The Definition of Points
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From: cbrown on 1 Apr 2007 00:22 On Mar 31, 8:38 pm, Tony Orlow <t...(a)lightlink.com> wrote: > cbr...(a)cbrownsystems.com wrote: > > On Mar 31, 5:33 pm, Tony Orlow <t...(a)lightlink.com> wrote: > >> Mike Kelly wrote: > >>> 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 > > > Funny; everyone else calls that "a transitive relation". > > Yes, is that unrelated? > It's certainly /necessary/ for "<" to be a partial order; but it's not sufficient. Just like it's necessary for my car to have gasoline to run; but not sufficient. A partial order is transitive; but not every transitive relation is a partial order. See: http://en.wikipedia.org/wiki/Partial_order It's just three simple rules, man. Sheesh! > > This ordering satisfies, for all x,y,z in S: if x<y and y<z, then x<z. > > > This is not what most people mean when they say "a linear ordering". > > Instead, it's an example of what people usually call a partial order. > > Okay, but that's an ordering that is based on some finite set of rules > regaring some finite set of points, which doesn't suffice to specify the > relationships between every pair of points. We can't say, from the > specified relationships, whether f<b or f<c or b<g or c<g. Right. That's why it's called a /partial/ order, and not a /total/ order; there are elements which are incomparable - i.e., they cannot be compared in the ordering. > That's why > there's a parallel route, and so the diagram is "nonlinear". It could be > all on a line, but there would be several possible ordering given the > stated relationships. > And sometimes, depending on the ordering, there is no particularly useful way to extend that ordering. Consider the subsets of {a, b, c}, ordered by inclusion. I can say that every subset A <= {a, b, c} in this ordering; and I can say that {} <= A for every subset A; but some subsets can;t be compared in this ordering; for example, {a,b} and {b,c}. That's a perfectly reasonable state of affairs; not every partial order is somehow "required" to be a particular canonical total order. > > See: > > >http://en.wikipedia.org/wiki/Partial_order > > >> Continuous means x<z -> Ey: x<y ^ y<z > > > Is "<" a partial order? a pre-order? a total order? Unless you > > specify, I might say that in a triangle, the third vertex is "between" > > any two given distinct vertices. > > Uh oh. Yup. If by "uh oh", you mean "he's actually asking me to /think/ about what the heck I'm /saying/". Uh oh, indeed! > > >> Discrete means not continuous, that is, given x and z, y might not exist. > > > So [0,1) u (1,2], with the usual ordering of the reals, is a > > "discrete" ordering? > > I appears to be the union of two discrete sets, mutually exclusive, and > without mutual continuity. I'm going to guess that what you wanted to say was "... the union of two continuous sets, ...". But who the heck knows? > I'd say if you break the real line into > (x,x+1] for xeZ, those are discrete partitions of R. > That's not a definition: it's an /example/. If I ask you to define "Chassian number", the response "3 is an Chassian number" is not a response which is a definition. > What I said was that a discrete order will have pairs of elements which > have no elements between them, whereas a continuous order will not. But, > I'm sure I'm wrong. :) > Not so much wrong as inconsistent to the point of incomprehensibility. There are /no/ two real numbers x, y in the set [0,1) union (1, 2] with x < y such that there exists no element z in [0,1) union (1,2] with x < z < y. 1 is /not/ an element of [0,1) union (1,2]. So when you say "discrete partions of R", you either mean something / different/ from saying "the ordering on that partition is discrete", or else you don't have a good sense of what you really mean yourself when you say "discrete". > >> For something to "occur", it must happen "at some time". > > > Does "1 + 5" "occur", i.e., happen, "at some time" different than when > > "2 + 4" "occurs"? > > "1 + 5" occurred earlier in that sentence than "2 + 4" did. A sentence > is a kind of sequence. Thanks for the example. > Yes, a sentence is indeed a kind of sequence. Because otherwise: "a a is of yes kind indeed sentence sequence" would mean the same thing. However, it is not a /definition/ of a sequence, nor of a sort of sequence. > >> 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. > > > Is "1+1" an "event" which "occurs" or "happens" at some "time"? When > > is that time? Has it already "occured"? > > I think I just saw it. Look! Up there.... > Funny. But it also shows that you can't actually answer: what "time" does 2^pi "happen" in the function f: R->R defined by f(x) = 2^x? > >> Oh gee, there has to be some word for it... > > > There almost certainly is; but as usual, it depends on what the /heck/ > > you're talking about. Perhaps the words "well-order" or "total order" > > actually already satisfy your requirements; or some particular proper > > subset of all non-isomorphic total orders satisfy your requirements. > > Or not. But how will you ever know if you refuse to /learn/ what these > > words refer to? > > > Cheers - Chas > > I guess not by asking around here. Geeze. > I can't give you proper directions to Springfield, until you tell me / which/ Springfield you actually want to go to. Cheers - Chas
From: Virgil on 1 Apr 2007 00:40 In article <460f1e93(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <460f00a0(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Look back. The nth is equal to n. Inductive proof holds for equality in > >> the infinite case > > > > Not in vN. > > I know that statement is not generally acceptable. I don't care. It's > true. Not in ZF or NBG, so just where does TO claim it is true? > Infinite-case induction has not been disproved It has not been proved, which is more to the point. And without a clear statement of what it says, it cannot be. > > And inductive proofs do not work that way. One can prove by > > induction that something is true for each natural, but that does not > > create any infinite naturals for which it is true. > > You can agree with your buddies these are the rules of your club, and > you can hang around the pickup drinking cans of beer and talking about > trucks and ladies, but somes of us gots interstates to travel, and, you > know, big cities to deliver to. The keep on truckin', since you don't seem to be good for anything else. > > Where an equality is proven true for all x>k, then any positive infinite > value is greater than any finite k, and the proof holds. Simple clear > logic. Refute, please. Refute what? So far you have produced nothing worth refuting.
From: Virgil on 1 Apr 2007 00:42 In article <460f1ef1(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <460f0317(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > > > >> There are not zero, nor any finite number of reals in (0,1]. > > > > There are every finite and more of reals in (0,1]. > > You mean any sequential ordering of the reals in (0,1] will contain > elements in finite positions, plus more. Same thang. Tru dat, yo. What I mean, and what TO misrepresents me to mean have nothing in common.
From: Virgil on 1 Apr 2007 00:53 In article <460f22e6(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > > > The obvious question is why haven't /you/ studied them; instead of > > making vague and uninformed statements about them (regardless of what > > you choose to call these ordered sets). > > > > The question is, "is there an acceptable term with which to refer to > such uncountable linearly ordered sets?" The set of real numbers, whether with or without infinitesimals is an uncountable linearly ordered set, but of course not discretely ordered. And I cannot believe that TO, who is usually quite inventive, if not always accurate, cannot create one. > > >>>> However, where every element of a set has a well defined > >>>> successor and predecessor, it's a sequence of some sort. Not necessarily. If a set is partitioned into two or more subsets each with such an order on it, but with no order between partitions, then the set itself is not even an ordered set even though every member has a well defined predecessor and successor.
From: Brian Chandler on 1 Apr 2007 00:54
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. 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 |