An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Virgil on 13 Sep 2006 16:47 In article <1158134484.026807.259290(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > David R Tribble schrieb: > > > David R Tribble schrieb: > > >> I really don't see where the physical limitation is for visualizing the > > >> elements or the set. More to the point, I don't see how the phyics > > >> of the real world have any limiting effect on abstract concepts. > > > > > > > mueckenh wrote: > > > What you are arguing is only the beginning of infinity, because you are > > > unable to see more than this little realm. Try to determine the natural > > > number which consists of the 10^100 digits following the first 10^1000 > > > digits of pi. Your unability is not due to lack of time. This is only > > > one of the infinitely many numbers which you cannot deal with (because > > > it is not a number). > > > > Why is it not a number? It has all the mathematical properties > > required of a number. Having "all its digits enumerated" is not a > > required property. > > It is the only important property of a number. It may be important for a numeral, but not for a number. Numbers do not have digits, but numerals sometimes do. Those who conflate the two have no sense of mathematics. > > > > Supposing that I can in fact determine all 10^100 digits of this > > number. What then? > > Then it would be a number, but you cannot determine these digits. > > > Does that number now magically possess > > an existence that countless others do not, i.e., has it suddenly > > become a "more real" number than, say, the next 10^100 digits > > of pi? > > Not necessarily suddenly. It has become existence for you. You know it. > > > > > Likewise, until I actually enumerate the digits of some specific > > number, are you saying it does not (yet) exist? > > Would you say that a poem I am going to make tomorrow is already > existing today? > > > If that's the > > case, does pi or e exist? > > pi and e do exist as ideas or as problems or as irrational proportios > but they do not exist as numbers. And they will never exist as numbers, > in contrast to numbers with 10^20 digits. > > Regards, WM
From: David R Tribble on 13 Sep 2006 17:04 Tony Orlow wrote: >> ....11111 binary (all bit positions finite) >> It is the successor to ....11110. Duh. I've already proven that this is >> a finite value, given that all bit positions are finite, and that >> therefore no place in that string can achieve an infinite value, and >> that any such number has predecessor and successor. The cute thing is >> that the successor to ...1111 is 0, and that ...1111 is essentially -1. :) >> >> I find it disappointing, but not surprising, that you don't understand >> such a simple proof, since it's contradictory to your education. I do >> find it annoying that you feel the right to disagree with it without >> understanding it. If you feel there is a problem with the proof, please >> state the logical error I made. > David R Tribble wrote: >> As I posted earlier, it leads to the conclusion that 0 > 0. > Tony Orlow wrote: > True, considering the number circle breaks the transitive nature of >, > except that it's only one way of looking at the number line. In computer > science, ....1111 is either -1 or the greatest possible value for the > register, depending on whether one is using signed or unsigned integers. > When we look at the number circle, we are assuming that one can never > make the transition from the largest positive to the largest negative, > since there is no end to reach. However, the number line can be viewed > as the number circle in the limit, as the number of bits increases > without bound. Whether ...1111 is viewed as the largest positive finite > integer or the smallest negative finite integer, its successor is either > the smallest infinite or 0. In computer science, the carry is lost, and > the successor is ...000. That's an overflowed for unsigned integers, but > not for signed integers. SO, it depends on the interpretation of the bit > string. Fixed-length binary computer notations are imperfect models of mathematical numbers. They are flawed for exactly the reason you state, that there are boundary cases where the usual rules of arithmetic fail to hold. For example, x+1>x for all x except x=2^m-1, which violates the mathematical rules for + and >. That's usually not a problem for most programs, but it is a huge deficiency for modeling mathematical truth. Computer numbers are nothing more that fixed-width fixed-point modulo 2^m number systems, which is a very very small subset of mathematical numbers. As an experienced software engineer, I repeat the warning uttered by mathematicians: Don't make the mistake of assuming that the concepts of programming apply to mathematics. Your alternative "way of looking at the number line" is not consistent with the rules of arithmetic, so we really can't accept what it is you think you're describing in mathematical terms. If the bitstring is "interpreted" to be a representation of a fixed-width modulo number, then it must be limited to the arithmetic rules of such a system, and obviously can't be used as a general representation for the set of naturals.
