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15th September 2020, 12:26 PM  #481 
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15th September 2020, 10:52 PM  #482 
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I follow your definition of set of finite sets (as the complementary set of set X) notated by you as N.
N is a general set of finite sets Code:
N = { ∅, > {{∅}}, > ... ↓ Nobijection {∅}, > {{∅}, {{∅}} }, > ... ↓ Nobijection {∅, {∅}}, > {{∅}, {{∅}}, {{∅}, {{∅}}}}, > ... ↓ Nobijection {∅, {∅}, {∅, {∅}}}, > {{∅}, {{∅}}, {{∅}, {{∅}}}, {{∅}, {{∅}}, {{∅}, {{∅}}}}}, > ... ↓ Nobijection ... ... } Code:
V = { ∅, ↓ Nobijection {∅}, ↓ Nobijection { ∅, {∅} }, ↓ Nobijection { ∅, {∅} , {∅, {∅}} }, ↓ Nobijection { ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} }, ↓ Nobijection ... } Please define the membership function of V. V is the set of all the finite sets of von Neumann set of ordinals iff 
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16th September 2020, 01:18 AM  #483 
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B.T.W I know that N members can also be infinite sets, but since we are talking about the minimal set satisfying the requirements of the Axiom of Infinity, only finite sets (which are the complements of set X) are the members of N.
N as a set of finite sets: Code:
N = { ∅, > {{∅}}, > ... ↓ Nobijection {∅}, > {{∅}, {{∅}} }, > ... ↓ Nobijection {∅, {∅}}, > {{∅}, {{∅}}, {{∅}, {{∅}}}}, > ... ↓ Nobijection {∅, {∅}, {∅, {∅}}}, > {{∅}, {{∅}}, {{∅}, {{∅}}}, {{∅}, {{∅}}, {{∅}, {{∅}}}}}, > ... ↓ Nobijection ... ... } Code:
V = { ∅, ↓ Nobijection {∅}, ↓ Nobijection { ∅, {∅} }, ↓ Nobijection { ∅, {∅} , {∅, {∅}} }, ↓ Nobijection { ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} }, ↓ Nobijection ... } Please define the membership function of V. V is the set of all the finite sets of von Neumann set of ordinals iff 
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That is also over the matrix, is aware of the matrix. That is under the matrix, is unaware of the matrix. For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video. 

16th September 2020, 06:00 AM  #484 
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And just what would your definition be, then?
By the way, my definition for nonfinite sets was based on a definition for relative cardinality. Are you now adopting my definition of cardinality, too? And just what did you mean by "(as the complementary set of set X) notated by you as N"? In particular, what is a "complementary set"? We are still all wondering, too, what you mean by "general set". 
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16th September 2020, 06:03 AM  #485 
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V is the set of all the finite sets of von Neumann set of ordinals iff (∅ ∈ V) AND (∀v ∈ V ( ( v ∪ { v } ) ∈ V )) AND (∀v ∈ V (v < V))

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16th September 2020, 06:18 AM  #486 
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That is also over the matrix, is aware of the matrix. That is under the matrix, is unaware of the matrix. For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video. 

16th September 2020, 06:42 AM  #487 
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16th September 2020, 07:48 AM  #488 
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No, I take a ride on it with you in order to find if it holds water.
In that case, finite sets' cardinality and X cardinality are complements, isn't it? For example, v < V are complementary terms, since both are needed in order to define that one is smaller than the other or that one is bigger than the other. Who is "We"? Do you need some group in order to discus with me? As for "general set" it is not relevant anymore since I wrote a new post, but you replied to the old one. Please look at the relevant one http://www.internationalskeptics.com...&postcount=483 and also at http://www.internationalskeptics.com...&postcount=485 . Thank you. 
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16th September 2020, 04:18 PM  #489 
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16th September 2020, 04:39 PM  #490 
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So, then, what are the definitions you a taking a ride upon? In your own words, please. Well, symbols, actually, are preferable.
Quote:
I have no idea what you meant by "complementary set", and I suspect you don't either.
Quote:
How about not making up terms in the first place?
Quote:
Nevertheless, I do read your posts in their entirety. If I choose to respond, I usually will limit myself to up to the first significant error, which is my right and habit. If the beginning of a post is garbage, there is seldom a reason to dignifying the trash after that point with any sort of response. You are welcome. 
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16th September 2020, 04:43 PM  #491 
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Who knows? This may become relevant at some point, too:
∃I (Ø∈I ∧ ∀x (x∈I ⇒ x ⋃ {x} ∈ I)) 
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17th September 2020, 12:06 AM  #492 
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That is also over the matrix, is aware of the matrix. That is under the matrix, is unaware of the matrix. For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video. 

