A question on a<b

  • Thread starter Organic
  • Start date
In summary: OrganicAs I have said before, this is not "normal" mathematics. It is not "abnormal" mathematics, either. It is not mathematics. It is a word game. In summary, the conversation discusses the concept of a 1-1 correspondence (map) between any two real numbers within a given set [a,b]. The conversation also explores the idea of "non-normal" elements that may exist between two "normal" numbers and how they could potentially be useful in mathematics. However, it is noted that this is not traditional mathematics and is more of a word game.
  • #1
Organic
1,224
0
Let us say that a and b are real numbers, and we have
[a,b] = {x: a <= x <= b}.

Can x be a 1 to 1 correspondence (map) between any two real numbers included in [a,b] ?


Organic
 
Last edited:
Physics news on Phys.org
  • #2
You probably need to elaborate a little.

In the expression you show, x is a dummy variable. It stands for anything.

If you want it to be a 1-1 correspondence, then you also have to define a way to compare such a thing with a real number, since you are supposed to look for those x objects that satisfy a <= x <= b (and you already stated that a and b are real numbers).

Also, if you do define a way to decide if a real number is "<=" than a 1-1 correspondence, you should change the notation you use for the set of 1-1corresps, since [a,b] already stands for a set of real numbers.
 
  • #3
Hallo ahrkron,

Since x can be a 1 to 1 correspondence between any two real numbers, is the set of all x maps is complete (have the power of the continuum)?


Organic
 
  • #4
Hi Organic,

You misunderstood my reply.

Your first post is flawed. It does not make sense as stated. You need to define properly the objects and relations you are using. Otherwise, no further comclusions can be drawn from it.
 
  • #5
Hi ahrkron,

May be you can help me to write it in a formal way.


The question is:


Theorem: 2^aleph0 < c


Proof:

Let A be the set of all negative real numbers included in (-1,0).

Let B be the set of all positive real numbers included in (0,1).

Let M be the set of maps (1 to 1 and onto) between any two single real numbers of
A and B sets.

Therefore |M| = 2^aleph0.

(0,1) = {x: 0 < x < 1 }, where x is a 1-1 correspondence between any two real numbers included in (0,1), and any x element has no more than 1 real number as a common element with some other x element.

Let T be the set of all x (1-1 correspondence) elements included in (0,1).

Therefore |T| = |M| = 2^aleph0.

B is a totally ordered set, therefore we can find x element between any two different real numbers included in (0,1).

Any x element must be > 0 and cannot include in it any real number.

Therefore 2^aleph0 < c (does not have the power of the continuum).

Q.E.D


A structural model of the above:
Code:
            set      set       set  
             A        M         B
             |        |         |
             |        |         |
             v        v         v
             !__________________!<---- set
             !__________________!<----  T members 
             !__________________!
             !__________________!   Any point is some real number 
             !__________________!   (A or B members).
             !__________________!   
             !__________________!   Any line is some 1-1 correspondence    
             !__________________!   (M or T members).
             !__________________!
             !__________________!





Am I right ?



Organic
 
Last edited:
  • #6
Doron, is there a reason for not posting this under the name you used before? The mathematician formerly known as "Doron Shadmi"?

The point of the first response by Organic, which you still haven't addressed is that it makes no sense to assert that x is a NUMBER, as you do in say 0<= x<= 1, and then state that x is a 1-1 correspondence: "(0,1) = {x: 0 < x < 1 }, where x is a 1-1 correspondence between any two real numbers included in (0,1)."

x is either a number or a 1-1 correspondence between numbers. It can't be both.
 
  • #7
Hi,

x is not a NUMBER but an element that exists between any two different real numbers included in (0,1).



Organic
 
  • #8
So, once again, you are not talking about "normal" mathematics and are not defining your terms. In "normal" mathematics, inequality with numbers is only defined FOR numbers. WHAT KIND of "element" exists between any two numbers but is not a number itself? What in the world do you mean by "element"?
 
