My Math Forum Full Paper - Conjecture 9-14-11

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February 20th, 2012, 11:39 AM   #51
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Re: Full Paper - Conjecture 9-14-11

Quote:
 Originally Posted by ChristopherM What? (I honestly have no idea what you are saying or why it is relevant to my work)
It directly addresses your oft-quoted "for any element 'j' in RI, 'j' is not a subset of the union of all 'g' in RI such that 'g' is not equal to 'j'." question.

Quote:
 Originally Posted by ChristopherM Function g works on finite strings too, and it is very easy to prove that for any finite string r, r is not a subset of the union of g(q) for ALL finite strings q where the cardinality of g(q) is equal to or less than g(r). The same holds for any infinitely large string r, where g(r) is not a subset of the union of all g(q) for ALL q where the cardinality of g(q) is equal to or less than g(r).
Perhaps the question you asked wasn't the question you intended. The answer to the one actually asked is "false".

 February 20th, 2012, 04:17 PM #52 Newbie   Joined: Feb 2012 Posts: 15 Thanks: 0 Re: Full Paper - Conjecture 9-14-11 I still have no idea what CR is talking about and when it comes to C-sheaves I agree with the definition of N but you'll have to explain how that is a definition of R or further what that definition of R has to do with my set RI. Proof that for any element 'j' of RI, 'j' is not a subset of the union of all 'g' in RI such that 'g' is not equal to 'j' is as follows: If I union two elements of RI together, then there are only two elements of RI that are a subset of the union (ie... the two sets that were unioned). If I union three sets, then there are only three sets that are a subset, and so on. If I union N elements of RI together, then there will be N elements of RI that are a subset of the union (again, the N sets that were unioned, not the N sets that were unioned and whatever additional ones you want to toss in there). CR, are you comprehending that g(0) was never part of the proof, that 10 could never be an element of a g(r) in RI, and that 0.01 and 0.11 could never coexist in an element of RI? Once two elements of RI differ they will differ forever more. Not many infinitely large sets of dyadic rationals constitute elements of RI as compared to all possible infinitely large sets of dyadic rationals. If I take the sequence/set A = {0.1, 0.01, 0.11, 0.001, 0.101, 0.011, 0.111, ... f}, then I get all the dyadic rationals in my set A and all elements of RI are subsets of the set A. If I consider two elements of RI, r = {0.01, 0.011, 0.0111, ...} and r' = {0.1, 0.11, 0.111, ...}, then which one becomes a subset of the sequence first? It would have to be r, as 0.01 appears before 0.11 in the sequence, 0.011 appears before 0.111, and so on, would it not? (This is a rhetorical question) It is a paradox that one can make C-sheaves argument and it is equally paradoxical that you can make my argument. Again, from the greatest of all paradoxes, "Both statements are equal." This is the definition of a point (a paradox). ... no using hyper-reals here (I do not believe in them).
February 20th, 2012, 05:21 PM   #53
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Re: Full Paper - Conjecture 9-14-11

Quote:
 Originally Posted by ChristopherM I still have no idea what CR is talking about and when it comes to C-sheaves I agree with the definition of N but you'll have to explain how that is a definition of R or further what that definition of R has to do with my set RI.
1. These are not definitions. My 'definition' of $\mathbb{N}$ uses $\mathbb{N}$ to index the union. Circular definitions suck.

That aside: do you accept the statement that $\mathbb{R}$ is equal to the union of open intervals $(-n,n) \subset \mathbb{R}$ for each $n \in \mathbb{N}$?

2. The reasoning is in the second half of that post. It's that stuff about how in every example (the two expressions within that post and the business with the $q^i$ from earlier) I am taking a union 'along' an infinite chain of nested sets that contains no maximal element.

The product of this infinite nested union, or 'least upper bound' of the $\{1,2,3...n]}$ in $\displaystyle \cup_{n \in \mathbb{N}} \{1,2,3...n\}$ is $\mathbb{N}$, but $\mathbb{N}$ is not equal to any $\{1,2,3...n\}$
The product of this infinite nested union, or 'least upper bound' of the $(-n,n)$ in $\displaystyle \cup_{n \in \mathbb{N}} (-n,n)$ is $\mathbb{R}$, but $\mathbb{R}$ is not equal to any $(-n,n)$
The product of this infinite nested union, or 'least upper bound' of the $g(q^i) \cap g(r)$ in $\displaystyle \cup_{i \in \mathbb{N}} \left( g(q^i)\cap g(r) \right)$ or alternatively $\displaystyle \cup_{q\in \mathbb{I}:q \neq r} \left( g(q)\cap g(r) \right)$ is $g(r)$, but $g(r)$ is not equal to any single $g(q) \cap g(r)$

3. Now go re-read my previous post and see if it makes sense to you now.

 February 20th, 2012, 05:37 PM #54 Newbie   Joined: Feb 2012 Posts: 17 Thanks: 0 Re: Full Paper - Conjecture 9-14-11 Footnote: I meant 'least upper bound' in the sense of the partial order relation $\subset$ on the class Set, not the order relation $<$ on $\mathbb{R}$
February 20th, 2012, 07:36 PM   #55
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Re: Full Paper - Conjecture 9-14-11

Quote:
 Originally Posted by ChristopherM Proof that for any element 'j' of RI, 'j' is not a subset of the union of all 'g' in RI such that 'g' is not equal to 'j' is as follows: If I union two elements of RI together, then there are only two elements of RI that are a subset of the union (ie... the two sets that were unioned). If I union three sets, then there are only three sets that are a subset, and so on. If I union N elements of RI together, then there will be N elements of RI that are a subset of the union (again, the N sets that were unioned, not the N sets that were unioned and whatever additional ones you want to toss in there).

Quote:
 Originally Posted by ChristopherM no using hyper-reals here (I do not believe in them).
No one here is.

 February 22nd, 2012, 07:13 PM #56 Newbie   Joined: Feb 2012 Posts: 15 Thanks: 0 Re: Full Paper - Conjecture 9-14-11 We are back to where we started then C-Sheaves. You feel statement #1 implies statement #2: 1) For each element rj in g(r), there exists a q such that rj is in g(q) implies 2) For all elements rj in g(r), there exists a q in such that rj is in g(q) I say statement #3 implies statement #4 using equal logic: 3) For each element g(q), there exists an rj in g(r) such that rj is not an element of g(q) implies 4) For all elements g(q), there exists an rj in g(r) such that rj is not an element of g(q) Statement #5 is proof of statement number #4: 5) For any q and r in I where q is not equal to r, g(q) is not equal to g(r), g(q) is not a subset of g(r), and g(r) is not a subset of g(q) ... The axiom of union says nothing about having to have an enumerable sequence, so you can stop trying to form them. The least upper bound of RI is not g(.111...), as g(.111...) is actually an element of RI whereas N is not an element of N. Further, if you are unioning elements of RI together, the least upper bound is not g(r), but r. The axiom of choice implies the well ordering theorem in ZF set theory and RI is put into one-to-one correspondence with the powerset of integers, so RI is well ordered. If RI is well ordered, then the greatest element of RI would be g(.111...). If you approach g(.111...) but refuse to equal it using an enumerable sequence (why not use all elements of RI?), then again, I do not readily accept your logic. You will have to do better. CR, put down the bag of potato chips, step away from the computer, and get some sunshine or something...
February 22nd, 2012, 11:50 PM   #57
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Re: Full Paper - Conjecture 9-14-11

Quote:
 Originally Posted by CRGreathouse
Amen.

ChristopherM: I give up.

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### lets see if the above conjectures are correct folding paper calculus

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