From: Virgil on 13 Sep 2006 17:05 In article <1158134793.927816.37420(a)d34g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > You are running in circles. You have to show that *using my definition* > > it should be in the list. > > > > > > If it does not hold you should be able to give an index position > > > > such that it is false. > > > > > > It is not in the list, although we cannot give an index position which > > > is responsible for that fact. All possible indexes are in the list. > > > > Yes, it is not in the list. So what is the problem? All possible indexes > > are in the list, so: > > (1) A{p = digit position} E{q = list item} {such that q indexes p} > > which you deny. > > That is not an argument. Correct would be to say: > A{p = digit position which can be indexed} E{q = list item} > But those positions are already in the list (by the axiom of infinity). Which version of the axiom of infinity is "Mueckenh" misreading? According to one version, all that the AoI says that there exists a set, S, such that (1) {} is a member of S and (2) if x is a member of S then Union(x,{x}) is also a member of S. That says nothing about any "digit positions" being in any list at all. So that "Mueckenh" is making up claims that are false. > > That is *not* the definition of your list. Your list contains *by > > definition* > > all finite sequences of 1. > > That is all natural numbers (in unary representation). > > > There is nothing in that definition that it > > contains all numbers that can be indexed and that can index. > > That is the same as all natural numbers (here in unary representation). Where does anything say that only natural numbers can be used to index anything AND that only natural numbers can be indexed by anything? > > > > Indexing and covering "by all list numbers" is equivalent. When in blazes does 'covering "by all list numbers"' mean? As far as I can see it means nothing at all. > > > > > > "Not to terminate" is not a property which can serve to distinguish a > > > number from others, because we can never observe the end (because it > > > does not exist) neither the non-end (because it is nothing but a > > > negation of an unobservable property). > > > > Again. We are arguing within the realm of the axiom of infinity. You > > are ignoring that axiom with that statement. With that axiom the set > > of natural numbers does exist, and so your list (which is also > > non-terminating) does exist, > > and no number does exist which has more digits than the list numbers > have. Meaningless dribble. Indexing, at least for countable sets, is performed by creation of a bijection from the set of all naturals or some initial subset of it to the set to be indexed. And any such bijection whatsoever is equally valid as an indexing function. > > > > Why? Your list is non-terminating, so it can index non-terminating > > numbers. > > On the other hand, each list element is terminating. > > Correct. The list is non-terminating, so it can index all existing > numbers. All are in the list by the axiom of infinity. But none is > non-terminating. So a non-terminating number cannot be indexed. According to "Mueckenh" above, a non-terminating numeral can be indexed and simultaneoulsy cannot be indexed. No ratoinal person wants any part of such a self-contradictory system.
From: David R Tribble on 13 Sep 2006 17:10 Tony Orlow wrote: > Wolfgang's and my position is that N is unbounded but finite, "Unbounded but finite" is a contradiction, meaning "not finite but finite". I'm sure you and Wolfgang think this double-think makes sense, but the rest of us don't.
From: Virgil on 13 Sep 2006 17:14
In article <1158134972.282052.7360(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1157836470.071804.254270(a)i3g2000cwc.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > In article <virgil-411702.16371406092006(a)comcast.dca.giganews.com> > > > > Virgil <virgil(a)comcast.net> writes: > > ... > > > A set of n elements is *a number*. It need no necessarily be a latin or > > > arabic symbol. > > > > Bizarre. What number is the set of all natural numbers? > > That does not exist. At least it is not described by n. III is a > representation of 3. "III" and "3" are both numerals representing the same number, but neither is anything more that a representation, neither is the number itself. > > > > > > I count with my fingers. Does that count? > > > > > > The positions of your fingers are numbers. And you know what to do in > > > order to get from 32 to 33. > > > > I know what is the successor of the finger positions for 32. The position > > the 32 is all fingers (including the thumb) of the right hand up and two > > fingers raised on the left hand. So I just raise the ring-finger of the > > left hand to get the next number. No addition involved at all. > > And in both cases, the set contains either seven or eight raised fingers. > > If you know the positions by heart, then you need no addition actually. > You had already used it or the person who devised that technique had > used it. But earlier or later your knowing by heart will end and you > will have to count +1. On the contrary, a successor operation has no need of any addition, but can be effected in a wide variety of ways. > > > > > > And I can go to 35 with my > > > > fingers, and I use my feet (not toes) to extend to 143. I never > > > > really > > > > managed to do it base 2, but 4 and 128 are interesting numbers in that > > > > case. > > > > > > Every position of your fingers and feet is an expression of a number. > > > > Perhaps. > > You cannot say yes? > > > > > Counting without numbers would be nonsense. A boy like Virgil may count > > > 1, 2, 3, 3, 3, ... > > > > a, b, c, d, e, f, ... > > The old Greek and othe cultures have just used their letters as numbers > too. Counting can be done by making tally marks or moving pebbles, for example, entirely without numbers, though we have become so sophisticated that we may have trouble realizing it. It is whether the tally marks or collected pebbles biject with the objects counted which is the issue. And no number need ever be mentioned or used. |