17th September 2020, 12:19 AM  #493 
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Ok, lets replace iff by if ... then.
p="(∅ ∈ V) AND (∀v ∈ V ( ( v ∪ { v } ) ∈ V )) AND (∀v ∈ V (v < V))" q="V is the set of all the finite sets of von Neumann set of ordinals" Code:
p → q T T T T F F F T T F F T If (∅ ∈ V) AND (∀v ∈ V ( ( v ∪ { v } ) ∈ V )) AND (∀v ∈ V (v < V)), then V is the set of all the finite sets of von Neumann set of ordinals. V is the requested set: Code:
V = { ∅, ↓ Nobijection {∅}, ↓ Nobijection { ∅, {∅} }, ↓ Nobijection { ∅, {∅} , {∅, {∅}} }, ↓ Nobijection { ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} }, ↓ Nobijection ... } Please define the (yes/no) iff membership function of V. V is the set of all the finite sets of von Neumann set of ordinals iff ... 
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That is also over the matrix, is aware of the matrix. That is under the matrix, is unaware of the matrix. For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video. 

17th September 2020, 03:37 AM  #494 
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17th September 2020, 03:37 AM  #495 
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V = {v : (∅ ∈ V) ∧ (∀v ∈ V ( ( v ∪ { v } ) ∈ V )) ∧ (∀v ∈ V (v < V))}

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That is also over the matrix, is aware of the matrix. That is under the matrix, is unaware of the matrix. For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video. 

17th September 2020, 04:03 AM  #496 
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That is also over the matrix, is aware of the matrix. That is under the matrix, is unaware of the matrix. For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video. 

17th September 2020, 04:33 AM  #497 
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17th September 2020, 04:34 AM  #498 
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17th September 2020, 04:37 AM  #499 
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17th September 2020, 05:22 AM  #500 
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That is also over the matrix, is aware of the matrix. That is under the matrix, is unaware of the matrix. For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video. 

17th September 2020, 05:39 AM  #501 
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delete ...

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That is also over the matrix, is aware of the matrix. That is under the matrix, is unaware of the matrix. For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video. 

17th September 2020, 06:53 AM  #502 
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"[T]he set of all sets that are not nonfinite sets"!! Phew, that's convoluted. No, the minimal set satisfying the requirements of the Axiom of Infinity, which I prefer to call N, is not the set of all finite sets. Informally, N has the empty set as a member, all successor that stem from the empty set as members, and nothing else. And for the avoidance of doubt, this informal description cannot be easily prettied up to make it a formal definition of N. 
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17th September 2020, 07:03 AM  #503 
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V = { v : (v is not nonfinite member of von Neumann set of ordinals) }
Here are some of the members of V: Code:
V = { ∅, ↓ Nobijection {∅}, ↓ Nobijection { ∅, {∅} }, ↓ Nobijection { ∅, {∅} , {∅, {∅}} }, ↓ Nobijection { ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} }, ↓ Nobijection ... } There is nobijection from ∀v∈V to ∀v∪{v} in V (where V order is irrelevant). Since ∀ ( v ∪ { v } ) ∈ V v∪{v} < V, V is not the cardinality of V. 
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That is also over the matrix, is aware of the matrix. That is under the matrix, is unaware of the matrix. For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video. 

17th September 2020, 07:13 AM  #504 
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In other words, Since V is not formally established, set X can't be nonfinite since V <= X (where V is the minimal set satisfying the requirements of the Axiom of Infinity) is not well defined, thanks to ∀.

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That is also over the matrix, is aware of the matrix. That is under the matrix, is unaware of the matrix. For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video. 

