
Re: Why are the real numbers wellordered?
Posted:
Sep 11, 2001 9:08 AM


In article <9nj6jk$of0$1@tron.sci.fi>, Tapio <hurmecom@dlc.fi> wrote:
>"Dave Seaman" <ags@seaman.cc.purdue.edu> wrote in message >news://9niqte$fol@seaman.cc.purdue.edu...
>OK, the confusing notation has been perhaps a reason why we had so much >problems and missunderstanding. I assume you mean now that the structure of >the number system would be something like this:
>...[w+2 sequences][w+1 sequences][w sequences= integers N(inf) and >N][decimal sequences][w+1 decimal sequences][w+2 decimal sequences] ...
Each of those is certainly a welldefined set of mappings. Is there some connection here to your claim that the reals are wellordered? The two problems I see are that (a) your digit strings are not the real numbers, and (b) your digit strings are not wellordered.
>>If I add 1+[9] >> >the result is 1(something) as the carry goes over aleph length of string. >> >1(something) cannot be after addition smaller than [9]. The digit 1 must >be >> >"behind the infinity" and therefore w(inf) = 1,000... = [0)1][0].
>> Then w(inf) is the (w+1)sequence a: (w+1) > D given by
>> a_k = 0, if k < w >> a_w = 1.
>Yes!! I assume now you have got the basic idea correctly.
If you have access to any of our discussions from several years ago, you may find that I was suggesting way back then that you were using (w+1)sequences of digits when talking about things like 0.999...8, but you have been resisting that characterization up until now. You do realize, I take it, that (w+1)strings do not map to real numbers or to integers in any natural way?
>> >> Remember, you can't use N(inf) in the definition of N(inf), because >that >> >> would make it a circular definition. Look at how other number systems >> >> are defined. The reals are defined in terms of the rationals, the >> >> rationals in terms of the integers, and the integers in terms of the >> >> natural numbers. In each case we define a new and more complex number >> >> system out of a simpler one. So it is with N(inf). If you want to >> >> describe what your numbers are in terms that people can understand, you >> >> need to describe them in terms of simpler kinds of numbers so as to >avoid >> >> circularity.
>Yes. I do understand your point. Therefore I set a question for myself. If >we define that the string is infinite instead of finite, then why should I >use in the sum (a_k)10^k kvalues in N as I have N(inf) in use? I think we >are in "half way", if we do not use the opportunity of infinite sequences >"infinite integers" for k in N(inf). Just like if we have k in N as soon as >N is available. I assume this need further consideration. Do you have any >constructive idea?
Once you have defined N(inf), you can use it to define something else, say N(hyperinf), that would have the kind of extended sequences you want. Or, you could define N(inf) directly to use ordinal sequences, which you are now doing already in the case of (w+1)sequences, thereby avoiding circularity.
But where is all this leading? You still have not answered these questions:
(1) What is the smallest member of Z? (2) What is the smallest member of N(inf)\N?
Note that your answer to (1) should be a member of Z, and your answer to (2) should be an wsequence. You cannot answer these questions, and this shows that R and N(inf) are not wellordered.
>> >I think I asked once about the circularity to define N as sum. But you >did >> >not answer, perhaps because you saw the analogy.
>> I have no idea what you are talking about.
>See above.
I still have no idea what you are talking about.
>>N can be defined from the >> Peano axioms or from the axioms of ZF. Neither approach makes any >> mention of sums until well after the natural numbers are already defined. >> There is no circularity. The notion that sums come first and numbers >> afterwards is entirely your own aberration.
>Not exactly, the key point is that the strings are first defined to be >infinite.
Huh? Thre is nothing in the Peano axioms and nothing in ZF about "strings".
>> This part of the discussion originated when I asked you for the smallest >> member of N(inf)\N.
>Yes.
>>There are two things wrong with your "[1(0]" answer:
>> 1. It is not a member of N(inf), since [1(0] is an >> (w+1)sequence and the members of N(inf) are all wsequences.
>No. I think you consider that the indexing of placeholders depends on N, but >not in N(inf)??? I cannot exactly follow your logic although above you >seemingly and finaly understood the structure of the system. Everything >within [] belongs to wsequence area.
Does your [1(0] represent a mapping a: w > D? If so, what are the values? For which k do you get a_k = 1? For which k do you get a_k = 0? It appears to me that all but one of the digits are 0, which makes it a member of N and therefore not a member of N(inf)\N. If you are not talking about an wsequence, then you do not have a member of N(inf).
>Actually we should talk about w_inf sequence areas w_inf +1, w_inf +2 etc. >Numbers are now infinite in N(inf) finite numbers (integers) are special >cases.
>> 2. You keep claiming that [1(0] = [9] + 1.
>No, no, there is missunderstanding or typo.
I was taking [1(0] to be an (w+1)sequence, but you seem to be saying that it is an wsequence. I don't understand how that can be, since it appears to have a leftmost (highestnumbered) digit and there is no largest member of w.
>[9]+1= [9]+[0)1]= [0)1][0] = 1,000.... where "," is the w+1 point of >reference according to your introduction.
I don't understand your notation, but I think you are talking about the (w+1)sequence given by a_w = 1 and a_k = 0 for k < w.
 Dave Seaman dseaman@purdue.edu Amnesty International calls for new trial for Mumia AbuJamal

