An uncountable countable set
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From: mueckenh on 12 Oct 2006 12:14 Dik T. Winter schrieb: > In article <1160310643.181133.6720(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > AXIOM OF INFINITY Vla There exists at least one set Z with the > > following properties: > > (i) O eps Z > > (ii) if x eps Z, also {x} eps Z. > > > > There are several verbal formulations dispersed over the literature > > without any "all". > > Perhaps. Does this mean that there are some x in Z such that {x} not in > Z? Or, what do you mean? I did not interpret what it may mean. I said that in verbal forms of the axioms the word "all" does not appear. I said: I never came across the word "all" in connection with this axiom. > > > In German: Unendlichkeitsaxiom: Es gibt eine Menge, > > die die leere Menge enthält, und wenn sie die Menge A enthält, so > > enthält sie auch die Menge A U {A} (oder die Menge {A}). > > Do you not think that that means that for all A in that set, also {A} in > that set? If not, why not? I wrote: > > The axiom of infinity does only state n+1 exists if n is given. It is > > realized by he numbers of my list. You replied: > That is *not* what the axiom of infinity states. The axiom states that > there exists a set that contains all the successors of its elements. And I reply: That is wrong. The axiom says: For all possible n in all possible cases: If n is a natural number, then n+1 is a natural number too. It does *not* say that it is meaningful to speak of *all* natural numbers. It does not say that the set N does actually or completely exists. It does not say that this infinity is finished, i.e., that all natural numbers can exist together. Well: "there is a set". But the meaning of these three words depends heavily on the interpretation. Therefore, the interpretation which usually is given cannot be free of arbitrariness, and it leads to contradictions. That is the "solid" foundation of present matheology. Regards, WM
From: mueckenh on 12 Oct 2006 12:17 Dik T. Winter schrieb: > > > You question whether "all x in N" does exist, apparently. Based on what? > > > > Based on the impossibility to index the positions of our 0.111..., > > False. > > > based on the vase, based on many other contradictions arising from "all > > x in N do exist". > > False. > > No proof given. No proof possible because every proof must be dismissed unless the game of set theory should perish. > > > > Yes, and that is false and not provable. > > > > A set containing all positions "up to position x" is a superset of a > > set containing "position x". > > By what rule? What do you *mean* by "a set containing all positions..."? > What do you *mean* by "a set containing...". I would state the the > set {1, 2, 3, 4} contains all positions up to position 4, but that the > set {4, 5} contains the position 4. But neither is a superset of the > other. A set containing all positions "up to position x" is a superset of a set containing just "position x". > > > A set containing "up to every position" defines a superset of set > > containing "every position". But "every position" cannot be a proper > > subset. Hence both sets are equivalent. > > Please first answer my question above, next, elaborate. A set containing all positions "up to every position" is a superset of a set containing just "every position". Regards, WM
From: mueckenh on 12 Oct 2006 12:18 Dik T. Winter schrieb: > In article <1160310273.517860.47380(a)m7g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > > Ok, but as we have agreement now, we can return to he main question: > > > > Why do you think that 0.111... with the index sequences 1,2,3,... or > > > > k+1,k+2,k+3 or -k, -k+1, -k+2, ... represents exactly *one* number > > > > only, as you asserted? > > > > > > Why do you still maintain that I think it represents a number? How many > > > times do I need to state that, without proper definition, it only is > > > a sequence of symbols that I on occasion call a "number". Because I have > > > not yet seen a definition of "number", and you have stated that you are > > > not able to give one... > > > > In the decimal system of current mathematics 0.111... = 1/9 and is a > > number without doubt. > > Yes, and in the octal system it is 1/7, and in the reversed 10-adics it > is -1/9. There are a host of systems where it has a meaning, but it has > no meaning if you do not specify the system. And there are also systems > where it has no meaning because there is no convergence (the 10-adics in > normal notation). > > > If omega does exist, then 0.111... has omega 1's. > > Yes. > > > But whatever. As a sequence of symbols, > > > igoring the "0.", it is in bijection with N. It also is in bijection > > > with {k+1,k+2,...