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27th January 2018, 09:37 PM  #2921 
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Unless redundancy is not ignored, for example:
By the ,so called, "established mathematics" [A] and [A,A] are addressed by the distinct pairs {(A,1),(A,2)} exactly because each pair defines different redundancy degree. A = {1,2,3,...} (the set of all natural numbers). If redundancy degree > 1, in case of, for example [A,A] there can be more than 1 degrees of freedom of the mapping between the As copies, as addressed in http://www.internationalskeptics.com...postcount=2864. A = A is simply the particular case of the pair {(A,1)}, so nothing is fundamental here. 
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27th January 2018, 09:48 PM  #2922 
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See? 2 may not be a member of the set containing 2.

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28th January 2018, 02:25 AM  #2923 
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As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help.  If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) 

28th January 2018, 07:57 AM  #2924 
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28th January 2018, 08:06 AM  #2925 
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jsfisher, why do you take only a part of my reply to your wrong argument in http://www.internationalskeptics.com...postcount=2922 about 2 as if it is not a member of a given multiset ?

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As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help.  If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) 

28th January 2018, 08:29 AM  #2926 
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My statement about 2 may not being a member of {2} was a bit of mockery. It is a clearly bogus claim you'd made in the past. In its larger form, you had claimed although 2 was in the set of even numbers, and in the set of prime numbers, and in the set of integers, and in so many other sets it either wasn't the same 2 in each case or it couldn't be in all of them simultaneously.
You continue this conceptual failure with your understanding of multisets, but that aside for a moment, the multisets [2, 2], [2, 2, 2], and [2] are all different, yet you continue to fly off on a tangent about the rest of us saying otherwise. Now, as for the reason those multisets are all different, it is simply a matter of definition for multisets. By definition, distinct members may appear more than once in a multiset. It has nothing to do with the tortured vocabulary invention you keep trying to force on us. I have found it best to limit your posts up to just the first bit of wrongness, so in responding, I clipped the quote at such a point. ETA: By the way, my post you linked to made no reference to multisets. 
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28th January 2018, 08:40 AM  #2927 
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28th January 2018, 08:52 AM  #2928 
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My phrasing was ambiguous and probably misleading for some, especially those for whom English was not their first language. I apologize.
The meaning I meant to convey, the meaning as originally expressed by Doronshadmi when he made the claim was: 2 is not necessarily a member of {2}. It could be a different 2, or maybe 2 was busy off doing something else. 
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28th January 2018, 02:03 PM  #2929 
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In other words, you wrote a reply that has nothing to do with http://www.internationalskeptics.com...postcount=2921, or in other words, another red herring of yours.
Exactly, they have different redundancy degrees among members' copies. And this is exactly the reason of why they have more than one degree of freedom, in case that at least two copies of infinite sets are mapped, as addressed in http://www.internationalskeptics.com...postcount=2864. Worng jsfisher, you are the one that forces the notion of A=A, which is the particular case of {(A,1)}, as if it the only possibility in order to do Math. 
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28th January 2018, 02:31 PM  #2930 
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No, 2 is an element. If this element is a member of some set, then this case may be written as {2} , {1,2,3} , {2, 4, 6, ...} , {{2}} etc. ad infinitum.
The simple fact in this case is that nothing forces 2 to be a member of some nonempty set, since 2 can also be taken as its own mathematical universe even if it is taken in terms of a "pure" set like {{},{{}}} (Von Neumann constucation) or {{{}}} (Zeremalo costruction). Moreover, if 2 is not taken in terms of a "pure" set, then 2 is not a member of any framework that is defined in terms of "pure" sets.  So, jsfisher your red herring is not going to save you from your own artificial restrictions. 
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28th January 2018, 02:39 PM  #2931 
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I do not use the conventional terms, and by doing so I discover the different degrees of freedom of the mapping between at least two copies of a given infinite set under a given multiset, as done here.
Generally, Mathematics is much more richer than any particular view, including my nonconventional view of the considered subject. 
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28th January 2018, 03:14 PM  #2932 
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28th January 2018, 03:32 PM  #2933 
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If you really think that, then you are truly missing the meaning of your own words. You see, phiwum wrote:
This is a perfectly correct statement. Mathematics doesn't need a Xerox machine. No matter how it is expressed, a single 2 is sufficient for all its needs. You then contradicted phiwum's statement with: (By the way, you might want to look up the meaning of that word. Its really meaning is rather amusing in light of how you continue to abuse it.) You continue to labor under a conceptual block that prevents you from accepting that a multiset can have, for example, 2 as a member of it 17 times without the need for any copies. And this all ties quite nicely to comments you have made in the past about 2 and its possible nonmembership in {2}. My post was completely topical and stood as a reminder to phiwum that you think copies are necessary. So, no, no red herrings were used in the making of my posts. 
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28th January 2018, 03:35 PM  #2934 
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Out of curiosity, Doronshadmi, have you figured out yet you do not need to appeal to multisets to define mappings from N to N? If you wanted to discuss mappings, you could have gone directly there.

