The Definition of Points
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From: Virgil on 20 Apr 2007 15:49 In article <46290fe2(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > MoeBlee wrote: > > On Apr 20, 9:24 am, Tony Orlow <t...(a)lightlink.com> wrote: > >> MoeBlee wrote: > > > >> So, defining N doesn't involve a successor relation between two > >> elements, as well as a member relation between an element and a set? > > > > The successor operation is DEFINED in terms of the membership > > relation. EVERYTHING in set theory is defined in terms of the > > membership relation. The only non-logical primitives of set theory are > > '=' and 'e' (and we could even define '=' in terms of 'e' if we want > > to set it up that way). There is NO formula of set theory that doesn't > > revert to a formula in the primitive language with just 'e' and > > '=' (or even just 'e') as the ONLY non-logical symbols. I've been > > telling you this for probably over a year now. Why don't you > > understand this? > > > > The particular successor relation for the vN ordinals is defined in > terms of 'e', but that's not the only model of successorship. Does TO know of any which are not definable in terms of e'? > Successors > don't necessarily have to be supersets. Does TO know of any which aren't, or at least cannot be modeled as such? > They are usually "greater" in > some sense, in which case subsets thereof can be formulaically compared > for infinite size, as long as the mapping preserves the same overall '<' > relationship between successive elements. You remember my beef about > not-quantitatively ordered mappings, don't you? > > >> When you define N, doesn't the rule E x e N -> E succ(x) e N define a > >> relation between two elements, as well as between those elements and N? But until you show that your definitions of succ(n) does not involve 'e', you have shown nothing. > > > > "Define a relation". From N of course we can define the successor > > relation on N. So what? That doesn't refute that every definition in > > set theory ultimately reverts to the membership relation. I'll say it > > YET AGAIN: > > Before you go on your rant, let me just say this. As a relation, I > specifically mean that each combination of inputs produces either a 0 or > 1. It's a truth table conception of what a relation is. That is not what a mathematical relation is. That is, at most, the characteristic function of a relation.
From: MoeBlee on 20 Apr 2007 15:58 On Apr 20, 12:09 pm, Tony Orlow <t...(a)lightlink.com> wrote: > MoeBlee wrote: > > On Apr 20, 9:24 am, Tony Orlow <t...(a)lightlink.com> wrote: > >> MoeBlee wrote: > > >> So, defining N doesn't involve a successor relation between two > >> elements, as well as a member relation between an element and a set? > > > The successor operation is DEFINED in terms of the membership > > relation. EVERYTHING in set theory is defined in terms of the > > membership relation. The only non-logical primitives of set theory are > > '=' and 'e' (and we could even define '=' in terms of 'e' if we want > > to set it up that way). There is NO formula of set theory that doesn't > > revert to a formula in the primitive language with just 'e' and > > '=' (or even just 'e') as the ONLY non-logical symbols. I've been > > telling you this for probably over a year now. Why don't you > > understand this? > > The particular successor relation for the vN ordinals is defined in > terms of 'e', but that's not the only model of successorship. In set theory, successor(x), whether x is a Von Neumann ordinal or ANYTHING ELSE, is defined in terms of 'e'. > Successors > don't necessarily have to be supersets. They are usually "greater" in > some sense, in which case subsets thereof can be formulaically compared > for infinite size, as long as the mapping preserves the same overall '<' > relationship between successive elements. You remember my beef about > not-quantitatively ordered mappings, don't you? This is another point that I explained to you at least half a dozen times before. There are two different senses of 'successor' in set theory: One is the successor operation (succcesor of x = su{x}) and the other is successor taken more generally with regard to different orderings. But in BOTH cases, successor reduces to membership since orderings reduce to relations reduce to ordered pairs reduce to unordered pairs reduce to 'e'. > > "Define a relation". From N of course we can define the successor > > relation on N. So what? That doesn't refute that every definition in > > set theory ultimately reverts to the membership relation. I'll say it > > YET AGAIN: > > Before you go on your rant, let me just say this. As a relation, I > specifically mean that each