The Definition of Points
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From: MoeBlee on 19 Apr 2007 13:33 On Apr 19, 7:13 am, Tony Orlow <t...(a)lightlink.com> wrote: > My problem with it is that it violates certain notions which are central > for finite sets, but discarded for infinite sets. 1+omega=omega>omega+1 Is that '+' ordinal addition or cardinal addition? And what exact ordering do you mean by '>'? > violates the principle tat a+b=b+a. I don't like that. Commutativity DOES hold for cardinal addition. It doesn't hold in general for ordinal addition, since ordinal addition has to do with structure. MoeBlee
From: MoeBlee on 19 Apr 2007 13:52 On Apr 19, 7:18 am, Tony Orlow <t...(a)lightlink.com> wrote: > Virgil wrote: > Can one partition an uncountable set into a countable set of countable > partitions? Only if one assumes Choice, right? A countable set of countable partitions? Maybe you mean a countable partition such that each member of the partition is countable. Anyway, with choice we prove that for an uncountable S there is no such partion of S as just described. You seemed to have gotten this backwards. We do NOT use choice to prove that an uncountable set can be partitioned into countably many countable sets. We use choice to prove we canNOT do that. > > Why not? All any well ordering of any set requires is that every > > non-empty subset of that set have a first member under that ordering. > > I thought it required that there be no infinite descending chains within > the well order. Does that not include an infinite descending chain of > limit elements? Come on! Read a damn textbook already! You can't do mathematics by just mis-gleaning various vague meanings from what certain terminology suggests to your imagination. > >> I just don't see that a well order > >> can be accomplished in principle on an uncountable set, whether one > >> declares an axiom to that effect of not. > > > There is a critical difference in mathematics between proving something > > exists and instantiating it. > > But, you don't PROVE it exists with Choice. You ASSUME it exists > axiomatically without any real justification that I can tell. No, we PROVE that choice entails well ordering. Of course, adopting choice as an axiom is or is not justified by one's intutions or arguments as to what should or should not be an axiom. But that choice entails well ordering is PROVEN. > > Given the AOC, one can trivially prove that for every set there exists a > > well ordering. Addressed to Virgil: Virgil, with all due respect (I don't want to get into another protracted argument about terms such as 'trivial'), I'm wondering whether perhaps you're thinking of the triviality of proving choice from well ordering. Yes, choice follows from well ordering with but a line or two of argument; but proving well ordering from choice is quite a bit more involved, usually requiring proving Zorn's lemma from choice and then well ordering from Zorn's lemma and/or using transfinite recursion with a fair amount of additional argument. > > Actually constructing an explicit well ordering may be a > > whole different ball of wax. Right. > It's plainly impossible for an uncountable set. Orlow, you say, "It's plainly impossible". What a willful ignoramus you are. > If you disagree with > that statement, by all means prove me wrong with a counterexample. Any uncountable ordinal is well ordered by the membership relation on that ordinal. Moreover, it is a theorem of ZFC that every set can be well ordered. Moreover, it is not required to show a counterexample just to note that YOU have not proved that there cannot be a well ordering on an uncountable set, which is to say that you have not proven the negation of the well ordering theorem. MoeBlee
From: David R Tribble on 19 Apr 2007 15:03 Lester Zick wrote: >> Tony, time for you to do a little work for yourself. I've already gone >> through this. You describe for me the mechanics of using binary truth >> values and I explain to you I'm interested in truth not binary truth >> values and how to ascertain truth in mechanical terms initially and >> not how to work with truth values mechanically once ascertained. > Tony Orlow wrote: > So, you are of the opinion that science can be performed without > collecting any data, doing experiments or studies, and ascertaining > truth from fact? Then you are in as much a religious limbo as the > Cantorian ball replicators. How far has this Ivory Tower approach gotten > you so far? If you get down the mechanics of deduction, then you can > consider more readily what underlying principles may be causing whatever > phenomena you're investigating. I've got to admit that this is one of the few guilty pleasures I get from reading sci.math: seeing two cranks arguing with each other over the meaning of "truth".
