An uncountable countable set
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From: Virgil on 30 Sep 2006 02:28 In article <451df438(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <451dd293(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Virgil wrote: > >>> In article <451d83c4(a)news2.lightlink.com>, > > > > > >>> So what balls remain in the vase at noon, oh waffler extraordinary? > >> For n balls inserted, balls n/10+1 through n remain at the end of any > >> iteration n. You specify the number of iterations, I'll give you the > >> sum. What was it? Aleph_0? > > > > All iterations executed before noon. > > And that would be how many? All of them.
From: Virgil on 30 Sep 2006 02:40 In article <451df5af(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <451dd5cf(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Virgil wrote: > > > >>> What Randy is avoiding is TO's fallacious insistence that there must be > >>> a last ball removed if one is ever to achieve a state where the balls > >>> are all removed. > >> > >> When the gedanken specifically states that only one ball is removed at a > >> time, what is fallacious about that statement? > > > > It omits that it is not just any ball which is removed. If one changes > > which one is to be removed, one changes the game and the result. > > Omitting an irrelevant detail does not make a statement fallacious. Violating one of the rules is not the same as omitting an irrelevant detail. And statements which force those violations are fallacious. > According to the gendanken, EITHER version, 10 balls are added, then one > is removed, and you repeat the process. Insufficient for the actual gedanken which requires numbered balls and specifies which numbers are moved in or out at each move. > Only one ball is removed at any > time, immediately preceding which 10 balls have been added. You had to > have -9 balls in your vase for that to have occurred. How does 10 - 1 come out -9? TO's arithmetic is getting as sloppy as his thinking. > >>> In a physical world that might be the case, but in an ideological one it > >>> need not be. > >> It is stated as a condition. Insert 10, remove 1, repeat. > > > > It is much more precise and detailed than that. > > Is that condition not part of the specification of the sequence of events? t is much more precise and detailed than that. > > > > >>> TO tries to change, or break, those rules, which is a form of cheating. > >> Oh, by rearranging a sequence and violating a specified order? No, that > >> was your trick. > > > > I followed exactly the sequencing defined by the problem. > > > > It is TO who tries to change or ignore those rules. > > No, when I expressed it as an infinite sequence, you tried to rearrange > terms to make it add up. When the two events, adding 10 and removing 1, > are coupled as one iteration, then you have a net gain of 9 per > iteration. Which I never denied. > You're playing "natural" labeling games based on the > unboundedness of the finite realm. It's hocus pocus. It follows the rules of the gedankenexperiment precisely and exactly. That TO is incapable of grokking what mathematicians regularly grok does not make it hocus pocus. It only points out TO's limitatains.
From: imaginatorium on 30 Sep 2006 04:25 Tony Orlow wrote: > imaginatorium(a)despammed.com wrote: > > Tony Orlow wrote: > >> imaginatorium(a)despammed.com wrote: > >>> Tony Orlow wrote: > >>>> imaginatorium(a)despammed.com wrote: > >>>>> Consider a (notional, theoretical, mathematical, not physical) x-y > >>>>> plane. That is, an area in which there is a point (0,0) in some > >>>>> particular place, an x-axis, y-axis, and points are identified by > >>>>> coordinates x and y, using (in normal maths) real values for these > >>>>> coordinates. Consider (for convenience) that this plane is embedded in > >>>>> a notional graphics application, with a "Fill" function. So if we draw > >>>>> the circle x^2 + y^2 = 49 (centre origin, (constant! Zick, be quiet!) > >>>>> radius 7), then click with the Fill function on the point (2,1), it > >>>>> fills the circle, and no paint spills outside that radius 7. > >>>>> > >>>>> Now suppose we have the graphs of x=2 and x=5. Vertical lines, > >>>>> extending up and down without limit. Suppose we click with the Fill > >>>>> function on the point (3, 4), what would you say happens? Obviously > >>>>> paint fills the vertical strip of width 3. Would you say that any paint > >>>>> was able to "spill" around the (nonexistent!) "top" of either of the > >>>>> graphs, and somehow fill more of the plane than this strip, or would > >>>>> you say we just get a (vertically) unbounded strip of blue? (Goddabe > >>>>> blue!) > >>>> I'd have to agree that it would fill the strip only. Proceed, but it > >>>> would be nice to know the context of the question. > >>> Ok, well just as a diversion: suppose you were on > >>> sci.comp.graphics.crank, and one of the residents produced a long, > >>> rambling argument, including mention of Planck's constant, twin-slit > >>> experiments and more, at the end of which was a claim that outside the > >>> strip would also be a very pale (ok "infinitesimally pale"!?) blue. How > >>> would you try to justify your claim that the blue fills the vertically > >>> unbounded strip only? > >> I'd have to see their fill algorithm to see what their malfunction is. > > > > Remember this is not a practical example. It can't be, because no > > actual physical computer can simulate a boundless x-y plane. > > > >>> Note that when discussing the behaviour of a real-world graphics > >>> program, within a bounded window, it's possible to discuss the > >>> paint-filling as a terminating procedure. With an unbounded strip, it > >>> obviously isn't. So I would say something like the following: for the > >>> paint to spill outside the vertical lines bounding the strip, there > >>> must be a path from a point inside to a point outside. But since the > >>> x-coordinate of the points on the path must go from (say) 4 to 6, at > >>> some point it must be 5; and that point must be a point on the > >>> boundary, so it would have crossed the boundary, and it's not allowed > >>> to cross the boundary, so this can't have happened. > >> Any general fill algorithm would probably leave some section of that > >> infinite strip un-blued. > > > > Really? We are talking about the (non-practical, not actually > > implementable, [what mathematicians call 'infinite']) notional plane, > > and a strip of width 3 that extends indefinitely up and down. An > > algorithm by definition terminates, and this strip doesn't, so > > obviously no algorithm could fill it. But there doesn't seem to be > > anything difficult in painting a strip that goes on forever, other than > > that the job can't be completed. Are you suggesting that somehow an > > algorithm that purported to paint the whole strip would just stop at > > some point? When the strip was a certain height? Even though it painted > > without end, it would reach the end? (Oh dear, this all seems rather > > familiar...) > > Yes, your imaginations don't seem to vary much. Meaning? You _can_ imagine that given an endless task, something could segue to "the infinite realm" and get to the end of the endless task? Hmm... > I am just suggesting that a general-purpose fill algorithm which could > fill any finite shape might not work on this shape. If it found a left > most point, filled that vertical line up and down until it hit the > boundaries of the shape, then went right to the next column, it would > never get to that next column, because it would never finish with the > first vertical column. In any discrete simulation of the area-filling task, this is true (for some shapes). But we cannot be talking about a discrete simulation, since we agree that the (to-be-painted-blue) sliver goes up without end, getting narrower and narrower without a minimum width; any discrete simulation would stop when the width was less than 1 pixel. I don't want to get bogged down too much in this "fill algorithm" stuff - we are not talking about real-world computation, but about a mathematical idealisation (which is what maths _is_ f'goodness sake). > If it started by finding a topmost or bottommost > point, it would never even get to filling one pixel. But yes, this is certainly true. > .... Of course you could > create an algorithm that worked its way back and forth a row at a time, > perhaps going both up and down, and continue filling the shape forever. Good. That's OK then. > >>> ---- back to the point ---- > >>> > >>> Now consider some other graphs: > >>> > >>> y=1/x, fill from the point (0, 0) - get blue lower left and upper right > >>> quadrants, plus filling out to the white lobes that almost fill the > >>> upper left and lower right quadrants. OK? (Graph is a hyperbola) > >>> > >>> Now consider the following two hyperbola-halves: > >>> > >>> y1 = -1/x (for negative x) > >>> y2 = -2/x (for negative x) > >>> > >>> Each of these is a lobe in the upper left quadrant, OK? > >>> > >>> Clicking on (-23, 34) would fill just the lobe formed by y2=-2/x (since > >>> this curve is always above and to the left of the other one); clicking > >>> in the lower right quadrant would fill three quadrants, and the area up > >>> to the y1 curve, leaving a (slightly larger) upper left white lobe. (I > >>> hope all this terminology is clear.) > >>> > >>> Would you agree that clicking on (-1, 1.5) fills the sliver between the > >>> two hyperbola lobes? > >>> (I say "sliver", though the area is infinite, since sum(1/n) doesn't > >>> converge - plus a bit of hand-waving.) > > > >
From: mueckenh on 30 Sep 2006 16:27 Tony Orlow schrieb: >Do I "misunderstand" that if you remove balls 1, then 11, then > 21, etc, that the vase will NOT be empty? This is an extremely good example that shows that set theory is at least for physics and, more generally, for any science, completely meaningless. Because the numbers on the balls cannot play any role except for set-theory-believers. The same issue we have with Tristram Shandy "who writes his autobiography so pedantically that the description of each day takes him a year. If he is mortal he can never terminate; but if he lived forever then no part of his biography would remain unwritten, for to each day of his life a year devoted to that day's description would correspond." (Fraenkel, Levy). When he is writing down always only the first on January, this assertion of Fraenkel and Levy is certainly false. And as absolutely nothing is changed with respect to speed and quantity if Tristram Shandy decides to follow another sequence, we see that this assertion is always false. Regards, WM
From: mueckenh on 30 Sep 2006 16:33