  • #9
Hi,

Any x is not a "normal" real number but a connector (a 1-1 correspondence element) between any two different "normal" real numbers.

No single "normal" real number has this property (to be a connector between some two other different "normal" real numbers).

Please do not forget that negative, irrational, complex and transfinite numbers where not "normal" elements, when they first discovered.

I think the important thing is to find if those "non-normal" elements
can be useful for the development of Mathematics Language, by opening new areas for research.

I think if we associate these "non-normal" elements with, so called, "normal" numbers, we shall obtain a new point of view on the abstract concept of a NUMBER.




Organic
 
Last edited:
  • #10
Well, they surely won't be useful if you don't DEFINE them.
 
  • #11
Please open the attached pdf file.
 

Attachments

  • cat.pdf
    89 KB · Views: 215
  • #12
This is far, far out.
To TD it goes.
 
  • #13
"Any x is not a "normal" real number but a connector (a 1-1 correspondence element) between any two different "normal" real numbers."

what is your PRECISE definition of "<"? so, when you say 0<x, where x is a connector, what precisely does that mean? certainly, this isn't the same < when we say 0<1. it would seem that your < must be an EXTENSION of the binary relation < as defined for real numbers. if it's not an extension, then i would think the whole thing falls apart, for then the two <'s have utterly different meanings even when confined to the concept of real numbers.

let's pretend i don't have adobe. what's a connector? if it's a bijection, what is the domain and range? and how is this bijection inserted into the continum in such a way that it makes sense to say it is between two real numbers?

this is not automatically bizarre because in nonstandard analysis, there are infinitely many hyperreals between two real numbers. each real number has a halo of hyperreals that are infinitesimally close to it.

i'd like to point out that he never said {x &epsilon; R: 0<x<1}, he just said {x: 0<x<1}; that could be a set of hyperreals or what not... clearly, more precision would be nice.

cheers,
phoenix
 
Last edited:
  • #14
Hi phoenixthoth,

First, thank you for your reply.

p and q are real numbers.

If p < q then
[p, q] = {x : p <= x <= q} or
(p, q] = {x : p < x <= q} or
[p, q) = {x : p <= x < q} or
(p, q) = {x : p < x < q} .

A single-simultaneous-connection is any single real number included in p, q
( = D = Discreteness = a localized element = {.} ).

Double-simultaneous-connection is a connection between any two different real numbers included in p, q , where any connection has exactly 1 D as a common element with some other connection ( = C = Continuum = a non-localized element = {.___.} ).

Therefore, x is . XOR .___.

Any C is not a "normal" real number but a connector (a 1-1 correspondence element) between any two different "normal" real numbers (D elements).

No single "normal" real number (a D element) has this property, to be a connector between some two different "normal" real numbers (D elements).

Between any two different arbitrary close Ds there is at least one C, and only C has the power of the continuum.


For more detailes please look at:


http://www.geocities.com/complementarytheory/CATpage.html


Please look also in this thread:

https://www.physicsforums.com/showthread.php?s=&threadid=6427


I'll be glad to get your remarks and insights.






Yours,

Organic
 
Last edited:
  • #15
This may help. Its his first post in his other thread.
In the attached address you can find A new approach for the definition of a NUMBER...
Today I went to the bank and tried to define a new definition of "exchange rate" but for some reason they weren't buying. Hmm, I don't understand why not...
 
