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 International Skeptics Forum Continuation Deeper than primes - Continuation 1/3*9

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 15th September 2020, 12:26 PM #481 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 22,920 Originally Posted by doronshadmi So N is a general set of infinitely many finite sets... Is it now? If only you would tell us what you mean by finite set (or its complement) they maybe we'd know what you mean. You will also need to define "general set" since you've added that to the lexicon. __________________ 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
 15th September 2020, 10:52 PM #482 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 Originally Posted by jsfisher Is it now? If only you would tell us what you mean by finite set (or its complement) they maybe we'd know what you mean. You will also need to define "general set" since you've added that to the lexicon. Originally Posted by jsfisher If only you would tell us what you mean by finite set 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 = { ∅, ---> {{∅}}, ---> ... ↓ No-bijection {∅}, ---> {{∅}, {{∅}} }, ---> ... ↓ No-bijection {∅, {∅}}, ---> {{∅}, {{∅}}, {{∅}, {{∅}}}}, ---> ... ↓ No-bijection {∅, {∅}, {∅, {∅}}}, ---> {{∅}, {{∅}}, {{∅}, {{∅}}}, {{∅}, {{∅}}, {{∅}, {{∅}}}}}, ---> ... ↓ No-bijection ... ... }``` V is the requested set Code: ```V = { ∅, ↓ No-bijection {∅}, ↓ No-bijection { ∅, {∅} }, ↓ No-bijection { ∅, {∅} , {∅, {∅}} }, ↓ No-bijection { ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} }, ↓ No-bijection ... }``` Please define the membership function of V. V is the set of all the finite sets of von Neumann set of ordinals iff __________________ 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. Last edited by doronshadmi; 16th September 2020 at 12:13 AM.
 16th September 2020, 01:18 AM #483 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 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 = { ∅, ---> {{∅}}, ---> ... ↓ No-bijection {∅}, ---> {{∅}, {{∅}} }, ---> ... ↓ No-bijection {∅, {∅}}, ---> {{∅}, {{∅}}, {{∅}, {{∅}}}}, ---> ... ↓ No-bijection {∅, {∅}, {∅, {∅}}}, ---> {{∅}, {{∅}}, {{∅}, {{∅}}}, {{∅}, {{∅}}, {{∅}, {{∅}}}}}, ---> ... ↓ No-bijection ... ... }``` V is the requested set: Code: ```V = { ∅, ↓ No-bijection {∅}, ↓ No-bijection { ∅, {∅} }, ↓ No-bijection { ∅, {∅} , {∅, {∅}} }, ↓ No-bijection { ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} }, ↓ No-bijection ... }``` Please define the membership function of V. V is the set of all the finite sets of von Neumann set of ordinals iff __________________ 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. Last edited by doronshadmi; 16th September 2020 at 02:14 AM.
 16th September 2020, 06:00 AM #484 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 22,920 Originally Posted by doronshadmi I follow your definition of set of finite sets (as the complementary set of set X) notated by you as N. And just what would your definition be, then? By the way, my definition for non-finite 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". __________________ 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
 16th September 2020, 06:03 AM #485 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 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|)) __________________ 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. Last edited by doronshadmi; 16th September 2020 at 06:36 AM.
 16th September 2020, 06:18 AM #486 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 Originally Posted by jsfisher And just what would your definition be, then? By the way, my definition for non-finite 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". Please look at http://www.internationalskeptics.com...&postcount=483 and http://www.internationalskeptics.com...&postcount=485 http://www.internationalskeptics.com...&postcount=482 is irrelevant. __________________ 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. Last edited by doronshadmi; 16th September 2020 at 06:20 AM.
 16th September 2020, 06:42 AM #487 Little 10 Toes Master Poster     Join Date: Nov 2006 Posts: 2,230 Originally Posted by jsfisher And just what would your definition be, then? By the way, my definition for non-finite 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". Originally Posted by doronshadmi I notice that you failed to answer jsfisher's questions. Is there a reason why? __________________ I'm an "intellectual giant, with access to wilkipedia [sic]" "I believe in some ways; communicating with afterlife is easier than communicating with me." -Tim4848 who said he would no longer post here, twice in fact, but he did.
