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12th December 2014, 08:53 AM  #281 
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The beginning of my neutral monist view thread can be seen in http://www.sciencechatforum.com/view...p?f=51&t=27823 .

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12th December 2014, 09:01 AM  #282 
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RickM Whenever you find that you are on the side of the majority, it is time to reform  (or pause and reflect). Mark Twain Use what language you will, you can never say anything but what you are. Ralph Waldo Emerson 

12th December 2014, 09:34 AM  #283 
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It is head AND heart, in case that you have missed it.
Please reply also to the first part of http://www.internationalskeptics.com...&postcount=280 (including its links). Also please reply to http://www.internationalskeptics.com...&postcount=277 and http://www.internationalskeptics.com...&postcount=281. Thank you, RickM. 
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As long as Comparison is impossible because of the imbalance of one's mind, 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 ) 

12th December 2014, 10:56 AM  #284 
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Hi Doron, Thanks for the offer, but I’m afraid my mathematical achievements if life are limited to two semesters of calculus and two semesters of math department statistics, so I think I’ll leave this one to the experts. I must admit, though, I’ve been following this thread for a few weeks now with some amusement, but is the storyline ever going to lead to some bedroom activity? 
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RickM Whenever you find that you are on the side of the majority, it is time to reform  (or pause and reflect). Mark Twain Use what language you will, you can never say anything but what you are. Ralph Waldo Emerson 

12th December 2014, 01:11 PM  #285 
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As long as Comparison is impossible because of the imbalance of one's mind, 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 ) 

14th December 2014, 09:57 AM  #286 
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"All is needed (and it is essential to my definitions) is to understand the actuality beyond the description, for example: Nothing is actually"  Doron Shadmi "But this means you actually have nothing."  Realpaladin  

14th December 2014, 02:43 PM  #287 
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"So let it be written. So let it be done." 

15th December 2014, 04:33 AM  #288 
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As long as Comparison is impossible because of the imbalance of one's mind, 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 ) 

20th December 2014, 01:23 PM  #289 
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"So this is Christmas, and what have you done? Another year over, a new one just begun..."
Doron, reflect on the fact that you have not managed to keep *any* new poster for more than a few days... 
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"All is needed (and it is essential to my definitions) is to understand the actuality beyond the description, for example: Nothing is actually"  Doron Shadmi "But this means you actually have nothing."  Realpaladin  

23rd December 2014, 08:02 AM  #290 
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Awww snap! Doron, seems like you have dawdled too long...
http://www.wired.com/2014/12/mathema...primenumbers/ Not only a new major discovery on primes, but also some words like 'infinity'. You were just too slow, Mr. Shadmi. 
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"All is needed (and it is essential to my definitions) is to understand the actuality beyond the description, for example: Nothing is actually"  Doron Shadmi "But this means you actually have nothing."  Realpaladin  

22nd April 2015, 12:22 AM  #291 
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If we use order and cardinality as significant factors of the mathematical concept of limit, then:
Given two numbers a and b such that a < b and given ordered sets S and K of numbers from a to b (where b is the limit): 1) If S=0, then S members can't close the gap between a and b. 2) If S>0, then S members can close the gap between a and b, but if K is a proper subset of S such that K<S AND k_{1}=s_{1}=a, then K members can't close the gap between a and b. According to the definition (2), no amount of K members completely covers S. 
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22nd April 2015, 04:12 AM  #292 
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Let's not. Let's just leave the welldefined meaning of 'limit' alone. If you feel compelled to introduce something new and different, start with a new and different definition for it and some new and different name. Then we can see what new and different utility your new and different concept might have. 
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24th April 2015, 10:44 PM  #293 
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http://www.internationalskeptics.com...&postcount=291 deals exactly with the failure of the attempt to define a mathematical concept like limit, by simply ignore the linkage among order and cardinality.
There is no need for new definitions here since cardinality and order are welldefined in this case. So if you have some detailed reply about http://www.internationalskeptics.com...7#post10613587 content, then please air your view. Moreover let's be more accurate about (2): 2) If S>0, then S members can close the gap between a and b, but if K is a proper subset of S such that K<S AND k_{1}=s_{1}=a AND K and S members are the same from a forward, then K members can't close the gap between a and b. According to the definition (2), no member of K is b, so no amount of K members completely covers S. 
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25th April 2015, 05:35 AM  #294 
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There is no "failure of the attempt to define a mathematical concept like limit." Limit is, in fact, defined very precisely and very completely.
Quote:
Moreover, whatever linkage you allege to exist you haven't shown. First things first: You would need to define with some mathematical detail what exactly you mean by this linkage to which you refer. History suggests that will never happen. Meanwhile, the meaning of limit remains unscathed by your onslaught. 
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25th April 2015, 02:01 PM  #295 
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As long as Comparison is impossible because of the imbalance of one's mind, 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 ) 