  • #16
russ_watters, no, it doesn't help. your comments remind me of those who didn't like the idea of irrational, transcendental, hyperreal, or complex numbers. gauss, as far as in know, invented a new kind of number and they were used in his PhD thesis. i say it doesn't help because it neither proves or disproves the existence of connector elements in such a way that they are naturally embeddable within R.

organic, a few things:

i'm still in need of a more precise definition of what .___. or "connector" means. i know you say it's a 1-1 correspondance (ie, bijection) element between two real numbers. one way to express functions is with notation like this (as we all know): f(x) = stuff. but another way is to represent them as sets of ordered pairs such that no two first-coordinates match (ie, no two different ordered pairs are present with the same first coordinate). so when you say x is a bijection between real numbers, a and b, let's say, that means that x is representable as the following set of ordered pairs (ie, function): { (a, b) }, the singleton containing the ordered pair (a, b). ok, fair enough. x = { (a, b) }.

so far, x is not any more of exotic of a number than the complex number (a,b) or a + b i. but exoticness is irrelevant as long as there is consitency. some people, like david hilbert, define mathematical existence as consistency. (and if physical existence also equals consistency, then, bingo, mathematical existence = physical existence. ha ha ha.)

here's the big question that will improve your theory for as it stands right now, there is a big gap. and here it is.

suppose y is a real number (to say "normal" real number is redundant) and x is a connector, x = { (a, b) }, where a and b are two real numbers. for x to be a connector, this implies a != b. (note that != is standard notation for unequal.)

this is the big question you must answer:

how can you extend the binary relation < so that a statement like

x < y

makes sense?

what do i mean by "make sense?"

well, at first glance, statements like { (1,2) } < 14 don't make sense.

here's how you might fix this. when a = b, then whatever your extended definition of < is should reduce to the definition of < for real numbers for when a = b, x is identified with the real number a. "identified" means that there is a bijection between the set of sets of the form { (a, a) } and the set of real numbers; this bijection is given by
f( { (a, a) } ) = a.
this is what i mean by saying that your definition of < meant for connectors and reals alike should be an extension of what < means for two real numbers.

doing this will, i think, embed the set of connectors within the set of real numbers, which is what i think you want to do.

then, once you have a definition of <, you can then talk about a connector being between two real numbers.

the first big theorem for you, once you have appropriately precise definitions, is an existence theorem. you need to prove the following, which is NOT clear to me:

let a < b be real numbers. there exists a connector x such that
a < x < b. (where < has been precisely defined to apply to connectors and reals.)

do that, and you've actually got your own pet theory on a new kind of number.

to explore a bit further, find the cardinality of the set of all connectors between a and b.

next, find applications!

the main thing about nonstandard analysis, what makes it useful, is that STANDARD analysis results can be obtained by working exclusively with hyperreals. I'm talking about nonstandard analysis for they also talk about there being things between real numbers. godel, well known for his incompletness thoerem, said that nonstandard analysis will be the analysis of the future. well, he's no analyst! go back to the corner, godel, and study your logic!

cheers,
phoenix
 
Last edited:
  • #17
Hi phoenixthoth,

First, thank you for your positive attitude.

Well, I have ideas, but I am not good in formal definitions.

So, if you can help me in this point, I'll be thankful to you.

In general I look on Math as a form of comnunication that can be developed through team work between persons, therefore I do not understand the aggression of some persons when I show them my ideas in a wrong "formal" way.

I think this forum will be a better place through positive attitude to each other's ideas.

OK..., more to the point.

I'll write my definitions again, and then I'll try to explain my ideas
to you the best I can (also sorry about my poor English).


p and q are real numbers.

If p < q then
[p, q] = {x : p <= x <= q} or
(p, q] = {x : p < x <= q} or
[p, q) = {x : p <= x < q} or
(p, q) = {x : p < x < q} .

A single-simultaneous-connection is any single real number included in p, q
( = D = Discreteness = a localized element = {.} ).

Double-simultaneous-connection is a connection between any two different real numbers included in p, q , where any connection has exactly 1 D as a common element with some other connection ( = C = Continuum = a non-localized element = {.___.} ).

Therefore, x is . XOR .___.

Any C is not a "normal" real number but a connector (a 1-1 correspondence element) between any two different "normal" real numbers (D elements).

No single "normal" real number (a D element) has this property, to be a connector between some two different "normal" real numbers (D elements).

Between any two different arbitrary close Ds there is at least one C, and only C has the power of the continuum.

