An uncountable countable set
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From: Virgil on 28 Sep 2006 13:28 In article <b1fd9$451b85a3$82a1e228$11085(a)news1.tudelft.nl>, Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > Virgil wrote: > > > In article <d12a9$451b74ad$82a1e228$6053(a)news1.tudelft.nl>, > > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > > > >>Randy Poe wrote, about the Balls in a Vase problem: > >> > >>>It definitely empties, since every ball you put in is > >>>later taken out. > >> > >>And _that_ individual calls himself a physicist? > > > > Does Han claim that there is any ball put in that is not taken out? > > Nonsense question. Noon doesn't exist in this problem. > What has noon to do with my question? According to my recollection of the problem, there was a specific time (prior to noon) at which any ball which had been put into the vase was taken out. Does Han dispute my recollection?
From: Virgil on 28 Sep 2006 13:33 In article <1159437062.473100.294820(a)k70g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > Several sets may all have the common property of being pairwise > > bijectable, but if any of their members are distinguishable from those > > of another set then the sets are equally distinguishable. > > Each one of the sets expresses, represents, and *is* the same > (cardinal) number. Then one apple and one orange are the same because they have the same cardinality. > This is but *one* possible method, or better "schema > of methods", to realize 3. And it is a typically bad one, as aer many of "Mueckenh"'s schemes. > > > > > > > > > You stated that you needed counting to determine the successor. That is > > > > false. The successor is defined without any reference to counting. > > > > > > The successor function *is* counting (+1). > > > > Not to those who can't count. Successorship does not require numbers, it > > only requires "next". > > How far would those who cannot count be able to find "the next"? Til they run out of markers. Possibly fingers or toes or pebbles or whatever.
From: Virgil on 28 Sep 2006 13:38 In article <1159437157.109258.97400(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1159186907.615747.304410(a)h48g2000cwc.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > 1/3 is a number, properly defined, for instance, by the pair of numbers > > > 1,3 or 2,6 or 3,9 etc. But 0.333... is not properly defined because you > > > cannot index all positions, you cannot distinguish the positions of > > > this number from those with finite sequences (and you cannot > > > distinguish them from other infinte sequences which could exist, if one > > > could exist). > > > > Def: 0.333... = lim_{n -> oo} Sum_{k = 1..n} 1/3^n > > Definitions (even correct definitions unlike this one) don't guarantee > existence (I used above "to be properly defined" but I meant "to > exist"). Example: The set of all sets is defined but is not existing. > > Regards, WM Def, corected: 0.333... = lim_{n -> oo} Sum_{k = 1..n} 3/10^k And it exists because there is a real number (actually a ratonal number) L = 1/3 such that for every epsilon greater than 0, there is a largest n such that | L - sum_{k+1..n}| >= epsilon.
From: imaginatorium on 28 Sep 2006 13:41 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...) > > ---- 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.) Do you agree, first of all, that there is an area between the two hyperbola lobes? I'm using this "paint-fill" thing in the hope that it clarifies what I'm getting at, but it isn't really relevant. Do you agree that an unbounded line (like the lobe of one hyperbola) divides this unbounded plane into two areas. Every point (p, q) is either _on_ the line, or in the upper-left lobe, or in the remaing three quadrants-plus-the-rounded-bit? Please say you see what I mean, or not. If necessary I can produce some graphics, but ASCII is a lot easier. > > Do you see a connection to the original problem? > > No. Would you care to be a little more explicit? OK, suppose you agree that between these two hyperbola we have painted an unbounded blue sliver, that goes on forever to the left, getting narrower and narrower, but never having zero width, and also goes on forever upward, getting narrower and narrower, but never having zero width. Suppose we replace the "Solid colour" paint, with paint that just draws horizontal lines at integral values of y. In other words, horizontal hatching. Where do you think the bits of horizontal hatching between the two hyperbol[damn, I was hoping to avoid the plural]ae are? Each hyperbola is always at a strictly positive y value, so the x axis can't be included in the hatching anywhere. Here are the relations again: y1 = -1/x (for negative x) y2 = -2/x (for negative x) So the lower hyperbola (can I just say "loop") goes through (-1, 1), and the upper one goes through (-2, 1). Therefore, I think the first bit of line goes between those two points - y=1, for -2<=x<=-1. OK? The second segment of horizontal hatching goes from (-1, 2) to (-0.5, 2) unless I'm much mista
From: Virgil on 28 Sep 2006 13:43
In article <1159437976.649538.246020(a)i3g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > Tony Orlow wrote: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > Mike Kelly schrieb: > > > > > > > >> It is not a proof. Division is not defined where either operand is an > > > >> infinite cardinal number. > > > > > > > > But you can conclude n / aleph_0 < 1 by inserting aleph_0 > n which is > > > > definied *if aleph_0 is a number in trichotomy with natural numbers*. > > > > > > > > You cannot have both, assert that aleph_0 is a number larger than any n > > > > but on the other hand prohibit that the inequality n < aleph_0 be > > > > utilized. > > > > > > > > Regards, WM > > > > > > > > > > I agree. If x is infinite, and that means greater than any finite, and > > > trichotomy holds, then 1/x is in [0,1], and is real, though > > > infinitesimal, of course. But, I am getting the feeling that set > > > theorists now don't want to claim that infinite is greater than finite. > > > > I think it would be more correct to say that set theorists don't now > > want, and have never wanted, to claim that some mathematical object is > > greater than another mathematical object without an /explicit > > definition/ of what is meant by "greater than" in the context of the > > discussion. > > There is only one common meaning of "greater than" in arithmetic: If > you subtract a from b, and the remaining is positive, then b > a. In certain arithmetics, that is not so. In the Peano arithmetic of positive naturals it was not always possible to subtract a from b. In the arithmetic of ordinal numbers a < b if and only if a is a proper subset of b, and subtraction in general is not defined. In > set theory we say set a bijects with a subset of b. In the same fashion > you can proceed, saying b/a > 1. There is no special definition of > division required to determine whether a ratio of trichotomic numbers > is larger or less than 1. > > Regards, WM |