25th April 2015, 02:07 PM  #296 
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You seemed to have left out something. Where is 'limit' anywhere in this?
Remember, you started this current arc with "If we use order and cardinality as significant factors of the mathematical concept of limit..." You haven't shown at all what that means. Instead, you are just asserting conclusions. 
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25th April 2015, 02:08 PM  #297 
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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. 

25th April 2015, 02:16 PM  #298 
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As long as Comparison is impossible because of the imbalance of one's mind, 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 ) 

25th April 2015, 02:39 PM  #299 
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As long as Comparison is impossible because of the imbalance of one's mind, 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 ) 

25th April 2015, 02:54 PM  #300 
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Is it now? You defined b to be a number and stipulated it to be greater than another number, a. So, it is to be a limit as well....
The limit of what? You understand, too, don't you, that having b as the limit of something isn't in any way dealing with the definition of 'limit' and its ties (imaginary or otherwise) to order and cardinality.
Quote:
But, even it you did, that would get you no where near tying limit to order and cardinality. 
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25th April 2015, 02:56 PM  #301 
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25th April 2015, 03:05 PM  #302 
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25th April 2015, 11:42 PM  #303 
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Already done in http://www.internationalskeptics.com...&postcount=277.

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26th April 2015, 04:47 AM  #304 
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Really? Well, since it isn't apparent you defined any terms or proved anything in that post from December, perhaps as a courtesy to the readers of today you would be so kind as to make a new post in which you define 'convergent series' and define 'limit' as you are using the terms.
Nothing else; just the definitions. That way we won't miss them buried among other clutter. You see, one of the problems is that given the meaning of 'convergent series' in Mathematics, having sets where "their members are used as convergent series" is gibberish. So, perhaps you could tell us what you really meant, then we can help translate it to something that isn't gibberish. (Just so you know, I'm expecting 'sequence' will be an important addition in untangling your word salad.) 
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26th April 2015, 06:56 AM  #305 
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Well, there is no problem to understand that there is a set that its members are used as the numbers of a given sequence, for example:
The members of set {0.09,0.009,0.9,0.0009...} (where order does not matter) are used as the numbers of sequence (0.9,0.09,0.009,0.0009,...) (where order matters) that its sum (also known as convergent series) approaches a given number, called the limit of that sequence. In case that S={0.09,0.009,0.9,0.0009...} and K is a proper subset of S such that K<S, that convergent series (which is based on the added K members) < the limit of the sequence that its numbers are S members. K<S is clearly shown in terms of the tower of power as given in http://www.internationalskeptics.com...&postcount=277. So according to what is written above, there is no problem to understand the following: b is the limit in the following definition: Given two numbers a and b such that a < b AND given sets S and K that their members are used as convergent series to b, if K is a proper subset of S such that K<S, then the sum of K members < b.  If we need N+1=N trick in order to conclude that 0.9_{10}+0.09_{10}+0.009_{10}+...=1 (as clearly shown in http://www.internationalskeptics.com...&postcount=277) then it is obvious that the sum of K members < b. 
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26th April 2015, 07:52 AM  #306 
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Ah! So after my hint and some quality time with Google, you are beginning to understand that 'sequence' may have been the term you wanted.
Quote:
For that matter, your whole set diversion suffers in that sequence elements need not be unique within the sequence.
Quote:
For a convergent series, the series is a single value.
Quote:
If you want a sequence at the base of all this, then make S a sequence. And make K a subsequence of S if you must. Also, you'll need to make sense of "the limit of the sequence that its numbers are S members." And then, when all that is resolved, you'll need to construct a proof for your conclusion. Proof, not mere assertion as you are wont to do. Finally, you'll need to show why the result (assuming you can establish the result) is at all interesting. It doesn't look like it will be interesting. 
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26th April 2015, 08:55 AM  #307 
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Thank you for your advice.
Since order does not matter, let S and K be mutlisets. The discussed elements are numbers. If the N+1=N trick is understood, then you immediately understand that this is a definition. It is not interesting if the noncomposed aspect of the tower of power line is excluded. If it is not excluded then one immediately understands the nonentropic nature of the discussed framework, which is essential for further development of living creatures like us. 
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26th April 2015, 09:07 AM  #308 
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You are just digging a hole. Put the shovel down. Your now multisets still do not determine the sequence. Take the short cut: Make S and K sequences.
Quote:
Quote:
Quote:

