 2 elements,  
and no more.
The powerset of a set with 2 elements has been shown having 4  
elements, and no more. (preceding post)
The powerset of a set with 3 elements has been shown having 8  
elements, and no more.
The powerset of a set with 4 elements has been shown having 16  
elements, and no more. (older post)

In math we like to abstract things. Let us look at the preceding line  
with all the words dropped! This gives

--- 0 ---- 1 ---
--- 1 ---- 2 ---
--- 2 ---- 4 ---
--- 3 ---- 8 ---
--- 4 ---- 16 ---

On the left, we see, vertically disposed, the natural numbers,  
appearing with their usual order.
On the right, we see, vertically disposed, some natural numbers, which  
seems to depend in some way from what numbers appears on the left. It  
looks like there is a functional relation, that is a function.

The notion of function is the most important and pervading notion in  
math, physics, science in general, and we will have to come back on  
that very notion soon enough.

The idea that there is a function lurking there, is the idea that we  
can guess a general law, capable of providing the answer to the  
general line:

   "The powerset of a set with n element has been shown having ?  
elements, and no more."

Can we determine ? from n. Surely it depends on n.

In this case, a simple guess can be made, by meditating on the  
sequence of numbers which appear on the right. Those are, when written  
horzontally:

1, 2, 4, 8, 16, ...

I guess you see what happens. Each number in that sequence is the  
double of the preceding one. So the next one can be obtained, by just  
continuing the multiplication by two.

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384,  
32768, 65536, 131072, ... that is,

1, 2x2, 2x2x2, 2x2x2x2, 2x2x2x2x2, 2x2x2x2x2x2, ...

We will write a possibly lengthy expression like 2x2x2x2x2x ... x2, as  
2^n, where n is the number of occurence of "2" in the expression.

So you can guess that in the "general line":

   "The powerset of a set with n elements has been shown having ?  
elements, and no more."

? does indeed depend on n, and is actually equal to 2^n.

Here is the general law, that we have guessed by experience (counting  
in the simple case) and generalized by intution:

The powerset of a set with n elements has been shown having 2^n  
elements, and no more.

When we found a law in such a way, we could ask if we couldn't prove  
it from facts we are already believing (or guessing).

Here, the theory is the intuitive basic  knowledge of logic, numbers  
and sets. And the question is really: can you prove, or justify, or  
explain WHY the powerset of a set of n elements has 2^n elements?

What happens which could explain why, when we add an element to a set,  
its powerset becomes two time bigger. is there a reason for that  
special happening. Can I convince myself that it has to be like that?

I let you think.

Hints. We have already see a "doubling scenario". Indeed, I often  
mention at the step 3 of the UDA, the iterated self-duplication. Some  
is cut and paste in Brussel, and copy at place W and M. Then both  
individuals come back, by train, in Brussels and both do the  
duplication experience again, and again, and again. The number of  
individuals grows like 2, 4, 8, ... that is 2^n, with n the number of  
iteration of the experience. After n iteration, each has written in  
its "first person diary" the sequence of places they have visit, which  
look like a string of W and M of length n.

No question? If everything is clear, I continue asap.

Oh! I have a question myself. If the law above is correct, it looks  
like 2^0 = 1. It is the laws applied to the first line---the case of  
the powerset of the empty set. Is it true that 2x2x2x...x2 is equal to  
1 in case the number of occurence of 2 is 0? How come?

Bruno


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




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[m.a.-2]

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

Bruno,

            I have no idea how to even begin to answer these questions. Have you given us the definitions we need to do so?     

                                                                                                                                                                                                 marty a.

 

 

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

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

>

> I let those interested to meditate on two questions (N is {0, 1, 2, 3, 
> 4, ...}):
>
> 1) What is common between the set of all subsets of a set with n 
> elements, and the set of all finite sequences of "0" and "1" of length 
> n.
> 2) What is common between the set of all subsets of N, and the set of 
> all infinite sequences of "0" and "1".
>
>
> http://iridia.ulb.ac.be/~marchal/

>
>
>
>
>
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[Bruno Marchal]

On 16 Jul 2009, at 15:17, John Mikes wrote:

 

I would not restrict 'a set' to what WE can conceive, or define now. (Not even within the 'math'-related domain).

Nor do I. 

I never say so. On the contrary we will see how different are sets when seen by machines and gods, but to explain this it is necessary we agree on elementary properties and definitions on sets, so that we can proceed.

Here by "we" you are free to take the  "we" by any entities (machines, humans, gods, or whatever).

the relations of the set-partners comes into play. Not only those which WE choose for 'interesting' to such set, but ALL OF THEM influencing the character of that "ONE".

