An uncountable countable set
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From: MoeBlee on 27 Oct 2006 13:12 Tony Orlow wrote: > I think Ross has a genuine > intuition that isn't far off with respect to what's controversial in > modern math. Surely fodder for a Jesse F. Hughes tagline. MoeBlee
From: David Marcus on 27 Oct 2006 13:17 Tony Orlow wrote: > David Marcus wrote: > > Tony Orlow wrote: > >> David Marcus wrote: > >>> Tony Orlow wrote: > >>>> Mike Kelly wrote: > >>>>> Now correct me if I'm wrong, but I think you agreed that every > >>>>> "specific" ball has been removed before noon. And indeed the problem > >>>>> statement doesn't mention any "non-specific" balls, so it seems that > >>>>> the vase must be empty. However, you believe that in order to "reach > >>>>> noon" one must have iterations where "non specific" balls without > >>>>> natural numbers are inserted into the vase and thus, if the problem > >>>>> makes sense and "noon" is meaningful, the vase is non-empty at noon. Is > >>>>> this a fair summary of your position? > >>>>> > >>>>> If so, I'd like to make clear that I have no idea in the world why you > >>>>> hold such a notion. It seems utterly illogical to me and it baffles me > >>>>> why you hold to it so doggedly. So, I'd like to try and understand why > >>>>> you think that it is the case. If you can explain it cogently, maybe > >>>>> I'll be convinced that you make sense. And maybe if you can't explain, > >>>>> you'll admit that you might be wrong? > >>>>> > >>>>> Let's start simply so there is less room for mutual incomprehension. > >>>>> Let's imagine a new experiment. In this experiment, we have the same > >>>>> infinite vase and the same infinite set of balls with natural numbers > >>>>> on them. Let's call the time one minute to noon -1 and noon 0. Note > >>>>> that time is a real-valued variable that can have any real value. At > >>>>> time -1/n we insert ball n into the vase. > >>>>> > >>>>> My question : what do you think is in the vase at noon? > >>>> A countable infinity of balls. > >>> So, "noon exists" in this case, even though nothing happens at noon. > >> Not really, but there is a big difference between this and the original > >> experiment. If noon did exist here as the time of any event (insertion), > >> then you would have an UNcountably infinite set of balls. Presumably, > >> given only naturals, such that nothing is inserted at noon, by noon all > >> naturals have been inserted, for the countable infinity. Then insertions > >> stop, and the vase has what it has. The issue with the original problem > >> is that, if it empties, it has to have done it before noon, because > >> nothing happens at noon. You conclude there is a change of state when > >> nothing happens. I conclude there is not. > > > > So, noon doesn't exist in this case either? > > Nothing happens at noon, and as long as there is no claim that anything > happens at noon, then there is no problem. Before noon there was an > unboundedly large but finite number of balls. At noon, it is the same. So, noon does exist in this case? -- David Marcus
From: David Marcus on 27 Oct 2006 13:20 Tony Orlow wrote: > David Marcus wrote: > > Tony Orlow wrote: > >> stephen(a)nomail.com wrote: > >>> What are you talking about? I defined two sets. There are no > >>> balls or vases. There are simply the two sets > >>> > >>> IN = { n | -1/(2^floor(n/10)) < 0 } > >>> OUT = { n | -1/(2^n) < 0 } > >> For each n e N, IN(n)=10*OUT(n). > > > > Stephen defined sets IN and OUT. He didn't define sets "IN(n)" and "OUT > > (n)". So, you seem to be answering a question he didn't ask. Given > > Stephen's definitions of IN and OUT, is IN = OUT? > > Yes, all elements are the same n, which are finite n. There is a simple > bijection. But, as in all infinite bijections, the formulaic > relationship between the sets is lost. Just to be clear, you are saying that |IN - OUT| = 0. Is that correct? (The vertical lines denote "cardinality".) -- David Marcus