Now I'll try to explain the above:

By p<q I mean that there exists some interval > 0 between any two different real numbers, which means that p or q never reach each other by definition (both of them are singletons).

You wrote:
here's the big question that will improve your theory for as it stands right now, there is a big gap. and here it is.

suppose y is a real number (to say "normal" real number is redundant) and x is a connector, x = { (a, b) }, where a and b are two real numbers. for x to be a connector, this implies a != b. (note that != is standard notation for unequal.)

this is the big question you must answer:

how can you extend the binary relation < so that a statement like

x < y

makes sense?

what do i mean by "make sense?"

well, at first glance, statements like { (1,2) } < 14 don't make sense.

I'll try to answer (hope I understand you):

a=p , b=q


Let us write the connctor between a and b as aCb.

So, x = aCb.

By x < y we mean that y is bigger than aCb range (I know that the word "range" or "domain" already have their meaning in Math languauge, so maybe instead of "range" I use "path-place").

If I draw it, it will look like this:
Code:
 a      b    y(a [b][i]D[/i][/b] element)
 .______.    .

If y = b[b][i]C[/i][/b]c ,where c > b then
 a      b    c
 .______.____.

If a = b then x is a D element (a "normal" real number).

As you see "<" "=" ">" works for both D and C elements.

D = {.} = a localized element.

C = {.__.} = a non-localized element (Double-simultaneous D places).

By (p, q) = {x : p < x < q} I mean that some C exists between a and b (aCb) where a > p and b < q, therefore x = aCb.

Hope I am understood.

Please tell me what do you think.

Thank you.


Organic
 
Last edited:
  • #18
Hi Organic.
Originally posted by Organic
Please open the attached pdf file.
from your file:
The most symmetrical and simplest content is Emptiness, which is represented by the empty set notation { } = content does not exist.
On top of this simplicity we can define two opposite types of symmetry contents, {__} and {._. .}.
Huh... really? However, you are not talking about "normal" mathematics. If I get you right, you are talking about symmetry, well, but if a set really has a symmetry - it has topology,
if a set has topology - it has signature, and if it has signature - it must have more than one element (may be uncertain nature)!
Now, just tell us - how do you construct the symmetry on empty set without signature in such a way that it makes sense to say it is a real symmetry? These (and many others) contradictions doesn't allow considering your math's invention...
 
  • #19
Hi Anton A. Ermolenko,

Why you are so aggresive, please take Math as some form of communication where persons share each other's ideas.

I'll be glad to get any remarks from you.


More to the point, I connect between simplicity and symmetry concepts.

The more you are simple the more you are symmetric.

The {} content is the most simple, therefore the most symmetric.

Please write to me on any part of my work, where you find contradictions.

You can find my work here:

http://www.geocities.com/complementarytheory/CATpage.html

Thank you.



Organic
 
Last edited:
  • #20
if there are any terms in the following post not clear, consult http://www.mathworld.com .

if x = { (a, b) }, which is the bijection with domain {a} and range {b}, is a connector between the real numbers a and b, and y is some other real number, then are you saying that < means the following:

x < y if b < y?

thanks.

cheers,
phoenix
 
  • #21
Originally posted by Organic
Why you are so aggressive…
Really? Perhaps, you’ve misunderstood...
please take Math as some form of communication where persons share each other's ideas.
Perhaps...
I'll be glad to get any remarks from you.
And I’m glad to see you’re glad...
More to the point, I connect between simplicity and symmetry concepts.

More you simple more you symmetric.

The {} content is the most simple, therefore the most symmetric.
In your case your definition isn’t right. It makes no sense to assert that your symmetry is a real internal symmetry. The space’s symmetry can’t have such definition because this kind of symmetry is internal, not external. Symmetry is not a senseless abstraction, only a simplest variety in the space on non-empty set can be more symmetric depending on simplicity. However, if I get you right, your precisely empty set is not (also can’t be) an element of any definite superset.

