The Definition of Points
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From: Brian Chandler on 1 Apr 2007 01:36 stephen(a)nomail.com wrote: > In sci.math Tony Orlow <tony(a)lightlink.com> wrote: > > stephen(a)nomail.com wrote: > >> > >> None of the options mention "size" Tony. What does "size" have > >> to do with a, b or c? > >> > > > Ugh. Me already tell you, nth one is n, then there are n of them. So > > easy, even a caveman can do it. Size is difference between. > > Brilliant Tony. Act like an idiot when backed into a corner. > Did you learn that trick from Lester? Don't think so. You think Lester is acting? Brian Chandler http://imaginatorium.org
From: Brian Chandler on 1 Apr 2007 01:43 ** Sorry, some confusion here *** Brian Chandler wrote: > Tony Orlow wrote: > > Brian Chandler wrote: > > > Tony Orlow wrote: > > > >> I'll give *you* a start, Brian, and I hope you don't have a heart attack > > >> over it. It's called 1, and it's the 1st element in your N. The 2nd is > > >> 2, and the 3rd is 3. Do you see a pattern? The nth is n. The nth marks > > >> the end of the first n elements. Huh! > > >> > > >> So, the property I would most readily attribute to this element Q is > > >> that it is the size of the set, up to and including element Q. > > > > > > Euuuughwh! > > > > Gesundheit! > > > > I seeee! Q is really Big'un, and this all jibes with my > > > previous calculation that the value of Big'un is 16. Easy to test: is > > > 16 the size of the set up to and including 16? Why, 1, 2, 3, 4, 5, 6, > > > 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 - so it is!! > > > Well, that's an interesting analysis, but something tells me there may > > be another natural greater than 16.... > > Indeed. So your "characterization" of Q isn't much use, because it > doesn't distinguish Q from 16. > > > > >> That is, > > >> it's what you would call aleph_0, except that would funk up your whole > > >> works, because aleph_0 isn't supposed to be an element of N. Take two > > >> aspirin and call me in the morning. > > > > > > I would call aleph_0 16, or I would call 16 aleph_0? > > > > > > > If you postulated that there were 16 naturals, that would be a natural > > conclusion. > > > > >> <snip> See above for a characterization of Q. > > > > > > Just to be serious for a moment, what do you understand > > > "characterization" to mean? In mathematics it usually implies that the > > > criterion given distinguishes the thing being talked about from other > > > things. But plainly your "characterization" applies perfectly to 16. > > > (Doesn't it? If not please explain.) What's more, even you agree on a > > > good day that there is no last pofnat - so your claim that Q is > > > somehow something "up to which" the pofnats go is not comprehensible. > > > > > > > > > > The set of all pofnats up to and including 16 constitutes 16 elements. > > The nth is equal to n. > > > > >>> So: > > >>> > > >>> Q has the property of being the last element in an endless sequence > > >>> Q has the property of nonexistence, actually > > >>> > > >>> Now it's your turn. > > >>> > > >> n has the property of being the size of the sequence up to and including n. > > >> > > >>>> Try (...000, ..001, ...010, ......, ...101, ...110, ...111) > > >>> Why? What is it, anyway? > > >> Google 2-adics. > > >> > > > > > > Yes, I know what the 2-adics are. You have written an obvious left- > > > ended sequence ...000, ...001, ... then two extra dots, a comma and an > > > obvious right-ended sequence ...101, ...110, ...111. Are you claiming > > > (perchance!) you have specified a "sequence" that includes all of the > > > 2-adics? In which case, which of ...1010101 and ...0101010 comes > > > first? > > > > Those are both right-ended, if you insist, though they both have > > unending strings of zeros to the right of the binary point. > > No: the elements (2-adics) are right-ended bitstrings, but I was > referring to the sequences you have included in your "Try" expression. > > ...000, ..001, ...010, ... > > is a left-ended sequence. At the left end is ...000, and I can > reasonably assume that the three dots on the right mean that after ... > 010 the sequence continues with ...011 then ...100 then ...101 and so > on, but on a good day even you can see that this sequence has no right > end. Of course, every element in this sequence has the property that > if I look sufficiently to the left in the bitstring I find that I have > reached the leftmost 