From: David Marcus on 27 Oct 2006 13:26 imaginatorium(a)despammed.com wrote: > Here's something I don't understand. I believe, Tony, that you think > that if every one of these pofnat-labelled balls is inserted one minute > earlier (so *informally*, instead of a "sliver" tapering to zero width, > we have an endless boomerang shape, with the width tending to 1 as you > go ever up the y-direction), then at noon no balls are left. Are you saying that for n = 1,2,..., we should let A_n = -1/floor((n+9)/10) - 1, R_n = -1/n ? > Presumably > because once all the balls are IN (at 11:59), there is only removal, > tick, tick, tick, ... and all are gone at noon. But why doesn't this > stuff about "noon being incompatible" apply here too? Is there a > *principled* way in which you determine which arguments apply at > particular points? (I'm sure it appears to most non-cranks here that > there isn't.) > > Note that in this scenario, at time noon- 1/n, there are, da-dah!, an > infinite number of balls in the vase. So the limit of the number of > balls in the vase at t approaches noon is infinity. Yet you (really?) > think that in this case the vase ends up empty? Do you have any sort of > *mathematical* argument for this (as opposed to intuition and > hand-waving?) -- David Marcus
From: David Marcus on 27 Oct 2006 13:40
Tony Orlow wrote: > David Marcus wrote: > > Tony Orlow wrote: > >> David Marcus wrote: > >>> You are mentioning balls and time and a vase. But, what I'm asking is > >>> completely separate from that. I'm just asking about a math problem. > >>> Please just consider the following mathematical definitions and > >>> completely ignore that they may or may not be relevant/related/similar > >>> to the vase and balls problem: > >>> > >>> -------------------------- > >>> For n = 1,2,..., let > >>> > >>> A_n = -1/floor((n+9)/10), > >>> R_n = -1/n. > >>> > >>> For n = 1,2,..., define a function B_n: R -> R by > >>> > >>> B_n(t) = 1 if A_n <= t < R_n, > >>> 0 if t < A_n or t >= R_n. > >>> > >>> Let V(t) = sum_n B_n(t). > >>> -------------------------- > >>> > >>> Just looking at these definitions of sequences and functions from R (the > >>> real numbers) to R, and assuming that the sum is defined as it would be > >>> in a Freshman Calculus class, are you saying that V(0) is not equal to > >>> 0? > >> On the surface, you math appears correct, but that doesn't mend the > >> obvious contradiction in having an event occur in a time continuum > >> without occupying at least one moment. It doesn't explain how a > >> divergent sum converges to 0. Basically, what you prove, if V(0)=0, is > >> that all finite naturals are removed by noon. I never disagreed with > >> that. However, to actually reach noon requires infinite naturals. Sure, > >> if V is defined as the sum of all finite balls, V(0)=0. But, I've > >> already said that, several times, haven't I? Isn't that an answer to > >> your question? > > > > I think it is an answer. Just to be sure, please confirm that you agree > > that, with the definitions above, V(0) = 0. Is that correct? > > Sure, all finite balls are gone at noon. Please note that there are no balls or time in the above mathematics problem. However, I'll take your "Sure" as agreement that V(0) = 0. Let me ask you a question about this mathematics problem. Please answer without using the words "balls", "vase", "time", or "noon" (since these words do not occur in the problem). First some discussion: For each n, B_n(0) = 0 and B_n is continuous at zero. In fact, for a given n, there is an e < 0 such that B_n(t) = 0 for e < t <= 0. In other words, B_n is not changing near zero. Now, V is the sum of the B_n. As t approaches zero from the left, V(t) grows without bound. In fact, given any large number M, there is an e < 0 such that for e < t < 0, V(t) > M. We also have that V(0) = 0 (as you agreed). Now the question: How do you explain the fact that V(t) goes from being very large for t a little less than zero to being zero when t equals zero even though none of the B_n are changing near zero? -- David Marcus |