Deeper than primes

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doronshadmi said:
Yes I know GOD-jsfisher ,the empty set exists because you declare that it exists. You just forgot the fact that X is empty because empty X is not a member of empty X (known as circular reasoning).
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Wrong. Please review http://www.internationalskeptics.com/forums/showthread.php?postid=6612258 and http://en.wikipedia.org/wiki/Circular_reasoning in details until you understand it. There can be no communication between us until you are willing to open your mind and get out of your :boxedin:.
 
doronshadmi said:
By using the elementary case of the diagonal method (the elementary case is done without bijection, by using the fact that the members of the set of irrational numbers are distinct, and this fact is used as the index of the diagonal method).
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So, in Doronetics, a "fact" can be used as an index into a sequence. How intriguing. Unfortunately, much like the set of irrational numbers, the set of all facts is uncountably infinite (since I can generate the fact "a is a real number" for every real number a).

I'd very much like to see how Doronetics uses uncountably infinite sets as indexes. Given the element at position Pi, for instance, which position comes next? Pi + 1? Pi + 0.1? Pi + 0.01...? Oh look at that, an infinite number of choices as to what the next element could be. It's almost as if the word sequence completely loses its meaning when you use an uncountable set as the collection of indexes...
 
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Heya, back... any earthshattering, mindblowing proof while I was out? I see a few avatars have changed, the subject of discussion is now more philosophical...

Great, looking forward to having some good banter with Doron.

*EDIT* Still having the problem with infinities...
 
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HatRack said:
So, in Doronetics, a "fact" can be used as an index into a sequence.
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HatRack, the order of some collection of distinct objects has no influence on the number of objects.

Let me show something cool.

The power set of, for example, {0,1,2} is {{},{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2}}.

Actually the power set of some set is 2^(the number of the distinct objects of that set), for example:
{
000 ↔ {}
001 ↔ {0}
010 ↔ {1}
011 ↔ {2}
100 ↔ {0,1}
101 ↔ {0,2}
110 ↔ {1,2}
111 ↔ {0,1,2}
}

So by using <0,1>^X we can define a bijection with any power set, such that the members of the set and the members of power set are constructed by the same rule of <0,1> symbols.

Please pay attention that if we use the diagonal method on any arbitrary set of X members (and in this case X=3), which are based on <0,1> symbols, we get the members of the power set that are not in the range of some X arbitrary members of that set, for example:

If the arbitrary members are:
{
111,
100,
101}
then the diagonal member of the power set that is not in the range of these X arbitrary members is 010.

We can change the X arbitrary members, but always we get some diagonal member of the power set of X, which is not in the range of the X arbitrary members.


By using the common rule of <0,1> construction, we are using the diagonal method on the set of ∞ distinct members, as follows:


{
111… ,
100… ,
101… ,
…
}

The diagonal member of the power set that is not in the range of set of ∞ distinct members, starts (in this case) with 010… <0,1> symbols, even if X=∞ (also in this case the members of the set and the members of power set are constructed by the same rule of <0,1> symbols).

Because the set and the power set share members that are based on the same rule and there are always members of the power set that are not in the range of the set (whether X is finite or not), then no set is complete exactly because every set has a power set and every power set has also power set etc... ad infinitum ...
 
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doronshadmi said:
Because the set and the power set share members that are based on the same rule and there are always members of the power set that are not in the range of the set (whether X is finite or not), then no set is complete exactly because every set has a power set and every power set has also power set etc... ad infinitum ...
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Through the power and insanity of Doronetics, we now know that no set is complete. It used to be just the infinite sets were incomplete, but Doron as revised his illogic to be more inclusive.
 
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doronshadmi said:
I know GOD-jsfisher...
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Really? And you surely think that you are the only one, right? These are the very basics of encoding, aren't they? Upper-case are added; lower-case are counted, no?
:rolleyes:

GOD - jsfisher = ?

Since the closest letter to '=' is 'r', then the most logical choice is

GOD - jsfisher = R \due to the letter identity r=r\

The proof that the logical choice is also the correct one relies on the rules of the basic encoding.

(G+O+D) - |jsfisher| = R

Since the cardinality of jsfisher is 8, then

(G+O+D) - 8 = R,

and due to the alphabetical order, G+O+D = 7+15+4 = 26. So you have

26 - 8 = R

Since R=18 in the alphabet, then

26 - 8 = 18

is the correct subtraction result, and the coincidence

GOD - jsfisher = R
GOD - jsfishe(r = R)

is good.