 16th September 2020, 07:48 AM #488 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 Originally Posted by jsfisher By the way, my definition for non-finite sets was based on a definition for relative cardinality. Are you now adopting my definition of cardinality, too? No, I take a ride on it with you in order to find if it holds water. Originally Posted by jsfisher 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"? Originally Posted by jsfisher Since finite and non-finite are complementary terms, defining one automatically defines the other. 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. Originally Posted by jsfisher We are still all wondering, too, what you mean by "general set". 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. __________________ 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. Last edited by doronshadmi; 16th September 2020 at 08:03 AM.
 16th September 2020, 04:18 PM #489 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 22,920 Originally Posted by doronshadmi 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|)) If only that were a way to define a set. Sadly, it is not. __________________ 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 Last edited by jsfisher; 16th September 2020 at 04:40 PM. Reason: Typo
 16th September 2020, 04:39 PM #490 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 22,920 Originally Posted by doronshadmi Originally Posted by jsfisher By the way, my definition for non-finite sets was based on a definition for relative cardinality. Are you now adopting my definition of cardinality, too? No, I take a ride on it with you in order to find if it holds water. So, then, what are the definitions you a taking a ride upon? In your own words, please. Well, symbols, actually, are preferable. Quote: Originally Posted by jsfisher 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"? Originally Posted by jsfisher Since finite and non-finite are complementary terms, defining one automatically defines the other. In that case, finite sets' cardinality and X cardinality are complements, isn't it? No, "finite set" and "non-finite set" are complementary terms. A set that is not finite is non-finite, and a set that is not non-finite is finite. Complements. I have no idea what you meant by "complementary set", and I suspect you don't either. Quote: Originally Posted by jsfisher We are still all wondering, too, what you mean by "general set". 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. This, then, is just another term you inserted in to a sentence because it sounded important, but in reality it has no meaning whatsoever. You used an invented term; you got caught; so now you run away from it. How about not making up terms in the first place? Quote: Please look at the relevant one http://www.internationalskeptics.com...&postcount=483 and also at http://www.internationalskeptics.com...&postcount=485 . Thank you. As has been pointed out to you before, your posts may seem really good when you compose them, but they are in fact a confused collection of misinformation, made-up and meaningless terms, and non sequiturs. 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. __________________ 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
 16th September 2020, 04:43 PM #491 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 22,920 Who knows? This may become relevant at some point, too:∃I (Ø∈I ∧ ∀x (x∈I ⇒ x ⋃ {x} ∈ I)) N = {x∈I : ∀y ((Ø∈y ∧ ∀z ((z∈y ⇒ z ⋃ {z} ∈ y))) ⇒ x∈y)} __________________ 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
 17th September 2020, 12:06 AM #492 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 Originally Posted by jsfisher No, "finite set" and "non-finite set" are complementary terms. A set that is not finite is non-finite, and a set that is not non-finite is finite. Complements. In other words, "finite set" is a complement of "non-finite set", and vice versa. Originally Posted by jsfisher I have no idea what you meant by "complementary set", and I suspect you don't either. Non-finite set is a complementary set of finite set, and vice versa. __________________ 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 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 Originally Posted by jsfisher If only that were a way to define a set. Sadly, it is not. 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 = { ∅, ↓ No-bijection {∅}, ↓ No-bijection { ∅, {∅} }, ↓ No-bijection { ∅, {∅} , {∅, {∅}} }, ↓ No-bijection { ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} }, ↓ No-bijection ... }``` 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 ... __________________ 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. Last edited by doronshadmi; 17th September 2020 at 01:23 AM.
