An uncountable countable set
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From: Tony Orlow on 27 Oct 2006 20:09 Lester Zick wrote: > On Fri, 27 Oct 2006 01:37:04 -0400, David Marcus > <DavidMarcus(a)alumdotmit.edu> wrote: > > [. . .] > >> It is interesting that when we try to ask Tony a question that doesn't >> mention balls or vases or time, his answer involves balls, vases, and >> time. I'm afraid to ask what 1 + 1 is because the answer might be "noon >> doesn't exist". > > So if the definition for "1+1" entails "1(x)+1(x)" "balls" don't lie > in the "domain of discourse" for "1+1"? Curious to say the least. > > ~v~~ 1+1=2, of course, and noon doesn't exist. I think that's what David wants to hear. It makes him feel good. :) Tony
From: Tony Orlow on 27 Oct 2006 20:13 Lester Zick wrote: > On Fri, 27 Oct 2006 00:07:16 -0400, Tony Orlow <tony(a)lightlink.com> > wrote: > >> MoeBlee wrote: >>> Tony Orlow wrote: >>>> I share your and Godel's concerns about point set theory >>> Oh how rich. How veddy veddy scholarly Mr. Orlow sounds when he says >>> such things, "I share Godel's concerns about point set theory." Too bad >>> Mr. Orlow doesn't know a single ding dang thing about Godel, or Godel's >>> concerns, or mathematical logic, or set theory, or point set topology, >>> or topology. >>> >>> MoeBlee >>> >> Wow, Lester's really getting under your skin, isn't he? He cracks me up. :) > > Of course a little levity, like motions to adjourn, is always in > order, Tony. Mathematikers take all this drollery way too seriously. > They get all huffy and self righteous when forced to actually explain > things they're used to outsourcing to books. Thanks for the comment. > > ~v~~ A motion to adjourn!! Now that's an idea! Thanks, Lester! Damn court!! Smiles, Tony
From: Virgil on 27 Oct 2006 20:18 In article <45429c69(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > cbrown(a)cbrownsystems.com wrote: > What changes the "number" of balls, if not "additions" and "subtractions"? Insertions and removals! > And, what if, for all t in [t1,t0), there are balls? Then balls can only > disappear entirely at t0. Does TO object to having a real function, f, such that f(x) = 0 if and only if x >= 0? > Sure, the question is put in terms proscribed by the constraints of the > problem. Nothing can occur at noon. It cannot become empty then, nor before. That is like saying f(x) = x cannot be 0 at x = 0 because it is not zero anywhere else. > That's what I was saying. Oy. Noon is not consistent with the problem. > The question is nonsensical given the parameters. The only parameter is time, and every time is measured from what TO claims does not exist. > > Yeah. Noon doesn't happen. Then no part of the problem can happen .. The only relevant question is "According to the rules set up in the problem, is each ball which is inserted into the vase before noon also removed from the vase before noon?" An affirmative answer confirms that the vase is empty at noon. A negative answer directly violates the conditions of the problem. How does TO answer?
From: Virgil on 27 Oct 2006 20:21 In article <45429dd1(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > cbrown(a)cbrownsystems.com wrote: > The universe is the product of numbers. :) > > TOEknee LEGparts is wrong! As Usual!! The universe might as easily be a sum of numbers quite as being a product of them, but seems to involve things which are not prurely numbers, too.
From: David Marcus on 27 Oct 2006 20:21
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? 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. > > 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. > > 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. > > 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? > > I'll consider answering that when you correct the errors above. Sorry. -- David Marcus |