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26th April 2015, 09:27 AM  #309 
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If I wish to use also cardinality as a factor of the discussed subject, I can't avoid mutisets, unless length can equivalently be used here, what do you think?
"the limit of the sequence that its numbers are S members" is b and the added members of S (which are all numbers) are sum=b, and as a result gibberish is the best you can get. Already done very simply by using your "Y = X union {[1,1]}" as seen in http://www.internationalskeptics.com...&postcount=277. Of course you will not do it, after all your framework arbitrarily excludes the noncomposed aspect of the tower of power line. 
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As long as Comparison is impossible because of the imbalance of one's mind, 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 ) 

26th April 2015, 10:14 AM  #310 
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I think you need to keep in mind that you are focusing on sequences. You keep trying to force sets as a vehicle towards that aim, but they are unnecessary and inappropriate. If the properties you seem to demand cannot be expressed directly for sequences, then perhaps you should not be including sequences at all.
One thing you are not allowed to sidestep, too, is that a sequence is equivalent to a function over a totally ordered, countable domain. Keep that in mind as you wave your hands at cardinality.
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Once you can finalize all the stuff that leads up to your conclusion, only then can we begin to consider its proof.
Quote:

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26th April 2015, 10:32 PM  #311 
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26th April 2015, 10:49 PM  #312 
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Some important corrections (which correct also http://www.internationalskeptics.com...ostcount=277):
1) The order of the added numbers (where all numbers are positive) does not matter. 2) Number 0 does not contribute anything to the sum of the added numbers, and so is the case of multiple added numbers of the same value, the fact that they have the same value also does not contribute any important thing to the sum of the added numbers. 3) According to (1) and (2) the added numbers into a given series are actually reduced into unique members of nonempty sets. So we can omit the usage of sequences, and use only the unique members of sets that are all positive numbers. By simplifying the framework into the sum of added positive unique numbers, I discovered that K (where K<S) is not a proper subset of S because K has at least one member which is not a member of S. Here is an example: {0.9_{10}, 0.09_{10}, 0.009_{10}, 0.001_{10}} is some set with cardinality 4 that its added members provide 1 by onestep, and it is obvious that 0.001_{10} is not a member of {0.9_{10}, 0.09_{10}, 0.009_{10}, ...} with cardinality N that its added members provide 1 by onestep. So K<S and K is not a proper subset of S, yet it does not change the validity of my argument, which is as follow: K or S must have smallest numbers > 0, such that the smallest number of S < the smallest number of K (such that the number of places of the smallest number of K < the number of places of the smallest number of S) and without those smallest numbers > 0, the sum of S or K members < the given limit. If K is finite cardinality, the number of places of the smallest number of K is finite, otherwise it is infinite according to the cardinality of the set. So we actually do not need the comparison between K and S in terms of set and proper subset, yet for all cardinality, there is the smallest number > 0 that the number of its places = the cardinality of its set. The invariant existence of such smallest number > 0, proves the inability to define the smallest number > 0 for all sets, and we can conclude that no collection of added numbers has the noncomposed property of 1dimensional space because this property is stronger than the cardinality of all sets that their members are added along it from number a up to b, where a<b. 
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27th April 2015, 01:27 AM  #313 
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Some corrections to my previous post:
Instead of "So we actually do not need the comparison between K and S in terms of set and proper subset, yet for all cardinality, there is the smallest number > 0 that the number of its places = the cardinality of its set." it has to be "So we actually do not need the comparison between K and S in terms of set and proper subset, yet for all cardinality, there is the smallest number > 0 that the number of its places at the right side of the dismal point is finite or infinite (and in case that it is infinite, the number of places is according to the cardinality of its set)." For example: Exactly as 0.001_{10} has finite places at the right side of the dismal point in case of {0.001_{10}, 0.9_{10}, 0.09_{10}, 0.009_{10}}, so is the case about 0.000...1_{10} in case of {0.000...1_{10}, 0.9_{10}, 0.09_{10}, 0.009_{10}, ...}. If one has no problem to understand that b is the result of infinitely many added numbers, one also does not have problems to understand that there is a number of the form 0.000...1_{10} that has infinity many places at the right side of the dismal point that is > 0, that actually enables the members of {0.000...1_{10}, 0.9_{10}, 0.09_{10}, 0.009_{10}, ...} to be added up to b (even if 0 is a member of some set, it is obvious that it does not contribute anything to get result b). So, without of numbers of the form 0.000...1_{10} the result of the added members of sets with cardinality N actually < b. The same principle holds with sets that their cardinality > N, but in this case we can't symbolize their members, because any string of symbols is limited to cardinality N.  Instead of "The invariant existence of such smallest number > 0, proves the inability to define the smallest number > 0 for all sets, it has to be "The invariant existence of such smallest number > 0 for each set, proves the inability to define the smallest number > 0 of all sets that their members are added up to b from a to b, where a<b," 
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As long as Comparison is impossible because of the imbalance of one's mind, 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 ) 