Just musing.

It is OK. The idea consists in simplifying the things as much as possible, and then to realize that despite such simplification we are quickly driven to the unprovable, unnameable, un-reductible, far sooner than we could have imagine. 

 

I may suggest (or: assume?) that instead of "despite" it would make more sense to write: "AS A CONSEQUENCE"

- think about it.

It could make sense. This would lead to finitism and or mechanism, which I am trying to share with you. But again, this is an anticipation, and can hardly been made precise if we don't train ourselves to think about those simple things before.

Bruno

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

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[Bruno Marchal]

On 16 Jul 2009, at 15:29, m.a. wrote:

Bruno,

            I have no idea how to even begin to answer these questions. Have you given us the definitions we need to do so?     

                                                                                                                                                                                                 marty a.

 

 

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

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

> 
> I let those interested to meditate on two questions (N is {0, 1, 2, 3,  
> 4, ...}):
> 
> 1) What is common between the set of all subsets of a set with n  
> elements, and the set of all finite sequences of "0" and "1" of length  
> n.
> 2) What is common between the set of all subsets of N, and the set of  
> all infinite sequences of "0" and "1".
> 

Read my last general post. I have to go now, you can wait for the answer. 

It is good to search without finding the answer, so as to better appreciate the answer.

When I will give the answer, ask yourself question like "how is it that I didn't think about that, how is it that I have not seen this or that, how is it that I was forgetting this or that ...

My last post should help a little bit,

Think about this is as far as you have some fun by asking yourself, and then stop.

Bruno

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

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[Bruno Marchal]

On 16 Jul 2009, at 16:06, Bruno Marchal wrote:

On 16 Jul 2009, at 15:29, m.a. wrote:

Bruno,

            I have no idea how to even begin to answer these questions. Have you given us the definitions we need to do so?     

                                                                                                                                                                                                 marty a.

 

 

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

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

> 
> I let those interested to meditate on two questions (N is {0, 1, 2, 3,  
> 4, ...}):
> 
> 1) What is common between the set of all subsets of a set with n  
> elements, and the set of all finite sequences of "0" and "1" of length  
> n.
> 2) What is common between the set of all subsets of N, and the set of  
> all infinite sequences of "0" and "1".
> 

Read my last general post. I have to go now, you can wait for the answer. 

It is good to search without finding the answer, so as to better appreciate the answer.

When I will give the answer, ask yourself question like "how is it that I didn't think about that, how is it that I have not seen this or that, how is it that I was forgetting this or that ...

My last post should help a little bit,

Think about this is as far as you have some fun by asking yourself, and then stop.

Ok, so now you have perhaps solve the riddle! The answer is No. I did not really provide all the definitions you could need!

I feel a bit sorry. Mathematicians are kind of summarizing the rest of the book in the exercises.

What are the sequences of "0" and "1" of length n? 

That is the question.

I give some examples of sequences of "0" and "1" of length 7:

  1111111, 0000000, 1010101, and  0100111

Here are some sequence of length three:  000, 101, 111.

Here are ALL examples of the sequences with length two:

00

01

10

11

Convince yourself that there are no other binary sequence of length 2. 

Here is an example of sequence of "0" and "1" of length 24:       

                        0111101010001000000000001

Here are the only two examples of binary sequences of length 1

0

1

And here is the only empty sequence, the one which length is 0.

You can't see it, of course. Even under a microscope! But there is one!

The problem "1)" was "What is common between the set of all subsets of a set with n elements, and the set of all sequence of "0" and 1"? 

You can search by yourself if the question is more clear, or look at the explanation below. You = anyone interested, except Kim and Marty, which have the obligation to train themselves, because they are victim of summer school (that happens in hot summers).

Take your time. The problem was certainly not clear if you did not know what is a so-called binary sequences, or sequences of "0" and "1". 

.

.

.

.

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

The answer is that they have the same number of elements. 

And that number is 2^n. 

Why?

Let us see, and, to fix the idea, consider what happens with the set {a, b, c}, which powerset has already just be computed in a preceding post:

The powerset of {a, b, c} is {{ }, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}. 

What is common between being a subset of {a, b, c} and a binary sequence of length 3?

You can see the link immediately by realizing that a subset of a set with n elements is entirely determined by the answers to n yes/no questions. With {a, b, c}, there are only three possible question? We can put them in the following order:

1)Does a belong to the subset?

2) Does b belong to subset?

3) Does  c belong to the subset? 

