Re: Non unique Universe
On 03 Jul 2009, at 19:07, Brent Meeker wrote:
Right. I have no problem with arithmetical possibilities,
provability,
etc. But without some defined scope the use of possible makes me
uneasy. In modal logic possible and necessary are just operators
that must be interpreted in some domain; just like some and for
all.
OK.
I've never seen { and } denoted accolades but I like it; they
are
more commonly called braces. I don't know a specific term for [
and ], I generally refer to them as square brackets.
OK. I think accolade is kind of formal hug, but in french it is a
punctuation mark see
http://fr.wikipedia.org/wiki/Accolade
So Accolade in fench is brace in english, I will try to remember, but
if you like accolades ... Natural languages are living things and I am
all for exchange of words between languages ...
Thanks for the info,
Bruno
http://iridia.ulb.ac.be/~marchal/
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Re: The seven step series
On 04 Jul 2009, at 04:31, m.a. wrote:
New comments in italics.
For example {1,2} INTERSECTION {2, 7} is equal to some set,
actually the set {2}. OK?..No!
Why
not the sets {1,2,7} if INTERSECTION means BOTH?
Ah, but the word both alone is ambiguous. You could say that the
UNION of two sets is the merging of BOTH set, and the intersection
is the given of the elements which are in both set. So the union of
{1, 2} and {2, 7} is {1, 2, 7}, which indeed merges BOTH sets. But
for computing the intersection, you must ask yourself, does this
*element* belongs to BOTH set? So, for the intersection of {1, 2}
and {2, 7}, you have to ask yourself the following question: does 1
belong to both set? well, the answer is NO. the 1 belongs to the
first set but not to the second, and so 1 does not belong to the
intersection. Does 2 belongs to both sets? The answer is yes. 2
belongs to {1, 2} and 2 belongs to {2, 7}. Does 7 belongs to both
sets, the answer is no, 7 belongs to the second set, but does not
belong to the first set, so 7 is not in the intersection.
Tell me if you are OK with this.
Not OK. You
previously defined UNION as one OR the other. Now you seem to be
giving me the same definition for INTERSECTION.
Let us take the set a = {1, 2} and the set b = {2, 3}.
Let us recall the definition (I abandon the capital letters because
they are ugly, and I feel talking louder!)
1) intersection
(a intersection b) = {x such-that (x belongs-to a) AND (x belongs-to
b) }.
So, some x, to belong to the intersection, has to belong
simultaneously to the two sets involved. Only when x is equal to 2, is
that condition verified.
2 belongs to (a intersection b) because 2 belongs to a, AND, 2 belongs
to b.
That condition is not verify for x = 1, nor for x = 3. 3 belongs to b,
but not to a. So 3 is not in the intersection. Nor 1, because it does
not belong to a.
So (a intersection b) = ( {1, 2} intersection {2, 3} ) = {2}.
2) Union
(a union b) = {x such-that (x belongs-to a) OR (x belongs-to b).
Does 1 belong to the union of a and b? That is do we have that 1
belongs-to (a union b)? With same a and b as above.
Let us see.
Does 1 belongs to a union b? Does 1 verify the condition written in
the definition? Do we have that (1 belongs-to a) OR (1 belongs to b))?
A proposition shaped like P OR Q is true in the case one or both of P
and Q is true. It is true that 1 belongs to {1, 2} OR to {2, 3}. A bit
like any number is odd or is not odd is always true. So 1 is in the
union.
2 is in the union, because it is true that 2 belongs to a or 2 belongs
to b. Indeed 2 belongs to both of them. And 3 is in the union too,
because iit belongs to one of them again, actually {2, 3}.
So (union b) = {1, 2, 3}.
OK?
Don't hesitate to tell me if it is not OK.
2°) And I give you a slightly longer exercise. Can you give me all
the subsets of the set {1, 2} ?. That is, can you give me all the
sets which are included in the set {1, 2} ? In case of doubt,
reread the definitions, reread the examples, and never panic! I
give you a hint: the set {1, 2} has four subsets. Can you find them?
{1
} {2} {1,2} {2,1} why not {3} ?
Not too bad. 3/4 correct:
{1} is included in {1, 2}. Indeed.
{2} is included in {1, 2}. Indeed.
{1, 2} is included in {1, 2}. Indeed.
{2, 1} is included in {1, 2}. Indeed, that is true, but you have to
remember what you have already agree on: the set {1, 2} is equal to
the set {2, 1}, so this is not a new solution. It is the preceding
one in disguised!
Why not {3}? {3} is not included in {1, 2} just because 3 does not
belong to {1, 2}. Reread the definition of inclusion. A is included
in B if all the elements of A belongs to B. OK?