 17th September 2020, 03:37 AM #494 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 22,920 Originally Posted by doronshadmi In other words, "finite set" is a complement of "non-finite set", and vice versa. No. The term "finite set" is the complement of the term "non-fnite set". __________________ 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
 17th September 2020, 03:37 AM #495 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 V = {v : (∅ ∈ V) ∧ (∀v ∈ V ( ( v ∪ { v } ) ∈ V )) ∧ (∀v ∈ V (|v| < |V|))} __________________ 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 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 Originally Posted by jsfisher No. The term "finite set" is the complement of the term "non-fnite set". A = "infinite set" ~A is (the complement of A) = "finite set" A = "finite set" ~A is (the complement of A) = "infinite set" The above holds only in case of excluded middle framework. __________________ 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. Last edited by doronshadmi; 17th September 2020 at 04:08 AM.
 17th September 2020, 04:33 AM #497 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 22,920 Originally Posted by doronshadmi Ok, lets replace iff by if ... then. Still not a set definition. You could not do this before; why to you suddenly believe you can do it now? __________________ 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
 17th September 2020, 04:34 AM #498 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 22,920 Originally Posted by doronshadmi V = {v : (∅ ∈ V) ∧ (∀v ∈ V ( ( v ∪ { v } ) ∈ V )) ∧ (∀v ∈ V (|v| < |V|))} Did you not notice you are defining the set V in terms of itself? ETA: ...not to mention the symbolic gibberish you've included nor the reference to cardinality. __________________ 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 Last edited by jsfisher; 17th September 2020 at 04:39 AM.
 17th September 2020, 04:37 AM #499 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 22,920 Originally Posted by doronshadmi A = "infinite set" ~A is (the complement of A) = "finite set" No, being a finite set is a property.A is a finite setThe complement of that property would then be:A is not a finite setand that is equivalent to:A is a non-finite set __________________ 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
 17th September 2020, 05:22 AM #500 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 Originally Posted by jsfisher That would be back when I defined a set X to be non-finite when |N| <= |X| (where N is the minimal set satisfying the requirements of the Axiom of Infinity). Since finite and non-finite are complementary terms, defining one automatically defines the other. Since N is the minimal set satisfying the requirements of the Axiom of Infinity, does it mean that it is at least the set of all sets that are not non-finite sets, which are the members of the von Neumann set of ordinals? __________________ 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. Last edited by doronshadmi; 17th September 2020 at 05:29 AM.
 17th September 2020, 05:39 AM #501 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 delete ... __________________ 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. Last edited by doronshadmi; 17th September 2020 at 07:03 AM.
 17th September 2020, 06:53 AM #502 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 22,920 Originally Posted by doronshadmi Since N is the minimal set satisfying the requirements of the Axiom of Infinity, does it mean that it is at least the set of all sets that are not non-finite sets, which are the members of the von Neumann set of ordinals? "[T]he set of all sets that are not non-finite 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. __________________ 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
 17th September 2020, 07:03 AM #503 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 V = { v : (v is not non-finite member of von Neumann set of ordinals) } Here are some of the members of V: Code: ```V = { ∅, ↓ No-bijection {∅}, ↓ No-bijection { ∅, {∅} }, ↓ No-bijection { ∅, {∅} , {∅, {∅}} }, ↓ No-bijection { ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} }, ↓ No-bijection ... }``` where ∀v ∈ V ( ( v ∪ { v } ) ∈ V ) There is no-bijection 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. __________________ 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. Last edited by doronshadmi; 17th September 2020 at 07:16 AM.
 17th September 2020, 07:13 AM #504 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 In other words, Since |V| is not formally established, set X can't be non-finite since |V| <= |X| (where V is the minimal set satisfying the requirements of the Axiom of Infinity) is not well defined, thanks to ∀. __________________ 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. Last edited by doronshadmi; 17th September 2020 at 07:20 AM.