} for every k. As {k+1,k+2,...} is in bijection with N. > > > > A bijection with N does not define the indexes such that there was only > > a unique sequence. Therefore, there is not, as you asserted, one unique > > number 0.111... . > > Oh. Whatever. Care to explain? 0.111... is the representation of a number, e.g., of 1/9 in decimal notation. And this representation is not unique, because the indexes are undefined. A bijection between two countable sets of indexes does not guarantee that the sets contain the same number of indexes. Therefore there are many different sets of indexes for the digits of the number 0.111.... Therefore there is no unique number 0.111... as you erroneously stated. With 0.111... this does not matter, but with 3.1415... we see that this number is not well defined. Regards, WM
From: Lester Zick on 12 Oct 2006 12:34 On 12 Oct 2006 08:28:14 -0700, "Ross A. Finlayson" <raf(a)tiki-lounge.com> wrote: [. . .] >Recently in this discussion about infinite sets and so on one of the >talking points about Cantor that has emerged is that he counts >backwards from infinity. This strikes me as a very curious spin on the problem of infinity. If one counts backward from infinity what is the first number counted and is it finite or infinite: for x=00-1 what is x? Certainly it has to be less than infinity but if 00+1=infinity then so should 00-1. On the other hand would 00-00=0? Maybe arithmetic operations aren't defined between finites and infinites any more than division between finites and zero. ~v~~
From: Lester Zick on 12 Oct 2006 13:06
On Wed, 11 Oct 2006 20:38:53 -0400, Tony Orlow <tony(a)lightlink.com> wrote: >Lester Zick wrote: >> On Wed, 11 Oct 2006 11:41:00 -0400, Tony Orlow <tony(a)lightlink.com> >> wrote: >> >>> Lester Zick wrote: >>>> On Tue, 10 Oct 2006 13:56:43 -0400, Tony Orlow <tony(a)lightlink.com> >>>> wrote: >>>> >>>>> Lester Zick wrote: >>>>>> Tony, I'm going to strip as much as seems unessential. >>>>>> >>>>>> On Mon, 09 Oct 2006 15:56:32 -0400, Tony Orlow <tony(a)lightlink.com> >>>>>> wrote: >>>>>> >>>>>>> Lester Zick wrote: >>>>>>>> On Sun, 08 Oct 2006 22:07:23 -0400, Tony Orlow <tony(a)lightlink.com> >>>>>>>> wrote: >>>>>> [. . .] >>>>>> >>>>>>>> As far as transcendentals are concerned, Tony, the only thing that can >>>>>>>> lie on a real number line in common with rationals/irrationals are >>>>>>>> straight line segment approximations. That's the only linear order >>>>>>>> possible. So either you give up transcendentals or a real number line. >>>>>>> The trichotomy or real quantity itself defines a linear order. Each such >>>>>>> value is greater than or less than every different value. Pi is >>>>>>> transcendental - is it less than or greater than 3? Is there any doubt >>>>>>> about that? >>>>>> No but that only applies to linear approximations for pi, Tony. It has >>>>>> nothing to do with the actual value of pi which lies off to the side >>>>>> of any real number line. Imagine if you will the linear order you >>>>>> speak of superimposed on a real number plane instead of a straight >>>>>> line (which isn't even possible either). Then pi has to lie off to one >>>>>> side of the straight line. So the metric for the straight line becomes >>>>>> a non linear variable instead of linear. Thus the fact that the value >>>>>> of pi lies between 3 and 4 doesn't show any value for pi and never >>>>>> will. All you wind uyp with is a linear metric which approximates pi >>>>>> which doesn't provide us with any real number line running along the >>>>>> lines of 1, 2, e, 3, pi. So the real number line metric is variable. >>>>>> >>>>> Pinpointing particular points on the line may require varying degrees