An uncountable countable set
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From: Dave L. Renfro on 12 Sep 2006 12:41 Tony Orlow wrote: [something I didn't read] If no one beats me to it, my post will be post 2006 in this thread (by chronological order) archived by google! Dave L. Renfro
From: Tony Orlow on 12 Sep 2006 12:42 stephen(a)nomail.com wrote: > imaginatorium(a)despammed.com wrote: > >> stephen(a)nomail.com wrote: >>> imaginatorium(a)despammed.com wrote: >>>> stephen(a)nomail.com wrote: >>>>> Tony's argument seems to be a upside-down inside-out version >>>>> of one of Zeno's paradoxes. >>>>> >>>>> ...111111 (in binary) = 1 + 2 + 4 + 8 + ... >>>>> >>>>> All the numbers on the right are finite, so the sum must >>>>> be finite. Apparently a sum can only be infinite if one of the >>>>> terms in the sum is infinite. An infinite sum of finite terms >>>>> cannot possibly be infinite in Tony's mind. >>>> Not entirely sure about this. The problem above is that we are >>>> restricting the string >>>> ...11111 >>>> so that every 1 is in a finite position. The position numbers never >>>> achieve true infinity. >>> Yes, which means that each term in the sum >>> 1 + 2 + 4 + 8 + ... >>> never achieves true infinity. Apparently because no single >>> term in the sum is infinite, the sum itself cannot be infinite. >>> If you had a 1 in an infinite position p, then the sum would >>> contain the term 2^p, which would be infinite, because p >>> is infinite, and the sum would be infinite. It certainly sounds like the inductive proof that no natural is infinite, since each is only 1 greater than the last, and incrementing a finite always results in a finite. That same argument structure could be applied to adding 2^n, which is always finite, producing a finite sum. Since this applies to each and every bit position indexed with a finite natural, it would seem it applies to each and every location within the string, no? There is no possible point in the string where it can become infinite. > >> Hmm, I don't think you're quite getting into the spirit of this. > > I am trying to figure out what Tony's misconceptions are. > Occassionaly I just make fun of them, but at other times > I try to see if I can figure out his intuition. Wolfgang and I see the same problem. You start with 0 and start enumerating the naturals using increment to generate the next successor. Increment is defined as the addition of 1. So, the first is 1, the second is 2, etc, etc. Where you have a set of consecutive naturals starting at 1, the nth is n, the sum of n increments, and the size of the set is always equal to the greatest value in the set. So, if you claim the size of the set is aleph_0, then that is also the greatest natural, and is therefore finite by definition. Wolfgang and I object to this being called infinite, since the identity relation between element count and value in the naturals makes it impossible for the set to be infinite without containing any infinite values. > >> In >> Tony's scheme (insofar as I understand it, and of course it appears to >> me to be as contradictory as it does to you), I believe you can count >> numbers 1, 2, 3, ... and so on, and go on for ever. > > Yes, I have the same understanding of Tony's scheme. Do you have a different scheme? > >> But at some foggy >> point (perhaps after about ever has elapsed) you find that you have >> arrived in the infinite zone, and are counting numbers which, in an >> infinite binary notation, have nonzero digits at least close to the >> (nonexistent) left end. One of these numbers is called Big'un, and if >> you write the sum > >> 1 + 2 + 4 + 8 + ... + 2^Big'un + 2^(Big'un+1) + ... > >> then it _is_ infinite (in Tony's scheme), because at least one of the >> numbers in the sum has achieved genuine infinity. > > This seems to be the same understanding I have. It's the difference between potential and actual infinity. Given N=S^L, the only way to have an actually infinite number of strings in your number system (N) is either to have strings of actually infinite length (L), or a set of symbols (S) which is actually infinite. Since you have a finite alphabet (0 and 1) and only allow finite strings, you cannot have an infinite set of strings in the set. > >> So in Tonyspeak, you have to quantify 1+2+4+... by saying what sort of >> zone the dots extend to. Precisely. You need to state a value range over which sets will be compared, as a variable which can assume infinite values like Big'un. > > In this particular case we are talking only