 17th September 2020, 07:34 AM #505 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 22,920 Originally Posted by doronshadmi V = { v : (v is not non-finite member of von Neumann set of ordinals) } (1) The ordinals defined by von Neumann do not form a set. (2) You haven't defined what you mean by "non-finite" 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. __________________ 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
 17th September 2020, 07:47 AM #506 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 22,920 Originally Posted by doronshadmi In other words, Since |V| is not formally established, set X can't be non-finite since |V| <= |X| (where V is the minimal set satisfying the requirements of the Axiom of Infinity) is not well defined, thanks to ∀. 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 well-defined. 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. __________________ 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, 01:27 AM #507 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 Originally Posted by jsfisher (1) The ordinals defined by von Neumann do not form a set. V = { v : (v is not non-finite 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 non-finite 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 no-bijection from v to v∪{v} ∧ ∀ (v ∪{v} ) ∈ V |v∪{v}| < |V| __________________ 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. Last edited by doronshadmi; 19th September 2020 at 02:26 AM.
 19th September 2020, 03:59 AM #508 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 Please ignore post #507. Originally Posted by jsfisher (1) The ordinals defined by von Neumann do not form a set. V = { v : (v is not non-finite 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 non-finite 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 no-bijection 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 = |∅|, | ↓ No-bijection | 1 = |{ ∅ }|, | ↓ No-bijection | 2 = |{ ∅, {∅} }|, | ↓ No-bijection | 3 = |{ ∅, {∅} , {∅, {∅}} }|, | ↓ No-bijection | 4 = |{ ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} }|, | ↓ No-bijection | v No-Bijection, order is irrelevant among all x members (where x=v or x=v∪{v}) and ∀(v∪{v}) ∈ V |v∪{v}| < |V| ... }``` which means that < (no-bijection) and = (bijection) are not gathered into <= in order to define |V| <= |X|, |V| <= |V|. Originally Posted by doronshadmi Originally Posted by jsfisher By the way, my definition for non-finite sets was based on a definition for relative cardinality. Are you now adopting my definition of cardinality, too? No, I take a ride on it with you in order to find if it holds water. Since |V| is not strictly established, the Cantorian notion of strict cardinality for set V, does not hold water. So once again, thank you ∀. __________________ 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. Last edited by doronshadmi; 19th September 2020 at 04:20 AM.
 19th September 2020, 04:00 AM #509 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 I like ∀ __________________ 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. Last edited by doronshadmi; 19th September 2020 at 04:02 AM.
 19th September 2020, 07:28 AM #510 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 22,920 Originally Posted by doronshadmi Please ignore post #507. V = { v : (v is not non-finite member of von Neumann set of ordinals) } = The set of all natural numbers in terms of sets. There is no such set as the von Neumann set of ordinals. You still haven't defined what you mean by non-finite (or finite, either will do) set. Quote: ...∀(v∪{ v}) ∈ V... Still don't understand quantification, do you. That is complete gibberish. __________________ 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, 07:47 AM #511 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 Originally Posted by jsfisher There is no such set as the von Neumann set of ordinals. V is not the von Neumann set of ordinals, bot only all of its members that are equivalent to all the natural numbers. Originally Posted by jsfisher Still don't understand quantification, do you. That is complete gibberish. 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) __________________ 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. Last edited by doronshadmi; 19th September 2020 at 08:23 AM.
 19th September 2020, 08:40 AM #512 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 V = { v : (v is not non-finite 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 non-finite 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 no-bijection 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 = |∅|, | ↓ No-bijection | 1 = |{ ∅ }|, | ↓ No-bijection | 2 = |{ ∅, {∅} }|, | ↓ No-bijection | 3 = |{ ∅, {∅} , {∅, {∅}} }|, | ↓ No-bijection | 4 = |{ ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} }|, | ↓ No-bijection | v No-Bijection, order is irrelevant among all x members (where x=v or x=v∪{v}) and ∀v(v∪{ v} ∈ V) ∧|v∪{v}| < |V| ... }``` which means that < (no-bijection) and = (bijection) are not gathered into <= in order to define |V| <= |X|, |V| <= |V|, in the first place. Since |V| is not strictly established, the Cantorian notion of strict cardinality for set V, does not hold water. So once again, thank you ∀. __________________ 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. Last edited by doronshadmi; 19th September 2020 at 08:45 AM.