Dik T. Winter schrieb: > > For instance: > > Does the set o all natural numbers include 0? In old Greek it did not > > even include 1. In future it may include even -1. > > Yes, indeed, that depends on your starting point with the natural numbers. > That does not make it "all natural numbers and some more natural numbers". There is a bijection *possible*, but that does not mean that this bijection is ruling the set number of numbers of sets. There is a bijection possible between the sets {1,2,3,...} and {11, 12, 13, ...} but the latter set obviously contains 10 elements more than the former, even though a bijection between {1,2,3,...} and {101, 102, 103, ...} is possible. The error of you is to believe that the possibility of a bijection rules the number o numbers. (There are exactly twice so much natural numbers than even natural numbers.) Therefore your assumption of a uniquely defined number 0.111... is wrong. > > Why not? Each and every number of the list terminates. That one is a number > that does *not* terminate. > > > If you think that 0.111... is a number, but not in the list, > > It is *you* who insists it is a number. In most of my communications > with you I put the word number in quotes, because it depends on how you > interprete it on whether it is a number or not. It is me who insists that it is not a representation of a number. > > > then > > you must be able to find a position which is different from those of > > the numbers of my list. > > No. It is sufficient to prove that for each number on your list there is > a position where it is different from that number. That does *not* > imply that there is a position where it is different from each number > of the list. You could come up with that argument for arbitrary numbers, but not for unary numbers. what you require is impossible. Either 0.111... is larger than any number of the list, then you have to give a position which is not covered by a list number or not. Take into account that also Cantor's diagonal argument cannot be executed in unary representation. The unary representation is capable of modelling rational numbers like 0.111 / 0.11111 and even some irrational numbers like sqrt(0.11). But it is not capable of modelling Cantor's diagonal argument. And it is in clear contradiction with your requirement of completely indexing but not covering 0.111... by the list numbers. > > > > > And why can't there be more than one number with > > > > infinitely many digits? You cannot answer these questions because > > > > already one infinite set is a contradiction. > > > > > > No, I can not answer this question because I have no idea what you mean > > > with a number with more than omega digits. Consider K = 0.111... . What > > > is K+1? Can you provide a definition (as I did for K)? > > > > k + omega is omega. And -k + omega is omega too. There is no well > > defined set. > > In what way is that an answer to my question? Do you understand that > 1 + omega = omega != omega + 1? > And (as far as I know) -k + omega is not defined for positive k. (With > the ordinals addition is defined only between ordinals.) Of course you can set up a bijection beween the sets k + omega = {-k, -k+1, -k+2, ... , 0, 1,2,3,...,} and -k + omega = {k+1, k+2, k+3, ...}. But that does not mean that both sets have the same number of elements. > > > > > > Which index distinguishes 0.111... from all the numbers > > > > of the list? You cannot answer? > > > > > > I can. None. > > > > The axiom extensionality tells us that two sets are different, if they > > differ in at least one element. If 0.111... differs from number n, then > > it differs from all numbers m < n. As 0.111... is different from each > > number of the list, it also differs from each one which is smaller than > > another one. As every number of the list is smaller than another one, > > 0.111... cannot be covered by all numbers. Hence, it cannot be indexed > > by all list numbers. > > Again, that last conclusion is not justified. On the contrary, your assertions that indexing is possible but covering is impossible, is completely unjustified and obviously wrong, as is easily seen by the unary representation. > > > > So the number can be indexed. > > > > It is curious. You could also assert something like "I can shout louder > > than anybody else but there is nobody who cannot shout louder than me". > > But it is impossible to try to exorcise your burnt-in anti-logicism. > > Apparently not with your burnt-in anti-logicism. I have *proven* that > it can be indexed, by the simplest of all possible proofs. Namely by > showing that there is no digit 1 at any position other than indicated > by a natural number, which *by your* definition makes the number > indexable. You have not *shown* that, but defined it, erroneously. But if you had shown it, then 0.111... was in the list, which also would have been wrong. > > > > > So we cannot answer which index > > > > distinguishes the many different infinite digit sequences 0.111... from > > > > each other. > > > > > > What different infinite digit sequences? I note that digit sequences are > > > countable, and so there is only one infinite digit sequence. > > > > I note that some sequences are finite, and so there is only one finite > > digit sequence? > > Is that really an argument? > > Not for finite sequences, but it is for infinite sequences. And you really think that the *possibility* of a bijection would in fact *guarantee* that the number of elements (or here: the number of digits of unary numbers like 0.111... ) must be identical? That is your firm belief? Regards, WM |