So you have found three subsets, among the four. Reading today's
explanations I think you could find the missing subset. I let you
search a little bit.
So just one exercise: what is the missing subset?
Is the missing subset { }?
Correct.
So the subsets of {1, 2} are { }, {1}, {2}, {1, 2}.
Could you find all subsets of {1, 2, 3}?
And now I give you an exercise which is so much easy that you could
panic, and so I will provide the solution. I have seen often that too
much easy question can make a student panic, and then the prey of out-
of place mockery, and useless loss of confidence.
The easy exercise: could you give me the set of subsets of {1, 2} ?
Solution: You already told me that the subsets of {1, 2} are { }, {1},
{2}, {1, 2}. So, the set of subsets of {1, 2} is
{ { }, {1}, {2}, {1, 2} }
OK? It is just the solutions you give me, enclosed by braces
(accolades) {, }. Look at the expression with a
Re: The seven step series
Bruno,
Can you provide definitions of belongs-to and included-in that
distinguish them from union and intersection?
Here we met a set of sets.
The set of subsets of a set, can only be, of course, a set of sets. The set
{2, 21, 14} is a set of numbers. The set { { }, {4, 78, 56} } is a set of sets.
It has two elements: the empty set {}, and the set of numbers {4, 78, 56}. Do
not confuse a number, like 24, and a set, like {24}, which is a set having a
number has elements. In particular it is the case that {4, 78, 56} belongs to
{ { }, {4, 78, 56} }. Take it easy, and meditate on the following exercise:
Which of the following are true
{3, 5} included-in {3, 5} True
{3, 5} belongs-to {3, 5}
{3, 5} included-in { {3, 5} }
{3, 5} belongs-to { {3, 5} }
Take your time,
Bruno
http://iridia.ulb.ac.be/~marchal/
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set incompleteness
Dear Bruno, I mentioned that I have something more on the 'set' as you (and all since G. Cantor) included it in the formulations. I had a similar notion about my aris-total, the definition of Aristotle that the 'total' is always more than the 'sum' of its components. Of course, at the time when A. thought about it, 'components' were only 'physical objects' included in an ensemble as individual and unrelated noumena. If you advance in our epistemic cognitive inventory to a bit better level (say: to where we are now?) you will add (consider) relations (unlimited) to the names of 'things' and the increased notion will exactly match the 'total' (what A was missing from the 'sum'). It will also introduce some uncertainty into the concept (values?) of a set. I see a similar situation with your ways writing of 'sets' (1,2,3...) - or: ( 1, 2, 3... ) neglecting the additional relations maybe expressed in the (neglected) commas, spaces, even the parentheses. All may mean something and that meaning gives completeness to the entire set beyond the 'factual' elements 1 2 3 . I don't know 'what', but for sure something well pertinent. In infinite sets such uncertainty may amount to infinite uncertainty. John M --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to everything-list@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
Re: set incompleteness
John,
On 04 Jul 2009, at 18:24, John Mikes wrote:
Dear Bruno, I mentioned that I have something more on the 'set' as
you (and all since G. Cantor) included it in the formulations. I had
a similar notion about my aris-total, the definition of Aristotle
that the 'total' is always more than the 'sum' of its components. Of
course, at the time when A. thought about it, 'components' were only
'physical objects' included in an ensemble as individual and
unrelated noumena.
We will see how we can do something similar with set. few
mathematicians are really interested in sets, but in sets together
with a structure (usually determined by operations and relations on
the set.
If you advance in our epistemic cognitive inventory to a bit better
level (say: to where we are now?) you will add (consider) relations
(unlimited) to the names of 'things' and the increased notion will
exactly match the 'total' (what A was missing from the 'sum'). It
will also introduce some uncertainty into the concept (values?) of a
set.
I am not sure that I understand.
I see a similar situation with your ways writing of
'sets' (1,2,3...) - or: ( 1, 2, 3... )
I guess you mean {1, 2, 3 ... }. { and } are standard, and ( and
) will be reserved for other things, like delimiter of expression,
like in (3+4), or the notion of couples (soon to be introduced).
neglecting the additional relations maybe expressed in the
(neglected) commas, spaces, even the parentheses. All may mean
something and that meaning gives completeness to the entire set
beyond the 'factual' elements 1 2 3 . I don't know 'what', but for
sure something well pertinent. In infinite sets such uncertainty may
amount to infinite uncertainty.
I don't see anything uncertain in most infinite sets. But this will be
scrutinized soon, or a bit later ... Some sets will appear more
complex than other, and *some* set will have uncertainties attached
to them, but to understand this we have to progress a bit more.
