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Mike Kelly
science forum Guru Wannabe

Joined: 30 Mar 2006
Posts: 119

Posted: Fri Jul 21, 2006 9:13 am    Post subject: Re: The List of All Lists

Russell Easterly wrote:
 Quote: "Patricia Shanahan" wrote in message news:y5Zvg.2660\$157.889@newsread3.news.pas.earthlink.net... Russell Easterly wrote: I will prove there exists a largest "effective" natural number. This is related to the Well Ordering the Reals thread, but I think you will find this proof unique. The list of all lists contains a largest "effective" natural number. Assume I am given a list of unique lists of natural numbers, A. By unique, I mean each list in A differs from every other list in A. Given two lists, A_i and A_j, we know these lists are unique and differ at some finite position, k. Define the effective number, k=F(i,j), as the first position where list A_i differs from list A_j. I can now find the largest effective number for a given j. For each j define maxj=max( F(i,j) for i=1 to j-1 ) as the largest effective number for all i less than j. Define maxn as the largest effective number. For each j, if maxj > maxn, let maxn = maxj. Even if every list in A is infinite, the largest effective number, maxn, will be finite. We can distinguish any two sets by only looking at the first maxn elements in the two lists. Suppose A_1 is N, and A_i, for every i>1, is N-{(i-2)}. A_1 = {0, 1, 2, 3, ... } A_2 = {1, 2, 3, 4, ... } A_3 = {0, 2, 3, 4, ... } A_4 = {0, 1, 3, 4, ... } etc. What is maxn? The last one. You haven't defined which one that is, but there must be one if every list differs from every other list.

Why?

--
mike.
Patricia Shanahan
science forum Guru Wannabe

Joined: 13 May 2005
Posts: 214

Posted: Fri Jul 21, 2006 10:10 am    Post subject: Re: The List of All Lists

Russell Easterly wrote:
....
 Quote: I proved in another thread that any non-empty proper subset of the set of all natural numbers that is transitive and well ordered must have a largest element. This means any inductive set that doesn't contain omega is finite. ....

Did you ever prove, rather than merely assume, that there is any
non-empty proper subset of the natural numbers that is transitive and
well-ordered?

If so, I missed it and would welcome a reference to the article.

Patricia
Patricia Shanahan
science forum Guru Wannabe

Joined: 13 May 2005
Posts: 214

Posted: Fri Jul 21, 2006 10:30 am    Post subject: Re: The List of All Lists

Russell Easterly wrote:
 Quote: "Patricia Shanahan" wrote in message news:y5Zvg.2660\$157.889@newsread3.news.pas.earthlink.net... Russell Easterly wrote: I will prove there exists a largest "effective" natural number. This is related to the Well Ordering the Reals thread, but I think you will find this proof unique. The list of all lists contains a largest "effective" natural number. Assume I am given a list of unique lists of natural numbers, A. By unique, I mean each list in A differs from every other list in A. Given two lists, A_i and A_j, we know these lists are unique and differ at some finite position, k. Define the effective number, k=F(i,j), as the first position where list A_i differs from list A_j. I can now find the largest effective number for a given j. For each j define maxj=max( F(i,j) for i=1 to j-1 ) as the largest effective number for all i less than j. Define maxn as the largest effective number. For each j, if maxj > maxn, let maxn = maxj. Even if every list in A is infinite, the largest effective number, maxn, will be finite. We can distinguish any two sets by only looking at the first maxn elements in the two lists. Suppose A_1 is N, and A_i, for every i>1, is N-{(i-2)}. A_1 = {0, 1, 2, 3, ... } A_2 = {1, 2, 3, 4, ... } A_3 = {0, 2, 3, 4, ... } A_4 = {0, 1, 3, 4, ... } etc. What is maxn? The last one. You haven't defined which one that is, but there must be one if every list differs from every other list.

Prove it. I didn't define which is the last one because there isn't one.

By the way, how are you getting on with MoeBlee's find-the-fallacy
challenge? I think the fallacy is a bit too subtle for you to have any
chance of finding it, but you may surprise me.

