An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Tony Orlow on 31 Oct 2006 10:54 David Marcus wrote: > Tony Orlow wrote: >> David Marcus wrote: >>> Tony Orlow wrote: >>>> David Marcus wrote: >>>>> Tony Orlow wrote: >>>>>> David Marcus wrote: >>>>>>> Tony Orlow wrote: >>>>>>>> David Marcus wrote: >>>>>>>>> Tony Orlow wrote: >>>>>>>>>> David Marcus wrote: >>>>>>>>>>> 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. >>>>>>>>>> Okay. >>>>>>>>>>> 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). >>>>>>>>>> I'll try. >>>>>>>>>>> First some discussion: For each n, B_n(0) = 0 and B_n is >>>>>>>>>>> continuous at zero. >>>>>>>>>> What??? How do you conclude that anything besides time is >>>>>>>>>> continuous at 0, where yo have an ordinal discontinuity???? >>>>>>>>>> Please explain. >>>>>>>>> I thought we agreed above to not use the word "time" in >>>>>>>>> discussing this mathematics problem? >>>>>>>> If that's what you want, then why don't you remove 't' from all >>>>>>>> of your equations? >>>>>>> It is just a letter. It stands for a real number. Would you >>>>>>> prefer "x"? I'll switch to "x". >>>>>> It is still related to n in such a way that x<0. >>>>>>>>> As for your question, let's look at B_2 (the argument is >>>>>>>>> similar for the other B_n). >>>>>>>>> >>>>>>>>> B_2(t) = 1 if A_2 <= t < R_2, >>>>>>>>> 0 if t < A_2 or t >= R_2. >>>>>>>>> >>>>>>>>> Now, A_2 = -1 and R_2 = -1/2. So, >>>>>>>>> >>>>>>>>> B_2(t) = 1 if -1 <= t < -1/2, >>>>>>>>> 0 if t < -1 or t >= -1/2. >>>>>>>>> >>>>>>>>> In particular, B_2(t) = 0 for t >= -1/2. So, the value of B_2 >>>>>>>>> at zero is zero and the limit as we approach zero is zero. So, >>>>>>>>> B_2 is continuous at zero. >>>>>>>> Oh. For each ball, nothing is happening at 0 and B_n(0)=0. >>>>>>>> That's for each finite ball that one can specify. >>>>>>> I thought we agreed to not use the word "ball" in discussing this >>>>>>> mathematics problem? Do you want me to change the letter "B" to a >>>>>>> different letter, too? >>>>>> Call it an element or a ball. I don't care. It doesn't matter. >>>>>>>> However, lim(t->0: sum(B_n| B_n(t)=1))=oo. Why do you >>>>>>>> conveniently forget that fact? >>>>>>> Your notation is nonstandard, so I'm not sure what you mean. Do >>>>>>> you mean to write >>>>>>> >>>>>>> lim_{x -> 0-} sum_n B_n(x) = oo ? >>>>>>> >>>>>>> If so, I don't understand why you think I've forgotten this fact. >>>>>>> If you look in my previous post (or below), you will see that I >>>>>>> wrote, "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." >>>>>> Then don't you see a contradiction in the limit at that point >>>>>> being oo, the value being 0, and there being no event to cause >>>>>> that change? I do. >>>>>>>>>>> In fact, for a given n, there is an e < 0 such that B_n(t) = >>>>>>>>>>> 0 for e < t <= 0. >>>>>>>>>> There is no e<0 such that e<t and B_n(t)=0. That's simply false. >>>>>>>>> Let's look at B_2 again. We can take e = -1/2. Then B_2(t) = 0 >>>>>>>>> for e < t <= 0. Similarly, for any other given B_n, we can find >>>>>>>>> an e that does what I wrote. >>>>>>>> Yes, okay, I misread that. Sorry. For each ball B_n that's true. >>>>>>>> For the sum of balls n such that B_n(t)=1, it diverges as t->0. >>>>>>>>>>> In other words, B_n is not changing near zero. >>>>>>>>>> Infinitely more quickly but not. That's logical. And wrong. >>>>>>>>> Not sure what you mean. >>>>>>>> The sum increases without bound. >>>>>>>>>>> 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). >>>>>>>>>
From: Tony Orlow on 31 Oct 2006 12:39 David R Tribble wrote: > Tony Orlow wrote: >> Apparently you are not aware of my >> position on the subject. Bijections alone do not prove equinumerosity >> for infinite sets. Cardinality is a rough measure of equivalence class, >> not a precise measure of the size of a set. In order to precisely >> compare such infinite sets of values, one must measure over a common >> infinite value range formulaically. Then we easily get that half the >> naturals are even, and other pleasant, intuitive notions. > > Consider the bijection > f(n) = n^2 + 2n - 1, for all natural n>1. > Obviously, this bijects all n>1 in N to some n>1 in N. > > n |f(n) > ---|---- > 1 | 2 > 2 | 7 > 3 | 14 > 4 | 23 > : | : > > Let's say that f maps from set N1 to set Nf. So now what does f > say about the "formulaic measure comparison" of the sizes of sets > N1 and Nf? How much "bigger" is N1 than Nf? > For n>=1, f(n)>n^2 because 2n-1>0, and f(n)<n^2+2n because -1<0. It's between log2(n) and something smaller, the inverse of n^2+2n. I was trying to think about the inverse of such functions at one point, and what the rules were, but I haven't finished that. Thanks for the reminder. Maybe you have a suggestion? Tony
From: Tony Orlow on 31 Oct 2006 12:40 MoeBlee wrote: > Tony Orlow wrote: >> Intuitionistic logicians reject that a false premise >> implies anything. > > Name such an intuitionistic logician and the work in which this > appears. > > Intutitionistic logic DOES have the principle > > For all formulas P, > > f -> P > > ('f', the 'falsehood' symbol is often a primitive of intuitionistic > (and certain formulations of classical) logic). > > Youv'e got it wrong, shooting of your big mouth on that which you know > nothing, as usual. > > MoeBlee > Yeah I misspoke. Sorry.