Haldegard, how long do you think you can live in Doron's head? I told you once what that takes. Egbert from the CB-platform departed from Doron's head after having been there for just six months, and he is still not finished with turning the Bible text into an opera. I wonder what your trip would be when you take a hike. No demon ever got out of Doron's head sane. If you think that you'll become the exemption and get the Medal of Uncommon Valor from your boss, then you are already crazy.
 
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doronshadmi said:
If a set and its power set share members that are based on the same rule, then no set is complete, as shown in http://www.internationalskeptics.com/forums/showpost.php?p=6704754&postcount=13565 .
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Because the set and the power set share members that are based on the same rule and there are always members of the power set that are not in the range of the set (whether X is finite or not), then no set is complete exactly because every set has a power set and every power set has also power set etc... ad infinitum ...
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Can you demonstrate the infinite expansion of a set via power its power sets using this example?

[latex]S=\{a,b\}[/latex]

P[latex]\powset{S}=\{\emptyset, \{a\}, \{b\}, S\}[/latex]
 
doronshadmi said:
If a set and its power set share members that are based on the same rule, then no set is complete, as shown in http://www.internationalskeptics.com/forums/showpost.php?p=6704754&postcount=13565 .
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(a) As a general rule, a set doesn't share members with its power set. Consider the example case you yourself offered. The set {0,1,2) and its power set have an empty intersection. No member of the set is also a member of the power set.

(b) What grand illogic you use. What to other sets have to do with the completeness of some set? Granted, you are no doubt using some bizarre meaning of completeness heretofore unknown in human existence, but since you will never articulate that meaning, it is obvious, once again, you are just making crap up to cover your ignorance.

(c) Your prior post you cite doesn't show what you claim it shows. It does, however, clearly show how vacuous your reasoning is.
 
doronshadmi said:
So by using <0,1>^X we can define a bijection with any power set, such that the members of the set and the members of power set are constructed by the same rule of <0,1> symbols.
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No, that's not correct. <0,1>^X is not only countable, but finite (since X is a finite number which specifies the number of 0's and 1's in each string by your demonstration). Hence, it cannot be put into 1-to-1 correspondence with provably uncountable sets (like the irrationals or its power set) by definition. Please open your mind up to every detail in http://en.wikipedia.org/wiki/Cardinality to understand why you are wrong.

doronshadmi said:
HatRack if you use X as a factor that determines X, you are using a circular reasoning.
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No, that's not correct either. Circular reasoning is when you assume a proposition A to prove A. Every set uses itself as a factor in its definition. For example, the set {1} is defined as the set such that 1 is its member and every other set X (including {1} itself) is not a member. This type of "self-reference" is inescapable in defining sets.

Please open up your mind to every detail in http://en.wikipedia.org/wiki/Circular_reasoning to understand why you are wrong. There can be no communication between us until you get out of your :boxedin:.
 
jsfisher said:
(a) As a general rule, a set doesn't share members with its power set.
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You have missed this:

Actually the power set of some set is 2^(the number of the distinct objects of that set), for example:
{
000 ↔ {}
001 ↔ {0}
010 ↔ {1}
011 ↔ {2}
100 ↔ {0,1}
101 ↔ {0,2}
110 ↔ {1,2}
111 ↔ {0,1,2}
}

The set above is the power set of the sets {000,001,010} , {111,011,110} , {101,010,111} , etc ...

As can be seen the diagonal object (which is a member of the power set) is constructed by the same rule of the members of these sets.

Since the set and the power set share members that are constructed by the same rule, we get exactly the case of Godel's first incompleteness theorem, where we define an object that has the properties of a given set (it obeys the rules of a give framework) but it is not a member of that set (but can't be proved within this framework).
 
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doronshadmi said:
{
000 ↔ {}
001 ↔ {0}
010 ↔ {1}
011 ↔ {2}
100 ↔ {0,1}
101 ↔ {0,2}
110 ↔ {1,2}
111 ↔ {0,1,2}
}

The set above is the power set of the sets {000,001,010} , {111,011,110} , {101,010,111} , etc ...
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No, that's not right. You claim that your set, let's call it A, is the power set of both {000,001,010} and {111,011,110}. But, the power set of {000,001,010} contains the member {000} and not {111}, and the power set of {111,011,110} contains the member {111} and not {000}. Hence, A cannot possibly be the power set of both of these sets.

This is really a simple concept, one of the first things they teach in elementary set theory. Please open your mind to every detail in http://en.wikipedia.org/wiki/Power_set to understand why you are wrong.
 