17th September 2020, 07:34 AM  #505 
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(1) The ordinals defined by von Neumann do not form a set.
(2) You haven't defined what you mean by "nonfinite" with regards to sets. (3) If you choose to accept the ordinals into the discussion, then since they are heavily founded in cardinality and many other concepts you abhor, you are implicitly accepting all those things. 
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17th September 2020, 07:47 AM  #506 
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In other words, you don't understand the definition I have provided for the comparisons of set cardinality.
A <= B if and only if there is an injection from A to Bis entirely welldefined. It's a definition, by the way. It is not subject to proof or disproof. Moreover, despite your fascination with bijections (which are absent from my definition), you have not in any way shown there is no set X meeting the criterion for V <= X. In fact, if X is the set V, itself, it is trivial to show V <= V. Also, if you are now back to shunning quantification, then your entire approach to an argument falls apart. It heavily relies on quantification. 
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19th September 2020, 01:27 AM  #507 
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V = { v : (v is not nonfinite member of von Neumann set of ordinals) } = The set of all natural numbers in terms of sets.
Since ∀ ( v ∪ { v } ) ∈ V v∪{v} < V, V is not the cardinality of set V. In other words, you don't have a basis for your X as a nonfinite set, and V <= V is not exceptional exactly because ∀ ( v ∪ { v } ) ∈ V v∪{v} < V. One asks: Is there a bijection among all V members? The answer is yes: Given any arbitrary member x in V (where V is the set of all natural numbers in terms of sets) the cardinality of x is x, which means that x is bijective with itself. But ∀ x ∈ V (where x=v or x=v∪{v}) there is nobijection from v to v∪{v} ∧ ∀ (v ∪{v} ) ∈ V v∪{v} < V 
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19th September 2020, 03:59 AM  #508 
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Please ignore post #507.
V = { v : (v is not nonfinite member of von Neumann set of ordinals) } = The set of all natural numbers in terms of sets. Since ∀(v∪{ v}) ∈ V v∪{v} < V, V is not the cardinality of set V. In other words, you don't have a basis for your X as a nonfinite set, and V <= V is not exceptional exactly because ∀(v∪{v}) ∈ V v∪{v} < V. One asks: Is there a bijection among all V members? The answer is yes: Given any arbitrary member x in V (where V is the set of all natural numbers in terms of sets) the cardinality of x is x, which means that x is bijective with itself. But ∀x ∈ V (where x=v or x=v∪{v}) there is nobijection from v to v∪{v} ∧ ∀(v ∪{v}) ∈ V v∪{v} < V Here it is: Code:
V = { > Bijection, order is irrelevant among all x members (where x=v or x=v∪{v}).  0 = ∅,  ↓ Nobijection  1 = { ∅ },  ↓ Nobijection  2 = { ∅, {∅} },  ↓ Nobijection  3 = { ∅, {∅} , {∅, {∅}} },  ↓ Nobijection  4 = { ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} },  ↓ Nobijection  v NoBijection, order is irrelevant among all x members (where x=v or x=v∪{v}) and ∀(v∪{v}) ∈ V v∪{v} < V ... }
Originally Posted by doronshadmi
Since V is not strictly established, the Cantorian notion of strict cardinality for set V, does not hold water. So once again, thank you ∀. 
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That is also over the matrix, is aware of the matrix. That is under the matrix, is unaware of the matrix. For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video. 

19th September 2020, 04:00 AM  #509 
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I like ∀

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That is also over the matrix, is aware of the matrix. That is under the matrix, is unaware of the matrix. For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video. 

19th September 2020, 07:28 AM  #510 
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19th September 2020, 07:47 AM  #511 
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V is not the von Neumann set of ordinals, bot only all of its members that are equivalent to all the natural numbers.
Still don't understand that ∀(v∪{v}) ∈ V v∪{v} < V, which means that V is not the cardinality of set V (the set of all natural numbers). If ∀(v∪{v}) ∈ V is a complete gibberish, it can be replaced by ∀v(v∪{v} ∈ V) 
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That is also over the matrix, is aware of the matrix. That is under the matrix, is unaware of the matrix. For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video. 

19th September 2020, 08:40 AM  #512 
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V = { v : (v is not nonfinite member of von Neumann set of ordinals) } = The set of all natural numbers in terms of sets.
∀v(v∪{v} ∈ V) ∧ v∪{v} < V, which means that V is not the cardinality of set V. In other words, you don't have a basis for your X as a nonfinite set, and V <= V is not exceptional exactly because ∀v(v∪{v} ∈ V) ∧ v∪{v} < V. One asks: Is there a bijection among all V members? The answer is yes: Given any arbitrary member x in V (where V is the set of all natural numbers in terms of sets) the cardinality of x is x, which means that x is bijective with itself. But ∀x ∈ V (where x=v or x=v∪{v}) there is nobijection from v to v∪{v} ∧ ∀v(v∪{v} ∈ V) ∧ v∪{v} < V. Here it is: Code:
V = { > Bijection, order is irrelevant among all x members (where x=v or x=v∪{v}).  0 = ∅,  ↓ Nobijection  1 = { ∅ },  ↓ Nobijection  2 = { ∅, {∅} },  ↓ Nobijection  3 = { ∅, {∅} , {∅, {∅}} },  ↓ Nobijection  4 = { ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} },  ↓ Nobijection  v NoBijection, order is irrelevant among all x members (where x=v or x=v∪{v}) and ∀v(v∪{ v} ∈ V) ∧v∪{v} < V ... } Since V is not strictly established, the Cantorian notion of strict cardinality for set V, does not hold water. So once again, thank you ∀. 
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19th September 2020, 08:45 AM  #513 
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There is no such thing as the von Neumann set of ordinals, so any reference to its members is meaningless.
Quote:
Please take some time out from your repetitious posts to define your terms. You need to explain what you mean by: finite or infinite or nonfinite setConcise, simple definitions. No need to wander off towards some agenda you are trying to push, just the definitions, simple and concise. And can we not agree that "A" is a notation that means "the cardinality of set A"? It is a definition and not subject to disproof. Statements like "V is not the cardinality of set V" are just dumb. 
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19th September 2020, 10:55 AM  #514 
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In that case N, known as the set of all natural numbers (such that all natural numbers are defined in terms of sets) does not exist, and you don't have the minimal nonfinite set to be based on.
Only if there are finitely many members (or no members at all) in A. Since you don't get that yet, let's simplify it for you: A is a finite set iff A is the cardinality of A. Since the term "nonfinite set" is the complement of the term "finite set", so are the following expressions: "A is the cardinality of A" (in case of finite sets) is the complement of "A is not the cardinality of A" (in case of nonfinite sets). 
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That is also over the matrix, is aware of the matrix. That is under the matrix, is unaware of the matrix. For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video. 