Patricia
Patricia Shanahan
science forum Guru Wannabe

Joined: 13 May 2005
Posts: 214

Posted: Fri Jul 21, 2006 10:33 am    Post subject: Re: The List of All Lists

Patricia Shanahan wrote:
 Quote: Russell Easterly wrote: ... I proved in another thread that any non-empty proper subset of the set of all natural numbers that is transitive and well ordered must have a largest element. This means any inductive set that doesn't contain omega is finite. ... Did you ever prove, rather than merely assume, that there is any non-empty proper subset of the natural numbers that is transitive and well-ordered?

That should be "infinite non-empty proper subset".

There are, of course, plenty of finite non-empty proper subsets of the
natural numbers, which are transitive, well-ordered, and have a largest
element.

 Quote: If so, I missed it and would welcome a reference to the article. Patricia
Rupert
science forum Guru

Joined: 18 May 2005
Posts: 372

Posted: Fri Jul 21, 2006 10:38 am    Post subject: Re: The List of All Lists

Russell Easterly wrote:
 Quote: "Rupert" wrote in message news:1153460812.700396.327670@b28g2000cwb.googlegroups.com... Russell Easterly wrote: "Rupert" wrote in message news:1153459847.203837.56200@75g2000cwc.googlegroups.com... Russell Easterly wrote: "Patricia Shanahan" wrote in message news:y5Zvg.2660\$157.889@newsread3.news.pas.earthlink.net... Russell Easterly wrote: I will prove there exists a largest "effective" natural number. This is related to the Well Ordering the Reals thread, but I think you will find this proof unique. The list of all lists contains a largest "effective" natural number. Assume I am given a list of unique lists of natural numbers, A. By unique, I mean each list in A differs from every other list in A. Given two lists, A_i and A_j, we know these lists are unique and differ at some finite position, k. Define the effective number, k=F(i,j), as the first position where list A_i differs from list A_j. I can now find the largest effective number for a given j. For each j define maxj=max( F(i,j) for i=1 to j-1 ) as the largest effective number for all i less than j. Define maxn as the largest effective number. For each j, if maxj > maxn, let maxn = maxj. Even if every list in A is infinite, the largest effective number, maxn, will be finite. We can distinguish any two sets by only looking at the first maxn elements in the two lists. Suppose A_1 is N, and A_i, for every i>1, is N-{(i-2)}. A_1 = {0, 1, 2, 3, ... } A_2 = {1, 2, 3, 4, ... } A_3 = {0, 2, 3, 4, ... } A_4 = {0, 1, 3, 4, ... } etc. What is maxn? The last one. There is no last one. She defined an A_i for every positive integer i. Prove it. She defined an A_i for every positive integer i. There is no last positive integer because for every positive integer i, i+1 is also a positive integer, showing that i is not the last positive integer. You can start by proving every natural number exists and is finite. Tell me what axioms I am allowed to use, and define "finite". Incidentally, I don't see why the burden of proof lies with me. She challenged you to give the value of maxn. It's your job to substantiate your claim that there is a last one. Then you can prove there isn't a finite number of finite numbers. Sure, I'll do that. Just tell me which axioms I'm allowed to use. Do you want me to do it in ZF? Sure, let's use ZF. You start by defining the set of all natural numbers as the intersection of all transfinite inductive sets. I want to talk about this intersection. I am assuming transfinite, inductive sets are transitive, well ordered and that every inductive set contains every natural number. Some inductive sets also include non-empty limit ordinals like omega. Obviously, omega will be in the intersection of two inductive sets that both contain omega. Since the set of natural numbers doesn't contain omega, there must be at least one inductive set that doesn't contain omega. I proved in another thread that any non-empty proper subset of the set of all natural numbers that is transitive and well ordered must have a largest element.

My apologies. This is actually correct.

 Quote: This means any inductive set that doesn't contain omega is finite.

However, this does not follow. The set of natural numbers is an
inductive set which does not contain omega, and it is not finite.

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