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28th January 2018, 11:01 PM  #2935 
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As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help.  If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) 

28th January 2018, 11:38 PM  #2936 
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You jsfisher, and phiwum are simply closed under the notion of A=A that is simply the particular case of {(A,1)}.
It has more than one meaning, so you have no case. It has 17 copies, but since your notion is restricted to {(A,1)}, you are not aware of it. Also http://www.internationalskeptics.com...postcount=2930 is not closed under your convectional notions. It is a red herring, and in both paths your restricted notions can't catch the red herring nor the fox.  Generally, Mathematics is much more richer than any particular view, including my nonconventional view of the considered subjects. 
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As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help.  If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) 

29th January 2018, 01:07 AM  #2937 
Penultimate Amazing
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Out of curiosity, jsfisher, have you figured out that multiset is generalization of the concept of set, such that the mappings from N to N (such that redundancy is ignored) is only the particular case of redundancy = 1 ({(N,1)}). If you wanted to discuss mappings you can't cover the considered subject by {(N,1)} particular case.
Jsfisher, under {(N,2)} ([N,N]) bijection is not the only possible mapping, but you are unaware of it since your notions of the considered subject are restricted to {(N,1)}. 
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As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help.  If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) 

29th January 2018, 03:52 AM  #2938 
Penultimate Amazing
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To those who follow the last discussion, jsfisher and phiwum understand, for example [A,A] in terms of {(A,1),(A,1)} and not in terms of {(A,2)} as I do.
In other words, their notion is restricted only to (A,1). 
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As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help.  If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) 

29th January 2018, 05:18 AM  #2939 
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Phiwum and I understand multisets in terms of how the concept is defined. You, Doronshadmi, don't bother with such formalities.
By the way, will you as some point tell us all what you had in mind with this (A,1) and (A,2) notation or will that be just another mystery of doronetics? 
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29th January 2018, 05:47 AM  #2940 
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Your understanding of, for example, [A,A] is understood in terms of {(A,1),(A,1)}.
Wrong, you and Phiwum don't bother to comprehend the formality of [A,A] in terms of {(A,2)}. No mystery. (A,2) has redundancy degree > 1, where (A,1) has redundancy degree = 1. Under redundancy degree > 1 two copies of infinite sets have more than 1 degrees of freedom of the mapping between them, as very simply addressed in http://www.internationalskeptics.com...postcount=2864. {(A,2)} is a mystery if observed in terms of {(A,1),(A,1)}. 
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As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help.  If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) 

29th January 2018, 06:03 AM  #2941 
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I think (A,1) represents the first A in the (apparently ordered) (multi)set, while (A,2) represents the second one. But I don't know which one of them is the original A and which one is the copy, or if the original A is stored in a safe place somewhere else, so nobody can put it in a set where it doesn't belong.