The poor empty subset { } is the one which get the answers: no, no, no.  

The less poor singleton {a} is the one which get the answers: yes, no, no.

Let us abstract, and let us write 0 in place of "no", and 1 in place "yes", 

------------------------------ { }  ------------------    000

------------------------------ {a}  ------------------   100

------------------------------ {b}  ------------------   010

------------------------------ {c}  ------------------   001

------------------------------ {a, b}  ---------------- 110

------------------------------ {a, c}  ------------------101

------------------------------ {b, c}  ------------------011

and the winner is:

------------------------------ {a, b,c}  ------------------111

So, there are as many subsets included in a set with n elements than there are binary sequences of length n.

And how many?

Suppose I have to invent a binary sequence of length 1.

Well, it is a binary sequence, so I have not much choice in the beginning. It is either 0 or 1. I can hesitate a long time, but I have only two possibilities. 0, and 1.

Suppose I have to invent a binary sequence of length 2.

I have two possibilities for the first digit, and for each such choice it remains two choice for the next, and last.

If I choose to begin with 0, I can still end with 1 or with 0:  01,  00. The possibilities of having 0 and 1 at a place does not depend of what happened before: so they multiply.

How many sequences of length 3? =

= (2 possibilities for the first digit) times (2 possibilities for the second digit) times (2 possibilities for the third digit)

= 2 times 2 times 2 = 8.

OK?

How many subsets of a set with three elements {a, b, c}.

If I have to "invent" a subset S

= (2 possibilities for the question "is a in the subset S") times (2 possibilities for the question "is b in the subset S") times (2 possibilities for the question "is c in the subset S")

= 2 times 2 times 2 = 8.

Question?

What about problem 2).?

I recall that N is the set of natural numbers {0, 1, 2, 3, ……… }. I let you think about the relation between subset of N, and infinite binary sequences.

Hint: the poor empty subset { } is the one getting the answer no, no, no, no, no, no, no, no, no, no, no, no, no, ....

Hot summer!

Bruno

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

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[m.a.-2]

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

Bruno,

             I don't know about Kim, but I'm ready to push on. I'm waiting for the answer to problem 2) see below. And could you please retstate that problem as I'm not sure which one it is? Thanks,   marty a.

 

 

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

From: marchal@...

To: everything-list@...

Sent: Thursday, July 16, 2009 3:56 PM

Subject: Re: The seven step series

On 16 Jul 2009, at 16:06, Bruno Marchal wrote:

What about problem 2).?

I recall that N is the set of natural numbers {0, 1, 2, 3, ……… }. I let you think about the relation between subset of N, and infinite binary sequences.

Hint: the poor empty subset { } is the one getting the answer no, no, no, no, no, no, no, no, no, no, no, no, no, ....

Hot summer!

Bruno

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

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[Brian Tenneson]

I found a paper that might be of interest to those interested in Tegmark's work.

http://arxiv.org/abs/0904.0867

Abstract
I discuss some problems related to extreme mathematical realism, focusing on a recently proposed "shut-up-and-calculate" approach to physics (arXiv:0704.0646, arXiv:0709.4024). I offer arguments for a moderate alternative, the essence of which lies in the acceptance that mathematics is (at least in part) a human construction, and discuss concrete consequences of this--at first sight purely philosophical--difference in point of view.

-Brian

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[Bruno Marchal]

On 20 Jul 2009, at 15:34, m.a. wrote:

Bruno,

             I don't know about Kim, but I'm ready to push on. I'm waiting for the answer to problem 2) see below. And could you please retstate that problem as I'm not sure which one it is? Thanks,   marty a.

 

Let us see:

> 
> 1) What is common between the set of all subsets of a set with n  
> elements, and the set of all finite sequences of "0" and "1" of length  
> n.
> 2) What is common between the set of all subsets of N, and the set of  
> all infinite sequences of "0" and "1".
> 

Let me restate by introducing a "definition" (made precise later). The cardinal of a set S is the number of elements in the set S.

The cardinal of { } = 0. All singletons have cardinal one. All pairs, or doubletons, have cardinal two.  

Problem 1 has been solved. They have the same cardinal, or if you prefer, they have the same number of elements. The set of all subsets of a set with n elements has the same number of elements than the set of all strings of length n.

Let us write  B_n for the sets of binary strings of length n. So,

B_0 = { }

B_1 = {0, 1}

B_2 = {00, 01, 10, 11}

B_3 = {000, 010, 100, 110, 001, 011, 101, 111}

We have seen, without counting, that the cardinal of the powerset of a set with cardinal n is the same as the cardinal of B_n.