HatRack said:
No, that's not correct. <0,1>^X is not only countable, but finite (since X is a finite number ...
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Wrong, X can be finite or infinite number.


HatRack said:
No, that's not correct either. Circular reasoning is when you assume a proposition A to prove A. Every set uses itself as a factor in its definition. For example, the set {1} is defined as the set such that 1 is its member and every other set X (including {1} itself) is not a member. This type of "self-reference" is inescapable in defining sets.
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You have a simple mistake, 1 is not a set, but can be a member of a set.

On the contrary set {} is one of the factors that are used to determine that set {} is empty , also because set {} is one of the sets that are not members of set {}.

If we are using a set to determine itself as a set, we are using a circular reasoning, because we are using the premise of the existence of set {} by using set {} (set {} is not a member of set {}) as a factor that determines the existence of set {}, as set {}.
 
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doronshadmi said:
Wrong, X can be finite or infinite number.
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If X is infinite in <0,1>^X, then <0,1>^X can only be interpreted to be the set of all sequences of 0's and 1's. So, let us construct any sequence of the elements of <0,1>^infinity:

n11 n12 n13 ...
n21 n22 n23 ...
n31 n32 n33 ...
...

Now, I can find an element n0 such that n0 = ~n11 ~n22 ~n33 ... (where ~ means flip 0 to 1 and 1 to 0). Thus, n0 is not in that sequence, and it follows by the diagonal argument that <0,1>^infinity is uncountable. So, this set cannot be used as indexes as you claim.



doronshadmi said:
You have a simple mistake, 1 is not a set, but can be a member of a set.
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The mistake is all yours. In set theory, 1 is traditionally defined to be the set {0}. Everything is a set in set theory, and you would know this if you ever bothered to learn it. Please obtain the book "Naive Set Theory" by Paul R. Halmos and carefully study it with an open mind, so that you may understand all of your blatant blunders and finally get out of your :boxedin:.
 
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HatRack said:
No, that's not right. You claim that your set, let's call it A, is the power set of both {000,001,010} and {111,011,110}. But, the power set of {000,001,010} contains the member {000} and not {111}, and the power set of {111,011,110} contains the member {111} and not {000}. Hence, A cannot possibly be the power set of both of these sets.
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In other words, you still do not get the fact that the power set of a set can be generalized to <0,1>^(the number of the distinct members of that set that are constructed only by <0,1> symbols), for example:

{
000 ↔ {}
001 ↔ {0}
010 ↔ {1}
011 ↔ {2}
100 ↔ {0,1}
101 ↔ {0,2}
110 ↔ {1,2}
111 ↔ {0,1,2}
}

where the set above is the power set of the sets {000,001,010} , {111,011,110} , {101,010,111} , etc ... that their members are constructed only by <0,1> symbols on the same level (no additional level of "{""}" is used, as used in the case of {{},{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2}) in order to define the 2^3=8 members of the power set of set {0,1,2}).
 
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doronshadmi said:
In other words, you still do not get the fact that the power set of a set can be generalized to <0,1>^(the number of the distinct members of that set that are constructed only by <0,1> symbols), for example:
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No, it cannot be. The power set of a set is the set of all that set's subsets. That is the definition that everyone who does mathematics agrees upon, and it will never change. If you wish to specify some other type of set, then give it a name that's not already taken.
 
HatRack said:
If X is infinite in <0,1>^X, then <0,1>^X can only be interpreted to be the set of all sequences of 0's and 1's. So, let us construct any sequence of the elements of <0,1>^infinity:

n11 n12 n13 ...
n21 n22 n23 ...
n31 n32 n33 ...
...

Now, I can find an element n0 such that n0 = ~n11 ~n22 ~n33 ... (where ~ means flip 0 to 1 and 1 to 0). Thus, n0 is not in that sequence, and it follows by the diagonal argument that <0,1>^infinity is uncountable. So, this set cannot be used as indexes as you claim.
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You are still missing it, the members of the set and the power set are constructed by the same rule.

Given any arbitrary order of the infinite members of a set that its members are constructed only by <0,1> symbols, by using the diagonal method you define members of the power set of this set, that their members are also constructed only by <0,1> symbols.


HatRack said:
The mistake is all yours. In set theory, 1 is traditionally defined to be the set {0}.
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Wrong, 1 is the cardinality of set {0} exactly as 0 is the cardinality of set {}.

The cardinality of a set is not a set.
 
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