Please write to me on any part of my work, where you find contradictions.
Well,
From your file:
Let {___} be Csim (C for Continuum).
Let {._. .} be Dsim (D for Discreteness).
If your set is exactly empty set, then it can’t has continuous or discrete structure because the definition of a continuity or discretization of set by the neighborhood of point, based on relation between elements of this exactly non-empty set, doesn’t allow it.

Thank you.
You’re welcome.
 
  • #22
Are the "connectors" you talking about simply intervals?
 
  • #23
"However, if I get you right, your precisely empty set is not (also can’t be) an element of any definite superset."

sure it can. let 0 denote the empty set. {0} is a superset containing 0. in fact, that is the definition of the number 1.
 
  • #24
no, connectors are bijections from a real number to another real number, if i understand.
 
  • #25
Hi phoenixthoth,
x < y if b < y?
Yes. and if x = aCb and y = cCd then
x < y iff c > b.


a,b,c,d are all "normal" real numbers (D elements, or singletons), where x and y, in this case, are C elements, or connctors.

So, in general I'm talking about singletons and/or connectors.

By the way, from the above example, x <= y iff b=c,
and there is no such a thing x <= y where c>a and c<b,
because in this case we get 3 connectors aCc, cCb and bCd.




Organic
 
Last edited:
  • #26
supppose we have p < q real numbers.

then p < { (a, 0.5(p+q) ) } < q for all a. this gives an infinite collection of connectors between any two real numbers.

do the set of connectors plus reals form a field? if not, since R is a field, i don't think you could call them numbers. http://mathworld.wolfram.com/Field.html

cheers,
phoenix
 
  • #27
phoenixthoth,

Please read my last post to you, beacuse I updated it, sorry.

I hope my updates helps.


Organic
 
  • #28
still wondering if the set of connectors plus reals is a field... that will decide whether or not connectors could be considered numbers, i think.
 
  • #29
Hi Hurkyl,

Thank you for reply.

Are the "connectors" you talking about simply intervals?
The connectors, which are C elements, close the interval between any two different "normal" real numbers, which are D elements (singletons).

Between any two different arbitrary close Ds there is at least one C, and only C has the power of the continuum.

By the way, I hope you have the time to help me in:

https://www.physicsforums.com/showthread.php?s=&threadid=6427

Thank you.


Organic
 
Last edited:
  • #30
D + C is a field.

You wrote:
supppose we have p < q real numbers.

then p < { (a, 0.5(p+q) ) } < q for all a. this gives an infinite collection of connectors between any two real numbers.
Let us say that b=0.5(p+q), then between a and b there is an infinite collection of D and C elements, which is a field.


Between any two different arbitrary close Ds there is at least one C, and only C has the power of the continuum.
 
Last edited:
  • #31
Hi Anton A. Ermolenko,


By using the empty set (with the Von Neumann Heirarchy), we can construct the set of all positive integers {0,1,2,3,...}:
Code:
0 = { } 

1 = {{ }} = {0}
               
2 = {{ },{{ }}} = {0,1}
  
3 = {{ },{{ }},{{ },{{ }}}} = {0,1,2}

4 = {{ },{{ }},{{ },{{ }}},{{ },{{ }},{{ },{{ }}}}} = {0,1,2,3}

and so on.
So, as you see we can use nothing but the {} to construct numbers, which are not empty.

The common building block -the simplest and invariant(=symmetric) element- is {}.


Organic
 
Last edited:
  • #32
that there are infinitely many of them isn't enough.

how is the arithmetic (addition/subtraction, multiplication/division), of connectors defined?

since these are functions with one element...

for example, what are { (a, b) } + { (c, d) }
and { (a, b) } * { (c, d) }?

if this is to be a field extension of R, then it would have to be the case that
{ (a, a) } + { (b, b) } = { (a+b, a+b) }
and
{ (a, a) } * { (b, b) } = { (ab, ab) }
since the connector { (a, a) } is identified with the real number a.

if it can be shown that if this has ring structure (again, see http://www.mathworld.com for def. of ring), can you show it is an integral domain (ie, no zero divisors)? what is the formula for the multiplicative inverse of a connector { (a, b) }?

a question for the long term is that if the connectors plus reals forms a field, then is it algebraically closed?

cheers,
phoenix
 
  • #33
phoenixthoth ,

You wrote:
...since the connector { (a, a) } is identified with the real number a.
Well {(a,a)} can't be but a D element, so there is no connactor (a C element, which exists only between two different Ds) in {(a,a)}.