And then we have seen that such cardinal was given by 2^n. 

You can see this directly by seeing that adding an element in a set, double the number of subset, due to the dichotomic choice in creating a subset "placing or not placing" the new element in the subset.

Likewise with the strings. If you have already all strings of length n, you get all the strings of length n+1, by doubling them and adding zero or one correspondingly.

This is also illustrated by the iterated self-duplication W, M. Mister X is cut and paste in two rooms containing each a box, in which there is a paper with zero on it, in room W, and 1 on it in room M. After the experience, the 'Mister X' coming out from room W wrote 0 in his diary, and the 'Mister X' coming out from room M wrote 1 in his diary. And then they redo each, the experiment. The Mister-X with-0-in-his-diary redoes it, and gives a Mister-X with-0-in-his-diary coming out from room W, and adding 0 in its diary and a  Mister-X with-0-in-his-diary coming out from room M, adding 1 in its diary: they have the stories 

00

01,

and then the Mister-X-coming-from room W, and with 1 written in the diary, similarly redoes the experiment, and this gives two more Misters X, having written in their diaries

10

11.

Obviously the iteration of the self-duplication, gives as result 2x2x2x2x...x2 number of Mister X. If those four Mister X duplicate again, there will be 8 of them, with each of those guys having an element of B_3 written in his diary. OK.

And we have seen that a powerset of a set with n elements can but put in a nice correspondence with B_n.

For example: The powerset of {a, b}, that is {{ }, {a}, {b}, {a, b}}, has the following nice correspondence 

00 .............. { }

01 .............. {a}

10 ...............{b}

11 ...............{a, b}

Each 0 and 1 corresponding to the answer to the yes/no questions 'is a in the subset?', is 'b in the subset?'.

Such a nice correspondence between two sets is called a BIJECTION, and will be defined later. What we have seen, thus, is that there is a bijection between the powerset of set with n elements, and the set of binary strings B_n.

And the second question?

What is common between the subsets of N, and the set of infinite binary sequences. An infinite binary sequence is a infinite sequences of "0" and "1".

For example: 00000000000..., with only zero is such a sequence. It could be the first person story of 'the mister X who comes always from room W. Or, if the zero and one represents the result of the fair coin throw experiment, it could be the result of the infinitely unlucky guy: he always get the head.

Another one quite similar is 1111111111111111111111111111..., the infinitely lucky guy.

A more 'typical' would be 1100100100001111110110101010001000... (except that this very one *is* typical, it is PI written in binary; PI = 11. 00100...). 

The link with the subsets of N? It is really the same as above, except that we extend the idea on the infinite.

A subset of N, that is, a set included in N,  is entirely determined by the answer to the following questions:

Is 0 in the subset?

Is 1 in the subset?

Is 2 in the subset?

Is 3 in the subset?

etc.

You will tell me that nobody can answer an infinity of questions. I will answer that in many situation we can.

Let us take a simple subset of N, the set {3, 4, 7}. It seems to me we can answer to the infinite set of corresponding question:

Is 0 in the subset?    NO

Is 1 in the subset?    NO

Is 2 in the subset?  NO

Is 3 in the subset?   YES

Is 4 in the subset?    YES

Is 5 in the subset?    NO

Is 6 in the subset?   NO

Is 7 in the subset?   YES

Is 8 in the subset?   NO

Is 9 in the subset?    NO

Is 10 in the subset?    NO

Is 11 in the subset?   NO

Is 12 in the subset?    NO

Is 13 in the subset?    NO

Is 14 in the subset?  NO

... (etc. and we answer NO for all remaining questions ...!)

So, in the same spirit as above we associate the infinite binary string 000110010000000000000000... to the subset {3, 4, 7}.

Even simpler: the empty set. First, is the empty set a subset of N?But we have seen that to verify if the empty set is a subset of any set, we have 0 verification to do: the empty set is included in any set. It correspond here to 

Is 0 in the subset?    NO

Is 1 in the subset?    NO

Is 2 in the subset?  NO

Is 3 in the subset?   NO

Is 4 in the subset?   NO

Is 5 in the subset?    NO

Is 6 in the subset?   NO

Is 7 in the subset?   NO

Is 8 in the subset?   NO

Is 9 in the subset?    NO

Is 10 in the subset?    NO

Is 11 in the subset?   NO

Is 12 in the subset?    NO

Is 13 in the subset?    NO

Is 14 in the subset?  NO

... (and we answer NO for all remaining questions ...!)

the empty set, seen as a subset of N, is determined by the sequence:

000000000000000000000000000000000000000000 ...

