The Definition of Points
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From: Lester Zick on 18 Apr 2007 19:47 On Wed, 18 Apr 2007 21:51:14 +0000 (UTC), stephen(a)nomail.com wrote: >That does not make any sense. There is no point in giving a nonsensical >answer, unless you are aiming to emulate Lester. Or intent on emulating yourself, Mike, Virgil, or any of the other gang of four horses asses of the apocalypse. ~v~~
From: Lester Zick on 18 Apr 2007 19:49 On Wed, 18 Apr 2007 14:55:43 -0400, 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? Uh, whatever. Next time pay attention the first time around. ~v~~
From: Tony Orlow on 19 Apr 2007 10:01 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. >>>>> -- >>>>> mike. >>>> 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? > >> 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'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. >> 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. >>>> So, >>>> I'm working on a better theory, bit by bit. 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. >>> Since any set theory definition of '<' is ultimately defined in terms of >>> 'e', why multiply root causes? >> It's based on the subset relation, which is a form of '<'. > > 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. > > -- > mike. > If you say so. tony.
From: Tony Orlow on 19 Apr 2007 10:02 Virgil wrote: > In article <462660f1(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Virgil wrote: >>> In article <46250ad6(a)news2.lightlink.com>, >>> Tony Orlow <tony(a)lightlink.com> wrote: >>> >>>> Virgil wrote: >>>> Yes, it can be that x<y and y<x and y<>x. >>> Then that is not an order relation. In any form of order relation "<" >>> allowed in mathematics (x<y and y<x) requires x=y. >>>>>>> For any in which "<" is to represent the mathematical notion of an >>>>>>> order >>>>>>> relation one will always have >>>>>>> ((x<y) and (y<x)) implies (x = y) >>>>>>> >>>>>> Okay, I'm worried about you. You repeated the same erroneous statement. >>>>>> You didn't cut and paste without reading, did you? Don't you mean "<=" >>>>>> rather than "<". The statement "x<y and y<y" can only be true in two >>>>>> unrelated meanings of "<", or else "=" doesn't have usable meaning. >>>>> TO betrays his lack of understanding of material implication in logic. >>>>> For "<" being any strict order relation, "(x<y) and (y<x)" must always >>>>> be false so that any implication with "(x<y) and (y<x)" as antecedent >>>>> for such a relation, regardless of conseqeunt, is always true. >>>> Oh, yes, well. Any false statement implies any statement, true or false, >>>> as long as you're not an intuitionist. If (x<y) -> ~ (y<x), then x<y ^ >>>> y<x is of the form P ^ ~P, or ~(P v ~P) which is false in classical >>>> logic, but not intuitionistically. There is debate on this topic. >>> I very much doubt that any intuitionist would say that for an order >>> relation,"<", on any set one could have x<y and y<x without having x=y. >>> >>>>>>> Is TO actually claiming that the irrationals form a subset of the >>>>>>> countable set NxN. >>>>>>> >>>>>>> That is NOT how it works in any standard mathematics. >>>>> Then why does TO claim it? >>>> I am not. I am saying it is a set equal in magnitude to the redundancies >>>> in NxN. >>> The set of "redundancies in NxN" must be a subset of NxN itself, so must >>> be countable, so that TO is saying that the set of irrationals is "equal >>> in magnitude" to a countable set. >> Well, actually, I mean N*xN*, including infinite naturals in N*. Sorry, >> I should have been more specific. > > At what point did N morph into N*? Everything up to now has been in > terms of plain N. > > And isn't doesn't N* surject to N*xN*, making N*xN* countable in R*? N* isn't countable.
From: Tony Orlow on 19 Apr 2007 10:06
MoeBlee wrote: > 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 > 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? TOEknee |