27th April 2015, 04:24 AM  #314 
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The ".000..." is used as a N size place value keeper that is inaccessible to "...1" that is at R size.
The same principle holds among R<P(R)<P(P(R))< ... ad infinitum, where no number of added values is accessible to the noncomposed property of the line in itself, which is beyond collections. Also in the case of sets that their members converge to a given number, known as the limit, these members can't be symbolized beyond cardinality N. Here is the standard definition (http://en.wikipedia.org/wiki/Sequenc..._convergence):
Quote:
for all ε > 0, there exists a natural number N in N such that for all n ≥ N, a_{n}  L < ε. So a_{n}  L must be = 0 in order to conclude that the sum over a_{n} = L. For example, where is the rigorous proof that the sum over (0.9_{10}, 0.09_{10}, 0.009_{10},...) = L = 1? 
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As long as Comparison is impossible because of the imbalance of one's mind, 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 ) 

27th April 2015, 06:30 AM  #315 
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Post #277 is from four months ago. We've all moved on since then, including you. You introduced something new with its own set of misused terminology and obtuse word constructions. Are you abandoning that new, or just saving it for a link four months from now?
That sequence of posts from four months ago got you nowhere you wanted to go. You cannot pretend now it did, nor can you expect anyone here to reexperience the failure from that far back so you can get nowhere with it once again. If you'd like to start anew, well, then you need to start anew. No silly links to different flavors of word salad. 
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27th April 2015, 06:47 AM  #316 
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The solution to this question is by constructing a new kind of number that is based on the mathematical fact that R>N.
In the case of (0.9_{10}, 0.09_{10}, 0.009_{10}, ...) the complement of 0.999..._{10} to 1 is the number 0.000...1_{10}, where the ".000..." part is used as a N size place value keeper that is inaccessible to "...1" that is at R size. "...1" at R size actually closes the gap between 0.999..._{10} and 1, and without it 0.999..._{10} = 1 is no more than an assertion. 
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As long as Comparison is impossible because of the imbalance of one's mind, 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 ) 

27th April 2015, 06:58 AM  #317 
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27th April 2015, 07:10 AM  #318 
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Please follow it from http://www.internationalskeptics.com...&postcount=314.

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As long as Comparison is impossible because of the imbalance of one's mind, 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 ) 

27th April 2015, 07:13 AM  #319 
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As long as Comparison is impossible because of the imbalance of one's mind, 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 ) 

27th April 2015, 07:43 AM  #320 
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