All singleton of number, like {0}, {1}, {2}, {3}, {4}, {5}, ...

will have the following "characteristic" sequences:

{0} ---- 1000000000000000...

{1} ---- 0100000000000000...

{2} ---- 001000000000000..

etc.

Doubleton will have characteristic sequences being strings having two occurences of 1, and thus an infinity of zero.

OK, for finite set, there is a time where all answers become "NO". 

But what about infinite sets? Can we still answer the infinity of question?

Of course, at least for "simple" infinite sets.

First, do we know infinite subset of N? yes, example the set of odd numbers. {1, 3, 5, ...} is included in {0, 1, 2, 3, ...}.

What is its characteristic sequences, that is the answer to the questions: is 0 in?, is 1in?, is 2 in?,  ...

Is 0 in the subset?    NO

Is 1 in the subset?    YES

Is 2 in the subset?  NO

Is 3 in the subset?   YES

Is 4 in the subset?   NO

Is 5 in the subset?    YES

Is 6 in the subset?   NO

Is 7 in the subset?   YES

Is 8 in the subset?   NO

Is 9 in the subset?    YES

Is 10 in the subset?    NO

Is 11 in the subset?   YES

Is 12 in the subset?    NO

Is 13 in the subset?    YES

Is 14 in the subset?  NO

... (and we answer YES and NO in that way for all remaining questions ...!)

So the set of odd numbers has the characteristic sequence:

{1, 3, 5, ...} ----------- 010101010101010101010101010101010101010101010...

and the even numbers:

{0, 2, 4, 6, ...} ---------- 101010101010101010101010101010101010101010...

And ... the prime numbers

{2, 3, 5, 7, 11, ...} ------- 00110101000101000101000100000101...

OK.

Do you see that for EACH subset of N we have a corresponding infinite binary sequence, and for EACH binary sequence we have a subset.

Exercise: gives the subset (= write some elements of the subset) corresponding o the sequence "PI"

1100100100001111110110101010001000...

The genius of Cantor has consisted in allowing himself the notion of "nice correspondence", bijection, to infinite sets, including infinite sets of infinite objects, like the subset of N and the set B_infinity, that is the set of infinite binary strings.

And here, I hope you have the feeling that indeed such a nice correspondence exists between the powerset of N, and B_infinity.

I let you think, and ask some questions. Bijection will occupy us for awhile. Unfortunately, to make them precise, I have to define what we have already met a lot, but yet not define, and which is the notion of function.

Functions are utterly important because they define the crucial notion of mathematical dependency, on which eventually the mathematical notion of computation will be a very peculiar particular case.

Bruno

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

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[Bruno Marchal]

Exercise: criticize the following papers mentioned below in the light of the discovery of the universal machine and its main consequences from incompleteness to first person indeterminacy. Think of the identity thesis. To be sure Tegmark is less "wrong" than Jannes.

Solution: search in the archive of this list where I have already explained this, or use directly UDA, or wait for what will (perhaps) follow.

I should send some of my papers on arXiv, but up to now, only logicians understand the whole "trick", so I have to better appreciated what physicians don't understand in logic, before making a version free of references to mathematical logical baggage. Logicians are not interested in mind, nor really matter, and physicians are still naïve on the link consciousness/reality, I would say.

To be sure Tegmark is closer than most physicists except perhaps Wheeler.

Also, Tegmarks' argument for mathematicalism is invalid (even with strong non-comp axioms). But I prefer to help you to understand this by yourself through the understanding of what a universal machine is, than trying a direct argument.

According of the part of UDA (or perhaps AUDA) you understand, you can already see the weakness of such direct mathematical approach. Note that comp makes physics much more fundamental, and separate it much clearly from possible geograpies. Above all comp does not eliminate the person, which Tegmark is still doing: the frog view is not yet a first person view, in the comp sense.

Interesting stuff, still. Thanks for the references.

Bruno

On 20 Jul 2009, at 19:44, Brian Tenneson wrote:

I found a paper that might be of interest to those interested in Tegmark's work.

http://arxiv.org/abs/0904.0867

Abstract
I discuss some problems related to extreme mathematical realism, focusing on a recently proposed "shut-up-and-calculate" approach to physics (arXiv:0704.0646, arXiv:0709.4024). I offer arguments for a moderate alternative, the essence of which lies in the acceptance that mathematics is (at least in part) a human construction, and discuss concrete consequences of this--at first sight purely philosophical--difference in point of view.

-Brian

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

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