An uncountable countable set
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From: Mike Kelly on 12 Sep 2006 15:15 Tony Orlow wrote: > 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. Two statements : 1) A finite string of 1s represents a (finite) natural number. 2) An infinite string of 1s represents a (finite) natural number. 1) doesn't imply 2). > > > > >> 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. What, even if there isn't a 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. Babble. > >> 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? Standard theory? What maths were you hoping to do with your new scheme that standard theory can't do, again? > >> 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... there are a finite number of strings of 1s? So I should be able to list them all in a finite amount of time... 1 11 111 1111 11111 111111 Wait, before I get into this, when am I going to finish? 1111111 11111111 111111111 1111111111 are you *sure* there a finite number of these things? How long till I get to the end? > >> 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. Define BigUn non-circularly. > > 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. Babble. > >> 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. Odd. Mo
From: Virgil on 12 Sep 2006 15:17 In article <4506dbd2(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David R Tribble wrote: > Wolfgang's and my position is that N is unbounded > but finite But since the power set of any finite set is finite, that makes the set of reals finite, and just about every set finite, including TO's ephemeral "Big'un".
From: Tony Orlow on 12 Sep 2006 15:26 Mike Kelly wrote: > Tony Orlow wrote: >> Mike Kelly wrote: >>> Tony Orlow wrote: >>>> Virgil wrote: >>>>> In article <44fe2642(a)news2.lightlink.com>, >>>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>>> >>>>>> Virgil wrote: >>>>>>> In article <44fd9eba(a)news2.lightlink.com>, >>>>>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>>>>> >>>>>>>> Dik T. Winter wrote: >>>>>>>>> In article <44ef3da9(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> >>>>>>>>> writes: >>>>>>>>> Your axiom uses things that are not defined. What is the *meaning* of >>>>>>>>> "x<z"? >>>>>>>> Geometrically it means that x is left of z on the number line. >>>>>>> And for someone standing on the other side of the number line would x be >>>>>>> on the right of z? >>>>>>> >>>>>>> And does the line stay horizontal as one moves around earth? Which way >>>>>>> is larger if the line ever goes vertical. And how does the "larger" work >>>>>>> at antipodes? >>>>>>> >>>>>> Silly questions. >>>>> In response to a silly definition. >>>>>>>> It means >>>>>>>> A y (y<x ^ y<z) v (x<y ^ y<z) v (x<y ^ z<y). That says about all it >>>>>>>> needs to, wouldn't you say? >>>>>>> Not hardly. >>>>>>> A y (y = x) v (y = z) v (y<x ^ y<z) v (x<y ^ y<z) v (x<y ^ z<y) >>>>>>> is a bit better but still insufficient. >>>>>> True, I should have specified y<>x and y<>z. I guess it's usually done >>>>>> using <= for this reason, eh? >>>>>> >>>>>>>>> > > That is not a definition, because it makes no sense. "The set of >>>>>>>>> > > naturals >>>>>>>>> > > is as large as every natural"? >>>>>>>>> > >>>>>>>>> > It is not larger than all naturals >>>>>>>>> >>>>>>>>> That is something completely different again. >>>>>>>> It's not LARGER than every finite. >>>>>>> Which natural(s) is it "not larger" than", in the sense of not being a >>>>>>> proper superset of that natural or having that natural as a member? >>>>>> ....11111 binary (all bit positions finite) >>>>> Unless that string has only finitely many bit positions as well as only >>>>> finite bit positions, it is not a natural at all, as it is then neither >>>>> the first natural nor the successor of any natural, and every natural >>>>> has to be one or the other. >>>> It is the successor to ....11110. Duh. I've already proven that this is >>>> a finite value, given that all bit positions are finite, and that >>>> therefore no place in that string can achieve an infinite value, and >>>> that any such number has predecessor and successor. The cute thing is >>>> that the successor to ...1111 is 0, and that ...1111 is essentially -1. :) >>> Does it not bother you that nobody else agrees with, or even >>> understands, your proof? >>> >> I find it disappointing, but not surprising, that you don't understand >> such a simple proof, since it's contradictory to your education. I do >> find it annoying that you feel the right to disagree with it without >> understanding it. If you feel there is a problem with the proof, please >> state the logical error I made. If the string is all finite bits, and >> none of them ever can possibly achieve an infinite value, then how can >> the string have an infinite value? There's nowhere in the string where >> that can occur. It's that simple. Grok it. > > 1) A finite string of 1s represents a (finite) natural number. > 2) An infinite string of 1s represents a (finite) natural number. > > 1) doesn't imply 2). > If the string is unbounded but finite, then 2) follows.