 19th September 2020, 08:45 AM #513 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 22,920 Originally Posted by doronshadmi V is not the von Neumann set of ordinals, bot only all of its members that are equivalent to all the natural numbers. There is no such thing as the von Neumann set of ordinals, so any reference to its members is meaningless. Quote: If ∀(v∪{v}) ∈ V is a complete gibberish.... It is. 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 non-finite set cardinality no-bijectionConcise, 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. __________________ 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, 10:55 AM #514 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 Originally Posted by jsfisher There is no such thing as the von Neumann set of ordinals, so any reference to its members is meaningless. 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 non-finite set to be based on. Originally Posted by jsfisher And can we not agree that "|A|" is a notation that means "the cardinality of set A"? 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 "non-finite 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 non-finite sets). __________________ 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. Last edited by doronshadmi; 19th September 2020 at 10:58 AM.
 19th September 2020, 11:01 AM #515 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 22,920 Originally Posted by jsfisher ...define your terms. You need to explain what you mean by:finite or infinite or non-finite set cardinality no-bijection Here, if these are of any use to you, are mine:Notation: |A| means the cardinality of set A Notation: N means the minimal set satisfying the requirements of the Axiom of Infinity. If the Axiom of Infinity is given by ∃I (Ø∈I ∧ ∀x (x∈I ⇒ x ⋃ {x} ∈ I)) then N = {x∈I : ∀y ((Ø∈y ∧ ∀z ((z∈y ⇒ z ⋃ {z} ∈ y))) ⇒ x∈y)} Definition: |A| ≤ |B| means there is an injection from A to B. For completeness:|A| = |B| means |A| ≤ |B| ∧ |B| ≤ |A| |A| < |B| means ¬(|B| ≤ |A|) Definition: A is an infinite set means |N| ≤ |A| Definition: No-bijection has no meaning of which I am aware. __________________ 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, 11:09 AM #516 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 22,920 Originally Posted by doronshadmi 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 non-finite set to be based on. 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: A is a finite set iff |A| is the cardinality of A. So, by definition, you will be restricting cardinality to only finite sets. You are building in a restriction a priori in order to get your desired conclusion. Rather self-serving, and not particularly useful. Nonetheless, what then is your definition for cardinality that self-limits somehow (without any reference to finite sets, since that would be circular)? __________________ 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, 11:15 AM #517 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 Syntactic correction: V = { v : (v is not non-finite 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 non-finite 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 no-bijection 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 = |∅|, | ↓ No-bijection | 1 = |{ ∅ }|, | ↓ No-bijection | 2 = |{ ∅, {∅} }|, | ↓ No-bijection | 3 = |{ ∅, {∅} , {∅, {∅}} }|, | ↓ No-bijection | 4 = |{ ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} }|, | ↓ No-bijection | v No-Bijection, order is irrelevant among all x members (where x=v or x=v∪{v}) and ∀v(v∪{v} ∈ V ∧|v∪{v}| < |V|). ... }``` which means that < (no-bijection) and = (bijection) are not gathered into <= in order to define |V| <= |X|, |V| <= |V|, in the first place. Since |V| is not strictly established, the Cantorian notion of strict cardinality for set V, does not hold water. So once again, thank you ∀ __________________ 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. Last edited by doronshadmi; 19th September 2020 at 11:31 AM.
 19th September 2020, 11:24 AM #518 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 Originally Posted by jsfisher Definition: No-bijection has no meaning of which I am aware. 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. __________________ 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. Last edited by doronshadmi; 19th September 2020 at 11:26 AM.
 19th September 2020, 11:29 AM #519 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 22,920 Originally Posted by doronshadmi 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. No one can "be aware of it" until you tell us what you mean by "no-bijection". 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 "no-bijection" is a mystery. Are you unable to tell us what you mean? Use your words. Examples are not definitions. __________________ 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 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 13,243 Originally Posted by jsfisher No one can "be aware of it" until you tell us ... 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. Last edited by doronshadmi; 19th September 2020 at 11:49 AM.

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