29th January 2018, 06:54 AM  #2942 
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As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help.  If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) 

29th January 2018, 07:31 AM  #2943 
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If (x,y) is meant as an ordered pair where x is an element and y is whole number representing multiplicity, then, sure, that is a way to embed the multiset concept into set theory. (It isn't perfect, but it is workable.)
E.g. { (2,3), (3,1), (5,1) } would be the multiset of the prime factors of 120 under this model. Doronshadmi, since no one has argued otherwise, why do you continue to argue as if we had? You have been doing that a lot of late, arguing against positions nobody has taken. 
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29th January 2018, 09:03 AM  #2944 
Penultimate Amazing
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If (x,y) is meant as an ordered pair where x is an element and y is any cardinal number > 0 representing redundancy among x, , then, sure, that is a way to embed elements in to multisets, where a multiset is a generalization of the concept of set (a set is simply a multiset with redundancy = 1).
jsfisher, you are still missing the notion of redundancy as defined, for example, in {(2,3)}, because your notion is restricted only to redundancy = 1 (defined in this case as {(2,1),(2,1),(2,1)}). Your restriction is in terms of notions and not in terms notations, exactly because whole your mathematical understanding is reduced into redundancy = 1. Here are your own words: and this is exactly the notion of redundancy = 1 (defined in this case as {(2,1),(2,1),(2,1)}). So jsfisher, until this vary moment you have no clue what is my argument in http://www.internationalskeptics.com...postcount=2864. Yet you continue your off topic replies even if you have no clue about my argument, which is not restricted to redundancy = 1. 
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As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help.  If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) 

29th January 2018, 09:42 AM  #2945 
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Nonsense. The multiset [A,A] is obviously represented in set theory as {(A,2)}. Duh.
But A had no referent here, unless you mean that for all sets A this is the case. Which is, of course, true. Thus, in [A,A], the single set A is represented twice, but it is nonetheless the same A. 
29th January 2018, 10:24 AM  #2946 
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Why do you say that? Other than your insistence on using the wrong term (multiplicity is the correct one), [2, 2, 2] and { (2,3) } are identical constructs (in terms of the context of this thread).
It is not clear why you brought up { (2,1), (2,1), (2,1) }. Just two more instances of you arguing against things no one else claimed. 
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29th January 2018, 09:19 PM  #2947 
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More of the same is not just the same.
A = {1,2,3,...} (the set of all natural numbers) [A,A] is more of the same, such that bijection is only one case among the As under [A,A], as very simply addressed in http://www.internationalskeptics.com...postcount=2864. 
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30th January 2018, 05:57 AM  #2948 
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That is an odd way to express it, but no one has argued to the contrary. Will all of your posts be striking out against strawmen?
Quote:
Mappings, on the other hand, require three things: a domain, a codomain, and a functional relationship from domain to codomain. Please stop conflating the two. 
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30th January 2018, 08:08 AM  #2949 
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I agree with you.
[A, A] is an example of more of the same, such that redundancy > 1. Since by convectional mathematics multiplicity is not understood in terms of redundancy, one wrongly understands, for example, {(A,2)} in terms of {(A,1),(A,1)} without being aware of his\her failure. Here is a concrete example: Since jsfisher is unaware of the redundancy in [2,2,2] he wrongly understands {(2,3)} (which is actually an expression that does not ignore the redundancy), in terms of {(2,1),(2,1),(2,1)} which ignores the redundancy in [2,2,2], where this reply very simply demonstrates jsfisher's nonawareness of the considered subject. If mappings is considered among the As under [A, A] such that A is an infinite set, then since [A, A] is more of the same (where redundancy is not ignored) bijection is only one case among the As under [A,A], as very simply addressed in http://www.internationalskeptics.com...postcount=2864. In other words, jsfisher, you continue your off topic replies, exactly because you understand {(A,2)} (redundancy is not ignored) in terms of {(A,1),(A,1)} (redundancy is ignored). 
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As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help.  If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) 

30th January 2018, 09:02 AM  #2950 
Penultimate Amazing
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As for phiwum, he wrongly thinks that I claim that [2,2] has two different members:
Originally Posted by phiwum
My argument simply does not ignore the redundancy > 1 in a given multiset, and in case that the redundancy > 1 is not ignored among the mappings between redundant infinite members, bijection is not necessarily the one and only one mapping, as very simply addressed in http://www.internationalskeptics.com...postcount=2864. 
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As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help.  If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) 

