The Definition of Points
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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
From: Virgil on 1 Apr 2007 00:57 In article <460f2439(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On the finite scale, it takes an infinite number of infinitesimals to > achieve measure. Are infinitesimal units ever equal to their reciprocals, as are all finitesimal units in the reals?
From: Virgil on 1 Apr 2007 00:59
In article <460f246a(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <460f1a41(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > >> > >> You have seen two apples, and three? > >> > > Are apples numbers? > > Naturals apply to objects. That is not an answer to what I asked. And is irrelevant to that question. |