combination of inputs produces either a 0 or > 1. It's a truth table conception of what a relation is. So what? We're talking about definitions in SET THEORY. We were talking about the axiom of infinity and things like that, in which the context is SET THEORY. For godsakes, OF COURSE we can't say that YOUR informal notions reduce to 'e'. No sane person would declare as to what primitives YOUR informal notions reduce to. > With successor, > the inputs are each a member of N, a natural, and certainly only one > cell in each row and one cell in each column will be a 1. That's a > relation from each natural to each other, only one of which is its > successor. There is also the relation of membership, which each of these > ordinals has with w. That is a separate relation. Each such natural we > define is a member of N, so we have 1's in the w column for all of them. > If we define other columns for other sets, like R, and include all their > members, then we will create new members, with 0's in the w column. Guess what? As to the POINT WE WERE DISCUSSING, I don't give a damn about your inputs and whatnot. I'm not talking about YOUR informal notions. I just said that in set theory successor (in ANY sense) reduces to 'e' as does EVERY predicate and operation in set theory. That's all. If you understand that, then good, we can more on. Meawhile, though, you'll be straight as to at least one thing about set theory: all defintions finally reduce to 'e'. > So, successor(x,y) and element(x,y) are two different relations, one of > order, and one of membership. But let's see what you say... > > The only non-logical primitives of set theory are '=' and 'e' (and we > > could even define '=' in terms of 'e' if we want to set it up that > > way). There is NO formula of set theory that doesn't revert to a > > formula in the primitive language with just 'e' and '=' (or even just > > 'e') as the ONLY non-logical symbols. > > Is '=' non-logical? It seems to me: Several months, I posted to you a very detailed post as to whether equality is considered logical or non-logical, as I especially recommended Quine's article on the subject. It is not important to me, in this context, whether '=' is considered a logical or non-logical symbol. Whether we regard '=' as logical or nonlogical, my remarks can be suitably adjusted . In the previous quote I just meant that 'e' and '=' are the only symbols other than the quantifiers, the connectives, and the variables. If '=' is regarded as logical, then that leaves 'e' as the only non-logical symbol. And, as I mentioned, even if '=' is regarded as non-logical, then, with a suitable set up, we can define '=' from 'e', so the point stands A FORTIORI that set theory reduces to 'e' (other than, of course, the logical apparatus of first order logic itself). > equals(a,b) = and(implies(a,b),implies(b,a)) > > '=' <> '<->' ? > > But of course, '=' means not '<>', but those are different arrows. Who gives a damn about your childish fingerpainting with math symbols? First learn some math and something about formal and symbolic languages why don't you? > > Over a year I've been telling you that over and over and over. But it > > seems that as far as you're concerned, such information is just a > > random collection of characters appearing on a computer monitor. > Repeating the same thing doesn't usually do much to change the conversation. Obviously! Indeed, how could anyone get your mind off the subject of your own inept and childish constructs, with which you are so infatuated, long enough for you to understand even the most basic things about mathematics, mathematical logic, and set theory? MoeBlee
From: MoeBlee on 20 Apr 2007 16:41 On Apr 20, 12:28 pm, Tony Orlow <t...(a)lightlink.com> wrote: > MoeBlee wrote: > > 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. > > Yes, alright. > > 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. > > In that case, since each member of the partition must be countable, so > as not to have any infinite positions within any one, I didn't say anything about "positions". I'm not going to talk about set theory as it comes out the other end of your reformulations of it. You keep doing that. Post after post after post, someone explains some set theory to you, and you just come back to post about it in terms of your own private terminology and personal informal notions. And I am not saying that it is not meaningful to understand and seek to understand mathematics in terms of one's own intuitions. But it is is LUDICROUS to not even learn the MOST BASIC things about mathematics yet still expect that that of which you don't even know anything about should still be explicable in terms of your own personal terminology