19th September 2020, 11:01 AM  #515 
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Here, if these are of any use to you, are mine:

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19th September 2020, 11:09 AM  #516 
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Your conclusion does not follow. There is no set of the von Neumann ordinals simply because it is too big to be a set.
Quote:
Rather selfserving, and not particularly useful. Nonetheless, what then is your definition for cardinality that selflimits somehow (without any reference to finite sets, since that would be circular)? 
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19th September 2020, 11:15 AM  #517 
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Syntactic correction:
V = { v : (v is not nonfinite member of von Neumann ordinals) } = The set of all natural numbers in terms of sets. ∀v(v∪{v} ∈ V ∧ v∪{v} < V), which means that V is not the cardinality of set V. In other words, you don't have a basis for your X as a nonfinite set, and V <= V is not exceptional exactly because ∀v(v∪{v} ∈ V ∧ v∪{v} < V). One asks: Is there a bijection among all V members? The answer is yes: Given any arbitrary member x in V (where V is the set of all natural numbers in terms of sets) the cardinality of x is x, which means that x is bijective with itself. But ∀x ∈ V (where x=v or x=v∪{v}) there is nobijection from v to v∪{v} ∧ ∀v(v∪{v} ∈ V ∧ v∪{v} < V). Here it is: Code:
V = { > Bijection, order is irrelevant among all x members (where x=v or x=v∪{v}).  0 = ∅,  ↓ Nobijection  1 = { ∅ },  ↓ Nobijection  2 = { ∅, {∅} },  ↓ Nobijection  3 = { ∅, {∅} , {∅, {∅}} },  ↓ Nobijection  4 = { ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} },  ↓ Nobijection  v NoBijection, order is irrelevant among all x members (where x=v or x=v∪{v}) and ∀v(v∪{v} ∈ V ∧v∪{v} < V). ... } Since V is not strictly established, the Cantorian notion of strict cardinality for set V, does not hold water. So once again, thank you ∀ 
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19th September 2020, 11:24 AM  #518 
Penultimate Amazing
Join Date: Mar 2008
Posts: 13,243

You can't be aware of it as long as you reject N as the set of all natural numbers in terms of sets, as given in http://www.internationalskeptics.com...&postcount=517.

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That is also over the matrix, is aware of the matrix. That is under the matrix, is unaware of the matrix. For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video. 

19th September 2020, 11:29 AM  #519 
ETcorngods survivor
Moderator Join Date: Dec 2005
Posts: 22,920

No one can "be aware of it" until you tell us what you mean by "nobijection". I know what it means to have a bijection between two sets. I know what it means for there to be no bijection between two sets. This "nobijection" is a mystery.
Are you unable to tell us what you mean? Use your words. Examples are not definitions. 
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A proud member of the Simpson 15+7, named in the suit, Simpson v. Zwinge, et al., and founder of the ET Corn Gods Survivors Group. "He's the greatest mod that never was!"  Monketey Ghost 

19th September 2020, 11:46 AM  #520 
Penultimate Amazing
Join Date: Mar 2008
Posts: 13,243

You simply can't say "I am not aware of it".
As long as you need some us in order to be aware of it, you don't say to yourself "I am not aware of it". Without this fundamental awareness, you actually unable to be aware of it. So first please leave this us ghost out of the framework, and read http://www.internationalskeptics.com...&postcount=517 to yourself by using your awareness, without be depend on any us ghost, at least in this first stage. 
__________________
That is also over the matrix, is aware of the matrix. That is under the matrix, is unaware of the matrix. For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video. 

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