From: Brian Chandler on 19 Apr 2007 15:22 Tony Orlow wrote: > Virgil wrote: > > In article <46266bb2(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Mike Kelly wrote: > >>> On 18 Apr, 06:17, Tony Orlow <t...(a)lightlink.com> wrote: > >>>> Mike Kelly wrote: > >>>> > >>>> < snippery > > >>>> but in my opinion doesn't produce believable results. > >>> Which "results" of cardinality are not believable? You don't think the > >>> evens and the naturals and the rationals are mutually bijectible? Or > >>> you do think the reals and the naturals are bijectible? Or did you > >>> mean you have a problem not with what cardinality actually is but with > >>> what you hallucinate cardinality to be? > >>> > >> As long as transfinite "equivalence" classes are referred to as such, > >> and not said to denote "equal" size or numerosity, then I have no problem. > > > > If we choose to define "size" and "equinumerosity' in terms of > > injections and bijections, then TO's problems are purely his own, and > > need not affect anyone else. > > My problem with it is that it violates certain notions which are central > for finite sets, but discarded for infinite sets. 1+omega=omega>omega+1 > violates the principle tat a+b=b+a. I don't like that. Hmm, you demonstrate again your total lack of mathematical understanding. One notion for a finite set is that if you arrange its elements in a row, labelled starting from '1', then there is a last element. Of course an infinite set "violates" this notion, which is a property of finite sets, and thus a property not held by infinite sets. Here's a suggestion. You have spent two years now flailing around - never mind about your own delusions of competence, you have plainly not convinced anyone of anything (you're surely not so desperate as to confuse clucking with Ross et al with achievement?) - and it's not likely that another five years of bashing your head against your miscomprehension of set theory is going to lead anywhere. So why not spend some time learning something different, such as the elementary theory of finite groups (something where everything can be pictured with a complete diagram on a real piece of paper)? Here's a taster. Make a cardboard letter 'L' and put it on your desk, right way up. Excusing my crude Asciifying (~ is supposed to be a top bar), try flipping it about the vertical axis, giving you a _|. Also, from |_, flip about the horizontal axis, making |~. Now try the effect of both flips - you should get ~|. Observe that however you combine these two flips, you only ever get four positions. Now identify each of these positions with the operation needed to get there from the base |_ position. Because these shapes are fiddly to type, call them as follows: |_ : 0 (zero / just a name, comes from arabic; means leaving the cardboard in the same position) _| : h (means flipping the cardboard horizontally) |~ : v (means flipping the cardboard vertically) ~| : r (means doing both flips, in either order, which has the same effect as rotating through 180 degrees) Now use the symbol '+' (just a symbol, comes from the middle ages, I think) to represent the combination of two of these operations. So for example v + h means the operation consisting of doing v then doing h. Observe that in face v + h = r. Carry on, and write out an addition table (for "addition" is the name we give to an operation represented by '+') for 0, v, h, and r. See whether it is true in this addition table that for any x, y (being from the set {0, h, v, r}) it is true that x+y=y+x. Ask yourself if this is because of some strange property of symbology (whatever that is, exactly) that means that any time you write the string of symbols "x+y=y+x" it just must be true. (You could ask Lester's "help" here, but I advise against taking any notice of it, since it will probably come whether you ask or not. Now make another piece of cardboard - square, with a circle on each side, and write the compass points NESW around one circle, so whichever way it is up there is a letter (right-way-up) at the top. With N at the top, flip the cardboard over horizontally (h above / slightly confusing, I know, because you are flipping _about_ the vertical axis), and write the points of the compass, this time as nesw in lowercase. Now we're going to add an operation - of turning the cardboard square through 90 degrees. Now make another addition table for the set {N, E, S, W, n, e, s, w}, where we identify each position name with the operation to get there from the N position. If we use 't' to represent turning anticlockwise (the mathematical "positive" direction) through 90 degrees, and h to represent the same as above, we can again represent any of there operations as (possibly empty, i.e. zero) combinations of t and h, as follows: N : 0 E : t S : t + t W : t + t + t n : h e : h + t s : h + t + t w : h + t + t + t (All this list really shows is the obvious, that NESWnesw can all be made from two basic elements ("generators"), h and t.) Again, investigate this addition table, and see whether it is always true that x + y = y + x within it. OK, then you can rant a bit, but it would be interesting to know whether one of these sets of operations is somehow more or less "correct", or "believable" or whatever than the other. Brian Chandler http://imaginatorium.org
From: MoeBlee on 19 Apr 2007 15:30
On Apr 19, 10:52 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > YOU have not proved that there cannot be a well > ordering on an uncountable set, which is to say that you have not > proven the negation of the well ordering theorem. Actually, even stronger than the negation of the well ordering theorem. Not only are claiming that it is not the case that every set can be well ordered, but you are claiming that NO uncountable set can be well ordered, which is stronger than just claiming that not every set can be well ordered. MoeBlee |