Bruno
http://iridia.ulb.ac.be/~marchal/
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Re: The seven step series
On 04 Jul 2009, at 15:17, m.a. wrote:
Bruno,
Can you provide definitions of belongs-to and
included-in that distinguish them from union and intersection?
Belongs-to and included-in are relations. Their value are true or
false.
1) (x belongs-to A) means that the object x belongs to the set A.
Examples: 3 belongs-to {3, 4}, because 3 is an element of the set {3,
4}, or, put in another way (which can be useful for later):
(3 belongs-to {3, 4}) = true
2) Similarly (x included-in y) is a relation bearing on sets, and (x
included-in y) = true, means that x is included-in y, and this means
that all elements of x are elements of y. You don't need to know more,
but if you want you can define (x included-in y) by
For any z ((z belongs-to x) - (z belongs-to y)). But I intended to
introduce - later, so don't worry.
Example {3, 4} is included in {3,5,4} because all elements of {3, 4}
are in {3, 5, 4}. We can write ({3,4} included-in {3,5,4}) = true.
Another example is ({3, 4} included-in {3,6,7,9,567}) is false,
because not all elements of {3,4} are in {3,6,7,9,567} (indeed 4
belongs to {3,4} and not to {3,6,7,9,567}.
Union and intersection are not relation, but operation.
({3,4} union {4,5}) is not equal to true or to false, it is equal to a
set: actually the set you get by the union of {3,4} with {4,5}, and
this is {3,4,5}.
Likewise, the intersection of {3,4} with {4,5}, that is ({3,4} union
{4,5}) is not true or false, but is equal to {4}.
So:
({3,4} included-in {3,4,5}) = true
({3,4} union {3,4,5}) = {3,4,5}
You see the difference. It is the difference between the brother of
Paul, which denotes a human. and Paul is greater than Julia, which
is true or false.
Or, on the numbers, less-than () is a relation, and addition and
multiplication are operation:
(3 7) = true
(3+4) = 7
(7 3) = false
(7*3) = 21
({ } included-in {3,4}) = true
({1,2} intersection {2,7}) = {2}.
Does this help?
Here we met a set of sets.
The set of subsets of a set, can only be, of course, a set of sets.
The set {2, 21, 14} is a set of numbers. The set { { }, {4, 78,
56} } is a set of sets. It has two elements: the empty set {}, and
the set of numbers {4, 78, 56}. Do not confuse a number, like 24,
and a set, like {24}, which is a set having a number has elements.
In particular it is the case that {4, 78, 56} belongs to { { }, {4,
78, 56} }. Take it easy, and meditate on the following exercise:
Which of the following are true
{3, 5} included-in {3, 5} True
This is correct.
{3, 5} belongs-to {3, 5}
{3, 5} included-in { {3, 5} }
{3, 5} belongs-to { {3, 5} }
Take your time,
Bruno
http://iridia.ulb.ac.be/~marchal/
http://iridia.ulb.ac.be/~marchal/
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set
Dear Bruno, thanks for the prompt reply, I wait for your further
explanations.
You inserted a remark after quoting from my post:
*
* If you advance in our epistemic cognitive inventory to a bit better
level (say: to where we are now?) you will add (consider) relations
(unlimited) to the names of 'things' and the increased notion will
exactly match the 'total' (what A was missing from the 'sum'). It
will also introduce some uncertainty into the concept (values?) of a
set.*
*I am not sure that I understand.*
***
Let me try to elaborate on that: What I had in mind was my 'interrelated
totality' view.
As you find it natural that 3 (!!!) and 4 () make 34 - if written
without a space in between - representing a quite different meaning - (not 7
as would be plainly decipherable: 3+4),
so all elements of a set carry relations to uncountable items in the
unlimited totality (even if you try to restrict the applicability into the
identified* { }* set. Nothing is excluded from the a/effects (relations)
of the rest of the world. No singularity or nivana IN OUR WORLD
Your 2+2=4 includes a library of conditions, axioms, relations, clarifiers,
just as e.g. the equation 4-2=2 includes the notion NOT in ancient Rome
(where it would have been '3')
So I referred to the tacitly included 'relations' (I use this word for all
kinds of knowables in connection with potential effects of other items)
implied in your technical stenography.
Since the relationally interesting items are unlimited, there is no way WE
(in our present, limited mind) could exclude uncertainty FOR 'ANY' THING.
Sets included. Occamisation of a set does not make it rigorous, just
neglects additional uncertainty.
Have a good weekend
John
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