and notions. > any well order on > an uncountable set would involve an uncountable number of such sets, > each with a first member, which would correspond to a limit ordinal of > some sort. I have no idea how you're deriving these statements of yours nor what you mean by your personal expressions such as "involve" and "correspond" and "of some sort". Again, you're asking me about set theory as you process it through your own terminology and mess of personal notions. > >>> 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. > > One can discuss ideas, when one already has some, and compare notes. You > expect me to abandon intuition that's guided me for most of my life, and > hop on the bandwagon, but that's pretty unlikely. So, I'd try nott o get > too frustrated. I didn't say that anyone has to abandon intutions or has to hop on a "bandwagon". I only said there that you need to get a damn textbook already! I just said that you need to have LEARN (in an organized and coherent way!) about the subject of set theory if you're going to talk about! It's amazing how UTTERLY RESISTANT you are to doing that, as you will give all kinds of responses to the matter of your needing to get a textbook, as long as those responses avoid yet again admitting that indeed you do need to get a damn textbook already! Sheesh, even if I were going to set myself up as a serious critic of an actual religion of some sort, I'd AT LEAST have familiarized myself with its basic tenets and writings even if to disagree with them! To do otherwise is just to be another ignoramus spewing on the Internet. > > 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. > > Then it is acceptable to have infinite descending chains of limit > elements, t'would seem. I ASKED you about THREE TIMES already to give me your definition of 'infinite descending chain'. Obviously, you prefer just to mouth math terms rather than to define them, understand them, and apply them in proofs. And, again, even with a definition of 'infinite descending chain' from you, it's not much use if you don't understand the axioms and the definitions and theorems previous to ('infinite descending chain') and how to prove things from them. > >>> 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. > > Thank you. No, really, you're being an ignoramus. You started out okay this time, saying why your intutions oppose well ordering of uncountable sets, which is fair enough. Then you asked how we prove well ordering from the axiom of choice, which is fair enough. Then I gave you references to find such proofs, and I explained to you that the proofs are not so simple that they can be condensed to a glib post and also that to understand these proofs you do have to understand the proofs that lead up to them, and, if you want to truly understand, that requries going back to the axioms to see the full sequence of proofs that leads to the proof of the well ordering theorem. THEN, you turn around to DECLARE that it is impossible to define a well ordering on an uncountable set. Just like that, SKIPPING learning about set theory and reading the relevant proofs, you just turn around to DECLARE with your arguments that consist of nothing more than you mouthing a bunch of math terminology that you can't define and do not understand. MoeBlee
From: Mike Kelly on 21 Apr 2007 06:51 On 20 Apr, 16:41, Tony Orlow <t...(a)lightlink.com> wrote: > Mike Kelly wrote: > > On 19 Apr, 15:01, Tony Orlow <t...(a)lightlink.com> wrote: > >> Mike Kelly wrote: > >>> On Apr 18, 6:55 pm, Tony Orlow <t...(a)lightlink.com> wrote: > >>>> Virgil wrote: > >>>>> In article <4625a...(a)news2.lightlink.com>, > >>>>> Tony Orlow <t...(a)lightlink.com> wrote: > >>>>>> Mike Kelly wrote: > >>>>>> < snippery > > >>>>>>> You've lost me again. A bad analogy is like a diagonal frog. > >>>>>> Transfinite cardinality makes very nice equivalence classes based almost > >>>>>> solely on 'e', but in my opinion doesn't produce believable results. > >>>>> What's not to believe? > >>>>> Cardinality defines an equivalence relation based on whether two sets > >>>>> can be bijected, and a partial order based on injection of one set into > >>>>> another. > >>>>> Both the equivalence relation and the partial order behave as > >>>>> equivalence relatins and partial orders are expected to behave in > >>>>> mathematics, so what's not to believe? > >>>> What I have trouble with is applying the results to infinite sets and > >>>> considering it a workable definition of "size". Mike's right. If you > >>>> don't insist it's the "size" of the set, you are free to do with > >>>> transfinite cardinalities whatever your heart desires. What I object to > >>>> are statements like, "there are AS MANY reals in [0,1] as in [0,2]", > >>>> and, "the naturals are EQUINUMEROUS with the even naturals." If you say > >>>> they are both members of an equivalence class defined by bijection, then > >>>> I have absolutely no objection. > >>> Then you have absolutely no objection. Good that you recognise it. > >>>> If you say in the same breath, "there > >>>> are infinitely many rationals for each natural and there are as many > >>>> naturals overall as there are rationals", > >>> And infinitely many naturals for each rational. > >> How do you figure? In each 1-unit real interval, there is exactly one > >> natural, and an infinite number of rationals. Which interval has one > >> rational and an infinite number of naturals? > > > No interval in the ordering on the real line. But there are orderings > > where there are infinitely many naturals between each pair of > > rationals. You seem hung up on the idea of orderings on reals/ > > rationals/naturals that aren't the standard one. Maybe because you > > think about everything "geometrically"? > > I think about quantity in terms of measure, yes. Where we have one atom > of space we have a point, without form. Where we have two, we have a > measurable difference, a line. Half of mathematics arose from > geometrical concerns, largely for dividing land and building buildings. > The other half arose from financial concerns, keeping books regarding > accounts. The one starts from the notion of points and lines, "real" > objects, and the other from the notion of a tally or count, in units. > That modern mathematics considers the second to supersede the first is, > I think, a mistake. Any theory that integrates mathematics should > satisfy both notions, no? Analytical geometry exists and is based on set theory so I don't know what your point is. > >>>> without feeling a twinge of > >>>> inconsistency there, then that can only be the result of education which > >>>> has overridden natural intuition. That's my feeling. > >>> Or, maybe, other people don't share the same intuitions as you!? Do > >>> you really find this so hard to believe? > >> Most people find transfinite cardinality "counterintuitive". Surely, you > >> don't dispute that. > > > I imagine most people have never heard of it. From what I can remember > > of freshman analysis some people had problems with it, some people > > didn't. I certainly don't think there was a clear majority who did. > > And I don't remember anyone taking 2 years to understand that all > > cardinality is really about is bijections. > > I never imagined it would take anyone two years to still not understand, > as I've repeatedly stated, that bijections alone are not sufficient for > proper relative measure of infinite sets. You can't explain what "proper relative measure" means. You may or may not have some coherent picture in your head but you've never explained it to anybody else. > When I say that, doesn't that sort of imply that I understand that > cardinality is based on bijection alone? Not when you continue to say things like "if aleph_0 is the size of N then aleph_0 is the largest natural, therefore set theory is bunk". Which you were saying just last week. > The mapping functions describe the relative size, as long as both > sets follow quantitative order. Otherwise it's slightly more complicated. Interesting. > >>>> I'd rather acknowledge that omega is a phantom quantity, > >>> By which you mean "does not behave like those finite numbers I am used > >>> to dealing with". > >> Or, does not fit into what I understand quantities to be. > > > OK, whatever. You don't think Omega is a "quantity". You haven't > > defined "quantity" in any unambiguous way but I'm perfectly happy to > > believe you don't think the set of all finite ordinals is a > > "quantity". So what? I don't think omega sounds much like a > > "quantity", either. So? This doesn't change anything about my > > understanding of set theory. It doesn't make me say silly things like > > "aleph_0/omega is a phantom". Who cares if they're called "numbers"? > > A raw quantity is a point on the real line. Points to the right > represent greater quantities than points to the left. When we add a > negative quantity, we indicate a point to the left of our starting > point, which is a different point on the real line, representing a > lesser quantity. If omega, or aleph_0, lies anywhere on this infinite > line containing all quantities, They don't. You think aleph_0 and omega are finite numbers. They aren't. If "quantity" means "real number" then of course omega and aleph_0 aren't "quantities". > then aleph_0-1<aleph_0. If a set size is > defined as "an integral quantity" of elements, It isn't. It's obvious that there is no "integral