1, and the remainder of the left-end-less > bitstring consists only of zeros. > > The remainder of the content of your "Try" is: > > ..., ...101, ...110, ...111 > > and this is a right-ended sequence. Again, starting from ...111 on the > right, I can see how to generate the next value on the left, and do > this indefinitely. But again, every element in this sequence has the > property that if I look sufficiently to the left in the bitstring I > find that I have reached the leftmost 0, and the remainder of the left- > end-less bitstring consists only of 1s. > > So if a mathematician wrote something resembling your "Try", it would > include all 2-adics that have an endless string of 0s or and endless > string of 1s to the left. > > But I surmise you claim to have included all of the 2-adics somehow, > so I'm asking you to explain how. I notice that the elements you have > named explicitly are all in "conventional numerical" order, or what we > might call "reverse lexicographical" order. *** No we mightn't. Since all of the elements have either an unending sequence of 0s or of 1s to the left, they are in normal "lexicographic" order extended in the obvious way; we don't need to look at the "left end" of the bitstring, since there isn't one - we look at the part to the left where the digits don't vary any more (i.e. ...0000xxx or ...1111xxx) put the ...000 ones before the ...111 ones, then sort the two-ended remainders of the bitstrings in the normal way. Anyway this does not give us a way to order ...0101010 and ...1010101. > > Note that for reverse-lexicographical order we are using the right end > of the bitstring, so for any element (except the last one ...111!) we > can find the successor by (the obvious extension of) normal binary > addition (to left-end-less strings). > > However, if you wish to claim that ...1010101 and ...0101010 are > somehow both included in this thing - in some sort of hiatus in the > middle of the central five dots, at least you need to say which one > comes first. > > > Which comes > > first, 01 or 10? I think I know. Which is greater, 0.10101010... or > > 0.010101...? > > Uh, yeah? The binary fraction 0.101010... is greater than the binary > fraction 0.0101010... but so what? > The binary fraction 0.100... is greater than the binary fraction > 0.0101111... but ...1110101 comes after ...0001 in (the comprehensible > parts of) your Try above. > > We have been through this before: to provide a *sequence* of two-ended > bitstrings is easy: you use the left end to start the lexicographic > orderings, and you use the right end to generate successors. Although > this is no proof that a set of one-ended strings cannot form a > sequence, it means you cannot rely on hand-waving to assure that it > does. > > Do you want to try again? > > Brian Chandler > http://imaginatorium.org
From: cbrown on 1 Apr 2007 02:56 On Mar 31, 8:44 pm, Tony Orlow <t...(a)lightlink.com> wrote: > cbr...(a)cbrownsystems.com wrote: > > On Mar 31, 5:45 pm, Tony Orlow <t...(a)lightlink.com> wrote: > >> Mike Kelly wrote: > > >>> When we say that a set has cardinality Aleph_0 we are saying it is > >>> bijectible with N. Are you saying it's impossible for a set to be > >>> bijectible with N? Or are you saying N does not exist as a set? > >>> Something else? > >> I have been saying that bijection alone is not sufficient for measuring > >> infinite sets relative to each other. > > > Since it is certainly sufficient for comparing sets by their > > cardinality, I can only ask: what do you mean by "measuring infinite > > sets relative to each other"? There are many, many ways of "measuring" > > one set against another; which do you have in mind? > > Let's examine what '<' means. By all means; that's the correct place to start. Can you provide a definition? > x<y ^ y<z -> x<z. So "<" is a transitive relation. Ok; we established that earlier. But you're being coy as usual: you haven't mentioned /all/ of the properties that you're /actually/ /thinking/ of. Consider me a total blank slate; an alien unfamiliar with your method of writing mathematics. If x < y, can it also be true that y < x? If x < y, then /must/ y > x? Can it be true that x < x? Given distinct x and y, /must/ it be the case that either x < y or y < x? Can I make it simpler for you? Suppose S is a set. Then "<" is a PARTIAL order on S means: (1) For all x, y in S, if x = y then not (x < y). (2) For all x, y in S, if a < b then not (b < a) (3) For