30th January 2018, 10:12 AM  #2951 
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30th January 2018, 02:02 PM  #2952 
Penultimate Amazing
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Mathematics is not a property of anyone or any group of people not in the past, not now and not in the future.
Your desperate quest to close it under dogmas actually transforms it into a religion. A more fruitful way is to reply in details to http://www.internationalskeptics.com...hp?p=12165528& , but currently it seems that your best reply is "you do not get to redefine Mathematics". 
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As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help.  If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) 

30th January 2018, 04:39 PM  #2953 
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30th January 2018, 07:17 PM  #2954 
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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. 

31st January 2018, 07:22 AM  #2955 
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As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help.  If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) 

31st January 2018, 09:45 PM  #2956 
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How about instead of making up words with your own private definitions, you work on using the correct terms?

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1st February 2018, 01:48 AM  #2957  
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Let's generalize the mathematical notion of strict membership (which is a particular case of Fuzzy logic).
Let E be a placeholder of any entity. Let # be a placeholder of any strict membership. Let (E,#) be a generalization of any strict membership of any entity. Multiset is a generalization of the concept of set. Any entity with strict membership 0 ( notated as (E,0) ), can't be defined but as E (it is strictly not a member of any multiset). Let [] (empty multiset) be a multiset of any entity with strict membership 0 ( notated as (E,0) ). Any entity with strict membership 1 ( notated as (E,1), can be defined as E = E (it is a strict and unique member of any multiset). Convectional mathematics is the mathematical framework of strict membership of the form (E,1), and it needs the axiomatic method in order to define [] in terms of (E,1), instead of simply deduce it in terms of strict membership 0 ( notated as (E,0) ). Let A = [1,2,3,… etc.] Let S = [A,A] S is a (E,2) mathematical framework, which enables mapping among infinite multisets that is not restricted to (E,1) mathematical framework, such that bijection is only a particular case among As under S, exactly because A is a strict but nonunique member of S (as very simply addressed in http://www.internationalskeptics.com...postcount=2864). Exactly as one can't claim that strict membership is the one and only one possible membership (since it is simply a particular case of fuzzy logic), one also can't claim that (E,1) is the one and only possible strict membership. 
 Mathematics is indeed deeper than primes. 

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As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help.  If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) 

6th February 2018, 08:54 AM  #2958 
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Here is some quote taken from Russell's article about the axiom of choice :
Quote:
The notion that defines a set as a collection of distinct members, is actually based on an act of selection. In order to be aware of it, all one needs is to use the notion of membership 0, such that exactly one element (out of infinitely many totally isolated and identical elements (for example: infinitely many identical socks that are not members of any set)) is chosen as a member of a given set (in this case the used notion is membership 1, such that, for example, {2} , {3,2} , {1,2,3,...} etc. are all cases of membership 1). {2,2} , {2,3,2} , {1,2,2,2,2,2,2,2,3,...} etc. are all examples of membership > 1, such that more than one element (out of infinitely many totally isolated and identical elements (for example: infinitely many identical socks that are not members of any set)) are chosen as members of a given multiset. If multiset is defined as a generalization of the concept of set, then: [] is the case that no element (out of infinitely many totally isolated and identical elements (for example: infinitely many identical socks that are not members of any set)) is chosen as a member of a given multiset. [2] is the case that one element (out of infinitely many totally isolated and identical elements (for example: infinitely many identical socks that are not members of any set)) is chosen as a member of a given multiset. [2,2] is the case that two elements (out of infinitely many totally isolated and identical elements (for example: infinitely many identical socks that are not members of any set)) are chosen as members of a given multiset. etc. ... ad infinitum. Convectional mathematics is simply the particular case of choosing exactly one element (out of infinitely many totally isolated and identical elements (for example: infinitely many yellow identical socks that are not members of any multiset, infinitely many rad identical socks that are not members of any multiset etc. ad infinitum)) such that infinitely many membership 1 multisets have no more than one yellow sock and one rad sock. 
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As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help.  If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) 

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