about the finite values. > So 1+2+4+... for all the finite values. According to Tony this > must be finite, because the sum of finite values, even an endless > sum of finite values, must be finite. Since the "endless" sequence is unbounded but only potentially infinite, the sum is potentially, but not actually, infinite. If you are allowed to have an infinite iteration in that sum, say step #Big'un, you will be adding an infinite value to the sum, 2^Big'un. So, they go hand in hand, the count and the value of the term in the sum. > >> I think. > > The point of my post is that Tony's intuition is sort of the > opposite of the intuition that makes Zeno's paradoxes so compelling. > The idea that an infinite number of finite numbers can "sum" to > a finite value is a bit counterintuitive. The idea that you can > keep adding to something, and that it keeps getting bigger, but > that it remains finite, has puzzled (some) people for millenia. > Tony has twisted this around, and insists that an endless (infinite) > sum of finite numbers must be finite. If no term has an infinite number of predecessors, then I don't consider it an actually infinite sum. > > Tony's intuition is based on the idea that numbers are processes. > ...11111111 = 1 + 2 + 4 + 8 + 16 + ... (all finite powers of 2) > If I add the numbers on the right one at a time, I will always > have a finite number: 1, 3, 7, 15, 31, ... Tony therefore > concludes that the right hand side must be finite. There > is no step at which it "becomes" infinite. This is true. Yes, there is no bit in that string where the sum becomes infinite. > Neither is there a step at which I have added all the numbers > on the right. ...11111111 equals the sum of all the numbers > on the right. So according to Tony's logic (if he was consistent), > there is no step at which the sum on the right becomes the value > on the left. So either the two are not equal, in which case > the fact that the right hand side is "finite" tells us nothing > about the value on the left. Or the right hand side "becomes" > infinite o
From: Tony Orlow on 12 Sep 2006 12:45 Dik T. Winter wrote: > In article <45005670$1(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > > Dik T. Winter wrote: > ... > > > But you stated "that you can specify which finite number of iterations" > > > etc. > > > That is something different. A number is computable when there is a Turing > > > machine such that, when given an arbitrary number n, it will calculate the > > > 1-st through n-th digit. There is nothing about a specification of a > > > finite number of iterations. > > > > Well, if you set up a Turing machine such that it will take an infinite > > number of operations to get to any finite digit, then you have probably > > not designed it well. > > You do not seem to understand. If you set up a Turing machine such that > given a number n as input calculates the first n digits of some number > can be very well designed. But can you proof that it will stop? Depending on what number you are calculating, you should be able to prove that it will reach a certain degree of accuracy in a certain number of steps. I'm not sure how to answer that question in such general terms.
From: Tony Orlow on 12 Sep 2006 12:46 Dik T. Winter wrote: > In article <45004b9b$1(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > > Dik T. Winter wrote: > ... > > > > So, while ...111 may not be considered a standard natural, I see no > > > > reason why it should not be considered, say, an extended natural. > > > > > > Why not use the proper name such numbers already have? 2-adics. > > > > No reason, except that 2-adics could possibly have infinite bit > > positions. > > They have not. I have yet to come across basic objects in mathematics that > have infinite posititions in their representation. That's why I invented the T-riffic numbers. It's about time. :) Tony
From: Tony Orlow on 12 Sep 2006 12:49
Virgil wrote: > In article <45005c7c(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > > >> The universe is always expanding at the speed of light. > > How does TO claim to know this? By what yardstick does he measure the > daily diameter of the universe to work out how fast it is expanding? The one that causes electromagnetic radiation to radiate in the first place. The farthest stars we see recede at close to the speed of light, and their distances are close to the distance they would be, given the age of the universe. So, the opposite end appears to be receding at such a speed. It just makes sense. :) |