quantity" that defines the size of infinite sets. > then all set sizes lie on > this line, and obey that principle. x-1<x, but aleph_0-1=aleph_0. > > So, whatever kind of "number" one may consider omega, or aleph_0, they > appear to have no relationship to actual set sizes, naturals, or reals. "Infinite things behave differently from finite things". Yes, they do. > >>>> and preserve basic notions like x>0 <-> x+y>y, and extend measure to the infinite scale. > >>> Well, maybe you'd like to do that. But you have made no progress > >>> whatsoever in two years. Mainly, I think, because you have devoted > >>> rather too much time to very silly critiques of current stuff and > >>> rather too little to humbling yourself and actually learning > >>> something. > >> That may be your assessment, but you really don't pay attention to my > >> points anyway, except to defend the status quo, so I don't take that too > >> seriously. > > > Great. Don't let me discourage you from your noble quest. I'll > > continue to glance at your posts when I see them and see if there's > > ever going to be anything of yours that actually interests me. And, > > yes, when you make what I consider amusingly egregious errors about > > set theory or other areas of mathematics I will "defend the status > > quo" by pointing out that you're being silly. > > As you wish. > > > >>> Every relation between objects in set theory is based on 'e'. > > >> If you say so. > > > I do say so. Does this mean you're going to stop claiming that > > relations in set theory aren't based solely on 'e'? > > No, not when sequences are defined using a recursively defined successor > function, which is a relation between two elements, as opposed 'e', a > relation between an element and a set. In set theory, successor is defined in terms of 'e'. >The combination of the two is what produces an infinite set, no? No. -- mike.
From: Mike Kelly on 21 Apr 2007 06:58
On 20 Apr, 17:13, Tony Orlow <t...(a)lightlink.com> wrote: > MoeBlee wrote: > > On Apr 19, 7:06 am, Tony Orlow <t...(a)lightlink.com> wrote: > >> MoeBlee wrote: > > >>> You just need to learn it theorem by theorem from the axioms. All it > >>> takes is an introductory set theory textbook, and all of them will > >>> prove the well ordering theorem for you. > > >> But, a well order cannot have infinite descending chains of any sort, so > >> if there are an uncountable number of limit ordinals within the order, > >> then there will exist an infinite descending chain within that sequence > >> of limit ordinals itself. How does one get around that? > > > Again, you're reasoning from vague ideas you have of what the > > terminology means. If you were to pin down these terms you use such as > > 'infinite descending chain' with the precise definitions, then tried > > to derive a formal contradiction, you wouldn't succeed to prove any > > such contradiction nor evidence of anything that we must "get around" > > here. Ironically, your response is right after I reiterated that you > > need to study this in a systematic and precise way, definition by > > definition, theorem by theorem. It is just more effort down the drain > > for me to untangle your confusions over terms of which you've not > > bothered to learn their precise definitions from primitives and > > axioms. > > > A well ordering doesn't preclude that a member of the field of the > > ordering may have infinitely many predecessors. What a well ordering > > does demand is that in any nonempty set of predeccessors, there is a > > least one, which doesn't preclude that there may be infinitely many > > "on top" (as we're "working downwards") of that least one. You don't > > even need to go to the uncountable; all you need is just one limit > > ordinal in the field of the well ordering to see the point: Just take w > > +1. The well ordering on w+1 by the membership relation has w itself > > as member of the field of the ordering. But to "travel from" w "to" > > the least member of w requires "travelling through" an infinte number > > of members of the domain of the field of the ordering. > > As it was explained to me, because every natural is finitely far from > the lest in w, there is no infinite descending chain of predecessors one > can define. In other words, between any two limit ordinals can only a > countable number of elements. Now, can a well order have an infinite > descending chain of limit elements? If so, then an uncountable set could > be partitioned into an uncountable set of countable partitions, and be > well ordered, without a doubt, and it would already have been done. There's a difference between having a limit ordinal as an element and having every element of a limit ordinal as an element. -- mike. |