all x, y, z in S, if a < b and b < c then a < c To make it a TOTAL order, we add the additional constraint: (4) For all x, y in S, x = y, or x < y, or y < x. Note: There is /nothing whatsoever/ in the above definitions to do with "+", "-", "subset of" and so on; or that S must be a subset of some well-known set like the naturals or the reals, or any other notion. "Is a (blood) descendant of" is a PARTIAL order on all people living and dead. If g is my father's father, and f is my father, and m is me, and I write m < f and f < g for "I am a descendant of my father" and "my father is a descendant of my grandfather", then it follows that m < g: "I am a descendant of my grandfather". But it would be bizzare in the extreme to then claim (as you sometimes do) that therefore m - g < 0 without first very clearly saying what "me - my grandfather is a descendant of 0" is supposed to mean. What person, living or dead, is m - g supposed to be? What person, living or dead, is "0"? Ordering and arithmetic functions like "+" and "-" /don't/ / automatically/ follow one from the other. > True for real > quantities, and true if '<' is taken to mean "is a proper subset of". Yes, but -1.5 is not a proper subset of pi, nor are the sets {a,b} and {a,b,c,d,e} real quantities. You have simply given examples of how the same symbol, "<" can have 2 /different/ meanings; in one case describing a total order, and in the other, a partial order. It is NOT the case that for any subsets A, B of S that either A = B, A < B, or B < A. That's why, in that case "<" is not a total order: it doesn't satisfy constraint (4); but it /does/ obey the other constraints. You need to have ALL FOUR constraints apply to get a TOTAL or linear order. Not just one, or two of them; ALL FOUR of them. When you say something like: x<y ^ y<z -> x<z. you are only mentioning /ONE/ constraint (3); and making us "guess" what other constraints you have in mind. > The proper subset is less than the whole, and the evens are half the > naturals. That's a very primitive result. > If by "primitive" you mean "naive", I agree. It is also a "primitive result" that if a truly, honestly, really, fair coin has come up heads 20 times in a row, it is more likely to come up tails on the next flip than heads. That doesn't make this primitive intuition actually / true/. 20 heads in a row comes up about once in a million trials of 20 flips. There are currently over 6 billion people living on the planet; so one can imagine that /thousands/ of actual /living/ human beings have been confronted with this disturbing situation. Honestly, even I'd be inclined to bet against logic; even believing that the coin was really, truly, cross-my-heart-and-hope-to-die fair. "The evens are half the naturals" does not follow /solely/ from the observation that the evens are a /proper subset/ of the naturals. You need a lot /more/ than that observation to make sense of the statement "the evens are half the naturals". Thus the need to be very /specific/ about what you mean when you say things in a mathematical sense. > >> Yes, NeN, as Ross says. I understand what he means, but you don't. > > > What I don't understand is what name you would like to give to the set > > {n : n e N and n <> N}. M? > > > Cheers - Chas > > N-1? Why do I need to define that uselessness? I find the set of naturals to be a useful concept, whatever you wish it were called. For example, I can say "if n is a member of (the set of naturals), then n has a unique factorization as a product of primes". Doesn't that seem at all useful to you? It's called the Fundamental Theorem of Arithmetic: http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic > I don't want to give a > size to the set of finite naturals... Fine. Then don't give it a size if you don't want to. But it seems rather self-defeating: wasn't your idea to put a consistent "measure" of "size" on /all/ sets? Cheers - Chas
From: cbrown on 1 Apr 2007 03:07 On Mar 31, 10:22 pm, step...(a)nomail.com wrote: > In sci.math Tony Orlow <t...(a)lightlink.com> wrote: > > > step...(a)nomail.com wrote: > > >> None of the options mention "size" Tony. What does "size" have > >> to do with a, b or c? > > > Ugh. Me already tell you, nth one is n, then there are n of them. So > > easy, even a caveman can do it. Size is difference between. > > Brilliant Tony. Act like an idiot when backed into a corner. So easy, even a caveman can do it. Cheers - Chas
From: Mike Kelly on 1 Apr 2007 06:40