Please look at the pdf that I sent to your personal email.

Then, please open and read my informal overview in:

http://www.geocities.com/complementarytheory/CATpage.html

I have a new point of view on the natural numbers, which are the simplest numbers in Math language.

In general I show that any natural number > 1 is several structural variations of the same quantity, where each structure is some tree-like element, constructed by an AND connective between C and D elements.

So any arithmetic operation must be found among those structures,
and there is no meaning to Cs XOR/OR Ds operations.


Please see if I am understood, and reply for any problem.


Thank you.

(By the way, I opned a thread in your forum at:

http://207.70.190.98/scgi-bin/ikonboard.cgi?;act=ST;f=2;t=48 )




Yours,


Organic
 
Last edited by a moderator:
  • #34
"However, if I get you right, your precisely empty set is not (also can’t be) an element of any definite superset."

sure it can.
Hi Phoenixthoth.
I still insist, it can’t.

let 0 denote the empty set. {0} is a superset containing 0. in fact, that is the definition of the number 1.
Let’s see. At first you’ve denoted an empty set as zero (but this denotation is not a number), then you’ve substituted this concept by zero (a number!).
The axiomatic theory of sets needs the concept of an empty set so that the result of any operation with sets is to be a set, too, but the results of all operations with sets form a class; this class is neither a set, nor a superset. In general, a class defines only the properties of some objects (which may be sets), not the objects themselves.
Zero (number) in fact can be an element of a numerical set of an algebraic system. However, it is a number – thus, an element of a numerical set.
But, if we keep on going this way, we’ll change the topic.
The idea I’d like to get across with in my message above is that an empty set can’t possesses symmetry as it has no elements.

To Organic:

In general, your ideas aren’t new extension of mathematics, but may have application in the computer science.
 
  • #35
I still insist, it can’t.


Let’s see. At first you’ve denoted an empty set as zero (but this denotation is not a number), then you’ve substituted this concept by zero (a number!).
The axiomatic theory of sets needs the concept of an empty set so that the result of any operation with sets is to be a set, too, but the results of all operations with sets form a class; this class is neither a set, nor a superset. In general, a class defines only the properties of some objects (which may be sets), not the objects themselves.
if you prefer, denote the empty set by e. then {e} is a superset of e containing e.

http://mathworld.wolfram.com/Zermelo-FraenkelAxioms.html

here, I'm using the unordered pair axiom (axiom 2) where I'm considering the pair e and e and saying {e, e} is a set. one can then use axiom 1 to show that {e, e} = {e}, so it is a set.




Zero (number) in fact can be an element of a numerical set of an algebraic system. However, it is a number – thus, an element of a numerical set.
in set theory, one defines zero to be the empty set. then it can be shown that all number systems can be built upon this.
But, if we keep on going this way, we’ll change the topic.
The idea I’d like to get across with in my message above is that an empty set can’t possesses symmetry as it has no elements.
what is the definition of symmetry, because whatever it is may be vacuously true of the empty set?

cheers,
phoenix
 

Similar threads

  • Other Physics Topics
Replies
16
Views
1K
Replies
8
Views
500
Replies
3
Views
3K
  • Other Physics Topics
Replies
1
Views
870
  • Other Physics Topics
Replies
1
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
1K
  • Math Proof Training and Practice
2
Replies
69
Views
3K
  • Calculus
Replies
8
Views
300
  • Differential Equations
Replies
1
Views
647
Back
Top