set set set
A M B
| | |
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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).
!__________________!
!__________________!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...In the attached address you can find A new approach for the definition of a NUMBER...
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.
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
.______.____.from your file:Originally posted by Organic
Please open the attached pdf file.
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,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 {._. .}.
Really? Perhaps, you’ve misunderstood...Originally posted by Organic
Why you are so aggressive…
Perhaps...please take Math as some form of communication where persons share each other's ideas.
And I’m glad to see you’re glad...I'll be glad to get any remarks from you.
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.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.
Well,Please write to me on any part of my work, where you find contradictions.
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.Let {___} be Csim (C for Continuum).
Let {._. .} be Dsim (D for Discreteness).
You’re welcome.Thank you.
Yes. and if x = aCb and y = cCd thenx < y if b < y?
The connectors, which are C elements, close the interval between any two different "normal" real numbers, which are D elements (singletons).Are the "connectors" you talking about simply intervals?
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.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.
0 = { }
1 = {{ }} = {0}
2 = {{ },{{ }}} = {0,1}
3 = {{ },{{ }},{{ },{{ }}}} = {0,1,2}
4 = {{ },{{ }},{{ },{{ }}},{{ },{{ }},{{ },{{ }}}}} = {0,1,2,3}
and so on.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)}....since the connector { (a, a) } is identified with the real number a.
Hi Phoenixthoth."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’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!).let 0 denote the empty set. {0} is a superset containing 0. in fact, that is the definition of the number 1.
if you prefer, denote the empty set by e. then {e} is a superset of e containing e.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.
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.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.
what is the definition of symmetry, because whatever it is may be vacuously true of the empty 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.
Huh... really? Let’s see. How do you define an empty set? The axiomatic theory of sets (ATS) defines it as: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 = { }
What it precisely means? Either an empty set is a subset of «1» (but «1» is not a natural number) or it is an element of «1». The ATS doesn’t let us define 1 (natural number) by non-number. Only a map correspond the elements of different nature to each other. And so on...1 = {{ }} = {0}
using axiom 5 inHuh... really? Let’s see. How do you define an empty set? The axiomatic theory of sets (ATS) defines it as:
("õ"="direct product", "<>"="not equal", "+"="direct addition")
AõA=0, B<>0, AõB=0, A+B=0: A=0.
However,
AõA=0, B<>0, AõB=0, A+B=0: A=0 or A="zero divisor", i.e. it is non-empty set.
let's look at the first statement and replace "empty set" by the phrase "concept X." i'll also rule out your double negative. it becomes this: without concept X mathematicans can't define concept X. i disagree with this statement. you can certainly define any concept X you like. whether concept X "exists" is another story...i.e. without a non-empty set no one from mathematicians can’t defines an empty set. And this is nature of an empty set (or ATS) – it is necessary to definition of operations with sets.
What it precisely means? Either an empty set is a subset of «1» (but «1» is not a natural number) or it is an element of «1». The ATS doesn’t let us define 1 (natural number) by non-number. Only a map correspond the elements of different nature to each other. And so on...
Excuse me, is e a subset of {e}, or e an element of {e}?if you prefer, denote the empty set by e. then {e} is a superset of e containing e.
Well, then how do you define zero divisor?Originally posted by phoenixthoth
using axiom 5 in
one can define an empty set to be such an x. then you can prove that all empty sets are equal, so it makes sense to give them all one notation.
the word set is undefined.
Originally posted by Anton A. Ermolenko
Excuse me, is e a subset of {e}, or e an element of {e}?
If e is a subset of {e}, then you’ll not define either {e} or e, because {e}=e, hence, {e, e}={e} imply e either empty set, or zero divisor. Without a non-empty set neither you, nor anybody else will not prove that e is empty set
If e is an element of {e}, then you’ll not prove that e is empty set, because in this case {e, e} always {e, e}.
Huh... ZF system of axioms? Well, but pair of "axioms" (subsets & foundation) are not precise axioms, only the schemes. I prefer the NBG-system of axiomsOriginally posted by phoenixthoth
if you prefer,
http://mathworld.wolfram.com/Zermelo-FraenkelAxioms.html
Huh... well, well, well... I like it. Then following your logic:Originally posted by phoenixthoth
element.
it's not about proving the set is empty. its emptiness is a postulation.
for any x, {x, x} = {x}
0 = { }
1 = {0}
n = {n-1}
n+1 = n U {n}
In the realOriginally posted by phoenixthoth
using axiom 5 in
0 = { }
1 = {0}
n = {n-1}
n+1 = n U {n}
Originally posted by Anton A. Ermolenko
Huh... well, well, well... I like it. Then following your logic:
0=1=2=3=4=5=6=7=8=9=...="infinity".
You really think so?
Exactly 0!=1, because 0! is number of permutations (there is an order!) {01}=1Originally posted by phoenixthoth
i don't see how you get 0 = 1 = 2 = ...
0 = { }
1 = {0}
two sets x and y are equal iff (a is an element of x iff a is an element in y).
0 is an element of 1 but 0 is not an element of 0 = { }.
therefore, it is not the case that a is an element of 1 iff a is an element of 0.
therefore, 0 != 1.
From the ZF system of axioms may be proved (almost) all mathematics. In spite of the fact that the actual number of axiom is equal to infinity (Z5 and ZF9 are not the axioms, only schemes), but there is no these: {0,0}={0}. Empty set (or zero element) defined as the result of these operations:Originally posted by phoenixthoth
!= means "does not equal."
i said, "therefore, 0 != 1." if ! is factorial, then it would be directly adjacent to the zero, which it is not.
if i meant ! as factorial, the statement 0! = 1 is a non-sequitor from my argument.
the statements above the conclusion demonstrated how the pair 0 and 1 don't fit the definition of set equality.
the conclusion was that the two sets are not equal.
i urge you to check out "elements of set theory" by enderton, "set theory" by stoll, or "axiomatic set theory" by suppes for all the details.
Originally posted by Anton A. Ermolenko
From the ZF system of axioms may be proved (almost) all mathematics. In spite of the fact that the actual number of axiom is equal to infinity (Z5 and ZF9 are not the axioms, only schemes), but there is no these: {0,0}={0}. Empty set (or zero element) defined as the result of these operations:
B != A, A õ A = A, A õ B= A, A + B = B: A=0.
Otherwise, how do you define a zero divisor or a nilpotent device?
setsOriginally posted by phoenixthoth
{0,0} = {0} is not an axiom but it can be proven from the axiom of extensionality.
what are the definitions of A and B?
x is a zero divisor if it is nonzero and there is a nonzero y such that x õ y = 0. [/qoute]
I want to see your definition of a zero divisor by ZF system axioms.
You've misunderstood. I define an empty set (or zero element). In other words, iif B != A, A õ A = A, A õ B= A, A + B = B, then A=0. If B != A, A õ A = A, A õ B= A, A + B != B, then A is a zero divisor. Your postulate {0,0}={0} doesn't allow defining a zero divisor.in your equations above, neither B nor A is conclusively a zero divisor because A = 0. in every ring, B õ 0 = 0; that doesn't make B a zero divisor since then, every element would be a zero divisor.
Originally posted by Anton A. Ermolenko
sets
You've misunderstood. I define an empty set (or zero element). In other words, iif B != A, A õ A = A, A õ B= A, A + B = B, then A=0. If B != A, A õ A = A, A õ B= A, A + B != B, then A is a zero divisor. Your postulate {0,0}={0} doesn't allow defining a zero divisor.
