The Definition of Points
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From: Virgil on 18 Apr 2007 15:42 In article <4626610e(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <46250ea8(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> In all fairness to Lester > > > > Why bother to be fair to one who is so compulsively unfair? > > To set a good example? But when you strive so hard in so many other other ways not to do so, why this one exception?
From: Mike Kelly on 18 Apr 2007 15:45 On Apr 18, 8:04 pm, Tony Orlow <t...(a)lightlink.com> wrote: > Mike Kelly wrote: > > On 18 Apr, 06:17, Tony Orlow <t...(a)lightlink.com> wrote: > >> Mike Kelly wrote: > > >> < snippery > > > >>> You've lost me again. A bad analogy is like a diagonal frog. > >>> -- > >>> mike. > >> Transfinite cardinality makes very nice equivalence classes based almost > >> solely on 'e', > > > Why "almost solely"? > > The successor relation is a relation not related to 'e' except in a > particular model of the naturals upo which the rest is built. What does this have to do with cardinality? Anyway, every relation in set theory is built up solely from 'e'. You can keep denying this if you like but it's just a rather bizarre thing to do. > >> 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. Ahh, progress. Well, nice to see you are not an incurable crank. Well done. > >> I'm working on a better theory, bit by bit. > > > You have never managed to articulate a problem with what cardinality > > actually says (all it says is which sets are bijectbile). You've never > > managed to explain the motivation behind your "better theory". I don't > > even know whether your ideas are supposed to be formulated in ZF or in > > something else, possibly a new foundation of your own devising(or, > > perhaps, never formalised at all?). FWIW I don't think it would be > > very hard to define, say, "density in the naturals" in ZF. I just > > don't see the motivation behind doing so. > > The idea is to create a method that works for comparing countable and > uncountable sets alike, and properly defining the relationship between > the two. Well you haven't really elucidated anything there. But you're free to keep trying. > >> I think trying to base > >> everything on 'e' is a mistake, since no infinite set can be defined > >> without some form of '<'. I think the two need to be introduced together. > > > But you are, in fact, completely mistaken here. Several people have > > told you this. You have some inability to acknowledge this. Oh well. No response :-[ > > PS you snipped a lot of my post; maybe you found it boring. I find > > your evasiveness quite interesting, though. What does cardinality > > claim to tell that it can't? > > I suppose it's more the way it's spoken about. Leave out "size" and > "equinumerous" and talk about "equivalence classes" and not "equality", > and I have no problem Great. > with the consistency of standard theory. Dunno what this has to do with consistency. -- mike.
From: MoeBlee on 18 Apr 2007 15:50 On Apr 18, 12:13 pm, Tony Orlow <t...(a)lightlink.com> wrote: > MoeBlee wrote: > > On Apr 17, 10:02 am, Tony Orlow <t...(a)lightlink.com> wrote: > > >> I am saying it's obvious that any countable set has a well ordering. It > >> is not obvious for uncountable sets. > > > It's not only obvious, but it's trivial to prove that every countable > > set has a well ordering. > > > So, just to be sure you understand, what is at stake with an axiom of > > countable choice is not at all that every countable set has a well > > ordering (since we don't need any choice principle to prove that) but > > rather that for any countable set there is a function that chooses > > exactly one member from every nonempty subset of that countable set > > (and actually, we need concern ourselves only with denumerable sets, > > since we don't need a choice principle to prove that for every finite > > set there is a function that chooses exactly one member from every > > nonempty subset of the finite set). > > > MoeBlee > > It's not trivial to prove any such thing for an uncountable set. Right. >That > requires full Choice, does it not? I'm pretty sure full choice is needed (not just dependent choice). > In fact, it was Virgil, I believe, > who said that was the MOTIVATION behind Zermelo's formulation of Choice, > so that ALL sets could be considered well orderable. Virgil may have said that, I don't know. Anyway, I said it also. > However, while I > see that a choice function produces a well ordering for any countable > set, I don't see that it does for an uncountable set. You MISSED my point COMPLETELY. We don't need choice to show that any countable set has a well ordering. Think about it for a minute, and you'll see how to prove it. As to how choice provides a well ordering on any set S, most basically (quite oversimplified), whenever we need to have a least element, we let the choice function pick it for us. Again, that is wildly oversimplified. But any set theory textbook will give you a proof. To supplement Enderton and Suppes, I recommend Stoll, since Enderton and Suppes use the axiom schema of replacement for certain of axiom of choice equivalence proofs while Stoll proves some of those theorems withOUT using the axiom schema of replacement. > Maybe I am confused about something. Can a well ordering include an > uncountable number of limit elements? If I'm not mistaken, yes. We just need to see that there are ordinals that have uncountably many members that are limit ordinals (I think that is right). > If so, then I can see how such a > well ordering could occur. Are limit elements which are an infinite > number of limit elements beyond the first what are called "inaccessible" > limit ordinals? I dont' think so, though I'm not expert on this. I know the definition of 'inaccessible cardinal' and just a few things about them, but I haven't yet studied them in depth or detail. I very much suggest you too learn the basics before getting off into inacessible cardinals and such. You run WAY ahead of yourself when you still don't even understand the nature of the membership relation in terms of set theory. > If this is the case, then I guess I can see that in this > sense, one could concoct a well ordering. But, no, then there would > exist an infinite descending sequence of limit ordinals withoin the well > order. So, that wouldn't work, right. 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. 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. MoeBlee
From: MoeBlee on 18 Apr 2007 15:55 On Apr 18, 12:50 pm, MoeBlee <jazzm...(a)hotmail.com> 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. I just want to emphasize something here: Your basic question about uncountability and well ordering is a GOOD question from you for a change. This suggests to me that you really do deserve to do YOURSELF the favor of learning the answers properly from a good textbook. MoeBlee
From: Virgil on 18 Apr 2007 17:25
In article <462669b3(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Lester Zick wrote: > > On Tue, 17 Apr 2007 13:33:39 -0400, Tony Orlow <tony(a)lightlink.com> > > wrote: > > > >>>> > >>>> Is there a set > >>>> of statements S such that forall seS s=true? > >>> No idea, Tony. There looks to be a typo above so I'm not sure exactly > >>> what you're asking. > > > >> I am asking, in English, whether there is a set of all true statements. > > > > No. There are predicates to which all true statements and all false > > statements are subject respectively but no otherwise exhaustively > > definable set of all true or false statements because the difference > > between predicates and predicate combinations in true or false > > statements is subject to indefinite subdivision. > > > > ~v~~ > > Uh, what? Just Zickism at its most opaque. |