From: Tony Orlow on 12 Sep 2006 15:29 Mike Kelly wrote: > Tony Orlow wrote: >> Mike Kelly wrote: >>> Tony Orlow wrote: >>>> Mike Kelly wrote: >>>>> Tony Orlow wrote: >>>>>> Virgil wrote: >>>>>>> In article <44fd9eba(a)news2.lightlink.com>, >>>>>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>>>>> >>>>>>>> Dik T. Winter wrote: >>>>>>>>> In article <44ef3da9(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> >>>>>>>>> writes: >>>>>>>>> Your axiom uses things that are not defined. What is the *meaning* of >>>>>>>>> "x<z"? >>>>>>>> Geometrically it means that x is left of z on the number line. >>>>>>> And for someone standing on the other side of the number line would x be >>>>>>> on the right of z? >>>>>>> >>>>>>> And does the line stay horizontal as one moves around earth? Which way >>>>>>> is larger if the line ever goes vertical. And how does the "larger" work >>>>>>> at antipodes? >>>>>>> >>>>>> Silly questions. >>>>>> >>>>>>>> It means >>>>>>>> A y (y<x ^ y<z) v (x<y ^ y<z) v (x<y ^ z<y). That says about all it >>>>>>>> needs to, wouldn't you say? >>>>>>> Not hardly. >>>>>>> A y (y = x) v (y = z) v (y<x ^ y<z) v (x<y ^ y<z) v (x<y ^ z<y) >>>>>>> is a bit better but still insufficient. >>>>>> True, I should have specified y<>x and y<>z. I guess it's usually done >>>>>> using <= for this reason, eh? >>>>>> >>>>>>>>> > > That is not a definition, because it makes no sense. "The set of >>>>>>>>> > > naturals >>>>>>>>> > > is as large as every natural"? >>>>>>>>> > >>>>>>>>> > It is not larger than all naturals >>>>>>>>> >>>>>>>>> That is something completely different again. >>>>>>>> It's not LARGER than every finite. >>>>>>> Which natural(s) is it "not larger" than", in the sense of not being a >>>>>>> proper superset of that natural or having that natural as a member? >>>>>> ....11111 binary (all bit positions finite) >>>>> That isn't a natural number, Tony. >>>>> >>>> Are you sure? Pay close attention. >>>> >>>> For any finite bit position n, it and all predecessors can only sum to a >>>> finite bit string value of 2^(n+1)-1. >>> OK. Any string 111....1111 with a finite number of 1s represents a >>> natural in binary. >>> >>>> Since there are only finite bit positions in the string, it can never achieve any infinite value at any position in the unending string of bits. >>> OK. Any string 111....1111 with a finite number of 1s represents a >>> natural in binary. >>> >>>> Therefore the value must be finite. >>> Why? You're supposed to be *demonstrating* that the string represents a >>> value, but you're *assuming* it instead. >>> >>> You've shown that any finite bit string of all 1s represents a finite >>> natural number. And concluded from this that an infinite string of 1s >>> represents a finite natural number. Why? Total non sequitur. >> At which bit does this string attain an infinite value? > > It doesn't. I never said it does so please STOP asking that question. I > say that 111..... doesn't represent a (natural) value at all. If it is a whole number (no fractional component), is finite, and has successor and precedessor, then ....1111 is a finite natural. Unless you have another kind of finite positive whole numbers. > >> All bits are >> finite, and have a finite number of predecessors. There is none where >> the string can ever become infinite in value, or even extent. > > Yeah, finite strings are finite and represent naturals, got it. > > Why does that imply that an infinite string represents a natural? Because no bit position is infinite so all points in the string are finite in value. > >>>> Furthermore, since any such number does have a predecessor and >>>> successor, in this case ....1110 and ...0000, respectively, it fits in >>>> the successorship model. The only concept this breaks is that 0 is now a >>>> successor as well, creating an infinite ring of successorship. Other >>>> than that, it works as a natural, and in fact, this is the way signed >>>> integers work in your very computer. >>>> >>>> So, while ...111 may not be considered a standard natural, I see no >>>> reason why it should not be considered, say, an extended natural. >>> And you think this is a better model of "natural numbers" than standard >>> ones? You think it accords better with the intuitive picture people >>> have of what "counting numbers" are? >>> >> Yes, in the infinite realm, it does. It's precisely the binary signed >> integer in the limit as the number of bits approaches oo. > > I laughed out loud here. You think anybody but you on the PLANET would > think that this is relevant to the intuitive, naive idea of what > "numbers" are? > Perhaps. Probably. There are a lot of people on the planet.
From: Virgil on 12 Sep 2006 15:30
In article <4506dbd2(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David R Tribble wrote: > > Tony Orlow <tony(a)lightlink.com> 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. What TO "does not see" in logic and mathematics is immensely greater than what little he does see. > > > > Which presumably requires some extended Peano axioms in order > > to exist? > > > > No, it just requires a loosening of the requirement that we only look at > the minimal set satisfying those axioms. There is such a minimal set satisfying the axiom of infinity, and that set is the set of naturals. There are other such sets possible as well , but all of then are necessarily limit ordinals, at least up to order isomorphism, which do NOT work the way TO tries to make them work. > I don't see that it contradicts > them as they stand. I repeat, what TO "does not see" in logic and mathematics is immensely greater than what little he does see. > Extensions of the naturals such as this also fit > into that model. Only limit ordinals. |