of >>>>> computation/construction. A natural requires only a finite number of >>>>> iterations of increment, a rational may require a division operation, >>>>> and an irrational or a transcendental like pi may require an infinite >>>>> computation to exactly pinpoint the value. That doesn't mean that value >>>>> doesn't exist as a point on the line. It just mans specifying it >>>>> requires an infinite process. >>>> It requires an infinite process only because it isn't on the line. The >>>> only thing an infinite process determines is how close the curve comes >>>> to a straight line without ever getting there. Every other rational/ >>>> irrational on the line can be located with finite processes because >>>> they are on the line. >>>> >>> If the line is defined by trichotomy, and 3<pi<4, isn't pi on that line? >> >> If by "trichotomy" you mean <=> on a straight line then no pi isn't on >> that line, Tony. Pi lies on a circular curve not on a straight line. >> Approximations for pi such as 3<pi<4 do lie on a straight line but >> only indicate how close circular arcs come to straight line >> approximations. >> > >Circular arcs approach the straight line in the limit as radius->oo, but >other than than, no, pi's a quantity, a distance from the origin That's >what a real number is. Not possible, Tony, unless you want to pretend circular arcs are congruent with straight lines. Pi is an exact measure on a circle. This is one reason I take issue with conventional classifications of transcendentals as irrationals. There is no "real" number line in formal terms and even Bob Kolker publicly admitted the point. No transcendental is perfectly congruent with any straight line segment. >>>>>>>>> Perhaps not, but if, say, y=1/x is both oo and -oo at x=0, and oo=-oo, >>>>>>>>> then the function is continuous in every respect, which is what we might >>>>>>>>> desire in such a fundamental algebraic relation. >>>>>>>> But for the division operation x never becomes zero. Which indicates >>>>>>>> that there can be no plus or minus infinity and no continuity. >>>>>>> Is 0 part of the continuum, or just another arbitrary "limit" >>>>>>> discontinuity? When you ask yourself, "If I divide this finite space >>>>>>> into individual points, how many will I have?", what answer do you get? >>>>>>> How much of the space between 0 and 1 does each real in that interval >>>>>>> occupy, and how many are there? >>>>>> Well points don't occupy any space at all, Tony. >>>>> Right, their measure is 0. >>>>> >>>>> Intersection ends in >>>>>> points but subdivision never ends in points. >>>>> In the limit, subdivision results in infinitesimal segments, where the >>>>> values of the endpoints cannot be distinguished on any finite scale. >>>> Oh I don't know about that, Tony. Certainly the first bisection in a >>>> series is exactly half the length of the starting finite segment. >>> Yes, and every finitely-indexed subdivision results in finite segments, >>> but that cannot be truly when the segment is infinitely subdivided. >> >> As far as that goes it never is. > >In essence, there is no end to subdivision. In the limit, the subsegment >is a point, or a pair of identical points. No it isn't a point because the limit is never reached. Infinitesimal subdivision is just a never ending process not a geometric thing or arithmetic result. Nor is there any difference between one point and two identical points. Either you have infinitesimals or nothing. >>>>> That's the difference >>>>>> between intersection and differentiation. Although not a natural zero >>>>>> certainly exists but the question is how it's used. Zero used as a >>>>>> limit is approachable but not reachable. Same with infinity. Zero used >>>>>> as a divisor is undefinable. Maybe you're trying to combine arithmetic >>>>>> and infinitesimal calculus operations in ways that are incompatible? >>>>>> >>>>> It's true that absolute 0 has an indeterminate reciprocal of oo, and >>>>> that regular arithmetic won't work on those. But, a specific unit >>>>> infinity lends itself to arithmetic quite nicely. >>>> So does zero. It just doesn't have any meaning when used for division. >>>> Nor does infinity when used to multiply. >>>> >>> But specifi |