On 1 Apr, 01:33, Tony Orlow <t...(a)lightlink.com> wrote: > Mike Kelly wrote: > > On 31 Mar, 16:46, Tony Orlow <t...(a)lightlink.com> wrote: > >> Mike Kelly wrote: > >>> On 31 Mar, 13:41, Tony Orlow <t...(a)lightlink.com> wrote: > >>>> Mike Kelly wrote: > >>>>> On 30 Mar, 18:25, Tony Orlow <t...(a)lightlink.com> wrote: > >>>>>> Lester Zick wrote: > >>>>>>> On Thu, 29 Mar 2007 09:37:21 -0500, Tony Orlow <t...(a)lightlink.com> > >>>>>>> wrote: > >>>>>>>>>> If n is > >>>>>>>>>> infinite, so is 2^n. If you actually perform an infinite number of > >>>>>>>>>> subdivisions, then you get actually infinitesimal subintervals. > >>>>>>>>> And if the process is infinitesimal subdivision every interval you get > >>>>>>>>> is infinitesimal per se because it's the result of a process of > >>>>>>>>> infinitesimal subdivision and not because its magnitude is > >>>>>>>>> infinitesimal as distinct from the process itself. > >>>>>>>> It's because it's the result of an actually infinite sequence of finite > >>>>>>>> subdivisions. > >>>>>>> And what pray tell is an "actually infinite sequence"? > >>>>>>>> One can also perform some infinite subdivision in some > >>>>>>>> finite step or so, but that's a little too hocus-pocus to prove. In the > >>>>>>>> meantime, we have at least potentially infinite sequences of > >>>>>>>> subdivisions, increments, hyperdimensionalities, or whatever... > >>>>>>> Sounds like you're guessing again, Tony. > >>>>>>> ~v~~ > >>>>>> An actually infinite sequence is one where there exist two elements, one > >>>>>> of which is an infinite number of elements beyond the other. > >>>>>> 01oo > >>>>> Under what definition of sequence? > >>>>> -- > >>>>> mike. > >>>> A set where each element has a well defined unique successor within the > >>>> set. > >>> So any set is a sequence? For any set, take the successor of each > >>> element as itself. > >> There is no successor in a pure set. That only occurs in a discrete > >> linear order. > > > What does it mean for an ordering to be "discrete" or "linear"? What > > does it mean for something to "occur in" an ordering? > > Linear means x<y ^ y<z ->x<z Linear means transitive? Very, very strange use of language. Not all transitive orderings are linear (by the normal definitions of those words). So why use that word? This is just going to confuse people. > Continuous means x<z -> Ey: x<y ^ y<z So the ordering on {a, b, c} of a < b, a < c, b < a, b < c, c < a, c < b is "countinuous"? Or were there some more conditions you didn't mention.. > Discrete means not continuous, that is, given x and z, y might not exist. So the set { [0,1] U [2,3] } with the usual ordering on the reals is a discrete ordering? > For something to "occur", it must happen "at some time". In a sequence, > this is defined as after some set of events and before some other > mutually exclusive set, in whatever order is under consideration. This paragraph is completely incoherent. You've used many undefined terms in an attempt to define something. I get the feeling you are *never* going to be able to explain your ideas. There has been a *lot* of work done on desribing various types of orderings. Maybe if you understood the common language used to describe orderings you'd have better luck getting people to understand your ideas. You're violently opposed to the idea of actually learning any math though.. > >>>> Good enough? > >>> You tell me. Did you mean to say "a sequence is a set"? If so, good > >>> enough. > >>> -- > >>> mike. > >> Not exactly, and no, what I said is not good enough. > > >> A set with an order where each element has a unique successor is a > >> forward-infinite sequence. Each can have a unique predecessor, and then > >> it's backward-infinite. And if every element has both a unique successor > >> and predecessor, then it's bi-infinite, like the integers, or within the > >> H-riffics, the reals. One can further impose that x<y ->~y<x, to > >> eliminate circularity. > > >> Good enough? Probably not yet. > > > So when you say "sequence" you're refering to a set and an ordering on > > that set? There are some conditions on the properties of the ordering. > > You're not, as yet, able to coherently explain what those conditions > > are. > > Explain away. What? I don't know what your ideas are. Nobody but you does. It's not even clear that *you* know exactly what you're trying to say, given that you're incapable of explaining your ideas to anyone else. -- mike. |