The Definition of Points
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From: Virgil on 20 Apr 2007 14:59 In article <4628e340(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > i is a number produced by extending completeness to the multiplicative > field, after introducing completeness to the additive field to produce > -1, like I said above. You do not have a field if either addition or multiplication is 'incomplete'.
From: Virgil on 20 Apr 2007 15:05 In article <4628e6bd(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > MoeBlee wrote: > > 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. Irrelevant. The point here is that in any well ordered set, between any limit member and any prior member there are infinitely many other members.
From: Tony Orlow on 20 Apr 2007 15:09 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. 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? >> 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? > > "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. 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. 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: equals(a,b) = and(implies(a,b),implies(b,a)) '=' <> '<->' ? But of course, '=' means not '<>', but those are different arrows. > > 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. > > MoeBlee > > Repeating the same thing doesn't usually do much to change the conversation. >> TOEknee
From: Virgil on 20 Apr 2007 15:10 In article <4628e964(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > There have been lots of objections, and a few valid points, but no major > flaws detected in what I propose. It's just not compatible with ZFC. TO simply chooses to ignore the many major flaws that have been pointed out to him in such miniscule snippets of his various mutually incompatible systems that have been presented here. > > >>> Get this through your head : every relation between objects in set > >>> theory is based on 'e'. It's really pathetic to keep mindlessly > >>> denying this. Set theory doesn't just "try to base everything on 'e'". > >>> It succeeds. > > > >> If you say so. > > > > No, not just "if he says so". He says so and he's RIGHT. And you can > > VERIFY for yourself just by reading a textbook already. > > > > MoeBlee > > > > So, defining N doesn't involve a successor relation between two > elements, as well as a member relation between an element and a set? All that is done in terms of 'e', so 'e' is 'e'nough to get N.
From: Tony Orlow on 20 Apr 2007 15:28
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, 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. >>> 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 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. > Then it is acceptable to have infinite descending chains of limit elements, t'would seem. >>> 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. > Thank you. >> 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 > True. |