An uncountable countable set
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From: Virgil on 12 Sep 2006 16:00 In article <4506e500(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > 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. > If the universe is already infinite, or very much larger that the parts we can see would indicate, then its apparent "expansion" may be an illusion, as all we can then say is that those parts of the universe we can see are receding from each other. There could be other parts condensing for all we know.
From: Mike Kelly on 12 Sep 2006 16:43 Tony Orlow wrote: > 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. What's a finite but unbounded infinite string? -- mike
From: Mike Kelly on 12 Sep 2006 16:47 Tony Orlow wrote: > 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. It isn't. Now what? >Unless you > have another kind of finite positive whole numbers. Nope. ...11111 isn't any kind of finite positive whole number. > >> 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. Yes, all finite bit strings are finite and represent naturals. Why does that imply that an infinite string represents a natural /at all/? -- mike.
From: Virgil on 12 Sep 2006 16:55 In article <450709c9(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > > 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. In standard mathematics there is no such thing as an unbounded finite string. TO's alleged system does not yet exist, so there are no unbounded finite strings in it either, other than as vaporware.
From: Virgil on 12 Sep 2006 16:59
In article <45070a8f(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > > 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. In standard mathematics, a sequence of digits with an endless subsequence of 1's does not represent any finite "whole" number. Since TO's alleged system does not yet exist, nothing can exist in it. Thus TO's allegations are self-contradictory in every extant system. |