From: Tony Orlow on 31 Oct 2006 12:45 MoeBlee wrote: > Tony Orlow wrote: >> I am beginning to realize just how much trouble the axiom of >> extensionality is causing here. > > Oh, now the axiom of extensionality. > > When you buy into Robinson's non-standard analysis you buy into the > axiom of extensionality, and all the other axioms of set theory, and > mathematical logic - the whole kit and kaboodle - including the axiom > of choice, ordinals, and uncountable cardinals, and all the > "transfinitology" (even if not with platonistic committments) you so > strenuously disclaim. > > MoeBlee > Dear Moe - When I say there are problems with the axiom of extensionality, I refer to the application of the fact that two sets, when viewed statically, contain the same elements. Yes, that means that, without regard to time or order or anything else, the sets are, theoretically, the same. I don't dispute that. But, I do dispute the application of that fact to the exclusion of specifically stated time constraints and their resulting definition of iterations. I am not saying that the axiom itself is wrong, as a definition. It just doesn't capture the essence of what's going on. Static sets are not sequences of arithmetical events. Do you disagree? TOE knee
From: Tony Orlow on 31 Oct 2006 12:48
David Marcus wrote: > Tony Orlow wrote: >> Randy Poe wrote: >>> Tony Orlow wrote: >>>> stephen(a)nomail.com wrote: >>>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>>>> stephen(a)nomail.com wrote: >>>>>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>>>>>> stephen(a)nomail.com wrote: >>>>>>> <snip> >>>>>>> >>>>>>>>> What does that have to do with the sets IN and OUT? IN and OUT are >>>>>>>>> the same set. You claimed I was losing the "formulaic relationship" >>>>>>>>> between the sets. So I still do not know what you meant by that >>>>>>>>> statement. Once again >>>>>>>>> IN = { n | -1/(2^(floor(n/10))) < 0 } >>>>>>>>> OUT = { n | -1/(2^n) < 0 } >>>>>>>>> >>>>>>>> I mean the formula relating the number In to the number OUT for any n. >>>>>>>> That is given by out(in) = in/10. >>>>>>> What number IN? There is one set named IN, and one set named OUT. >>>>>>> There is no number IN. I have no idea what you think out(in) is >>>>>>> supposed to be. OUT and IN are sets, not functions. >>>>>>> >>>>>> OH. So, sets don't have sizes which are numbers, at least at particular >>>>>> moments. I see.... >>>>> If that is what you meant, then you should have said that. >>>>> And technically speaking, sets do not have sizes which are numbers, >>>>> unless by "size" you mean cardinality, and by "number" you include >>>>> transfinite cardinals. >>>> So, cardinality is the only definition of set size which you will >>>> consider.....your loss. >>>> >>>>> In any case, it still does not make any sense. I am not sure >>>>> what |IN| is for any n. IN is a single set. There is only >>>>> one set, and it does not depend on n. In fact, there isn't >>>>> an n specified in the problem. Yes I used the letter n in >>>>> the set description, but that does not define an entity named 'n'. >>>>> >>>> There most certainly is an 'n'. The problem describes a repeating >>>> process, each repetition of which is indexed with a successive n in n, >>>> and during each repetition of which ball n is removed. What do you mean >>>> there's no n??? >>> The ORIGINAL problem. This is a new one, inspired by >>> the original, but it is one with no balls, no vases, no >>> time steps, no iterations. Just a definition of two subsets >>> of the natural numbers, one called IN and one called >>> OUT. >>> >>> The definition of the set IN does not include a definition >>> of something called IN(n). >>> >>> You are being asked to characterize these two subsets. >>> >>> - Randy >>> >> Set-theoretically, they are the same set, by the axiom of >> extensionality. That doesn't mean the axioms of ZFC account for all the >> information in the problem. There is the combining of +10 and -1 in each >> iteration n at time t=-1/n, a coupling that is not being addressed by >> your method. You are considering the two sets statically, outside of >> time, as completed, but for any n, the max of in is 10 times the max of >> out. Just because these both have some limit at oo, even though they >> don't reach it, doesn't mean they reach it at the same time. They DON'T >> reach it, and if they did, if noon occurred, in would reach it in 1/10 >> the time as out. But,there is no such ending to the finites, and so the >> set-theoretic approach using the set N is bogus at its core. > > It is remarkable that you seem to be unable to answer any post without > mentioning the ball and vase problem. Why is this? Are you afraid that > if you do, you will be trapped into an inconsistentcy? > > Is the following an accurate description of what you are saying? > > 1. You agree that (given the definitions above) IN = OUT and that |IN > \OUT| = 0. > > 2. You don't agree that, in the ball and vase problem, the number of > balls in the vase at noon is |IN\OUT|. > 2 |