An uncountable countable set
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From: Virgil on 6 Sep 2006 01:57 In article <44fe1b09(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > MoeBlee wrote: > > Tony Orlow wrote: > >> Indeed, or we may NOT allow such things, if we adopt a rule of the form > >> x = S: A P P(x) = P e S. > > > > We don't need to go to second order to disallow urelements. The plain > > axiom of extenstionality disallows urelements. You don't even > > understand the axiom of extensionality. You're overflowing with > > opinions about set theory but don't even know what the axiom of > > extensionality says. > > Yes, I do. You think you are such an expert on what I don't know, but > that's only because you can't think beyond what you do know. http://en.wikipedia.org/wiki/Axiom_of_extension In axiomatic set theory and the branches of logic, mathematics, and computer science that use it, the axiom of extensionality, or axiom of extension, is one of the axioms of Zermelo-Fraenkel set theory. In the formal language of the Zermelo-Fraenkel axioms, the axiom reads: For all A, for all B( (A = B) <==> for all C (C \in A <==> C \in B) ) or in words: Given any set A and any set B, A is equal to B if and only if, given any object C, C is a member of A if and only if C is a member of B. (It's not really essential that C here be a set but in ZF, everything is. See Ur-elements below for when this is violated.) This means that "properties" of a set are irrelevant to its identity, as only its membership counts. > > What I am talking about is not covered by the axiom of extensionality, > which does not address the notion of properties defining objects or sets > whatsoever, now, does it? it scuttles any notion of propeties defining a set, as it requires that only members define any set. > It simply says that two sets whoich contain > the same objects are the same set. How does that address the > relationship between sets, elements, and properties? It says that, in determining a set, only membership is relevant and properties are irrelevant. > > No. What I don't understand is why you think you wouldn't benefit from > > understanding what other people (such as those who have studied the > > subject and written books on it) have come up with so that you can > > contribute or even dissent on an informed basis rather than flounder in > > your ignorance. > > If all you can produce is insults, what's the point? If all TO can produce is dissent on an uninformed basis, what is his point? > You answer "No" to > the general notion that properties of the elements define the set. Can > you give any coherent reason why, besides my not having read the same > books as you? In all of our set theories we will have something equivalent to the axiom of extentionality, which says that only membership counts. > > > > >> That's nice. Is there a difference between an abstract object like a > >> set, and its definition? The abstract idea of a set and of membership in one are undefined terms in set theory, though e have a pretty fair idea of how they work. > > > > That a different definition may be of a different set does not imply > > that the definition IS the set. > > The set consists of the elements within it and nothing more. The > elements are distinguished from all other objects in the universe by > some set of properties. Do you have a counterexample? Unless those members are non-set ur objects, each of them is also determined only by its members, and so on down until one gets to empty sets. > > No, in this latest matter, it's more like you can't distinguish the > > picture on the screen from the buttons on the remote control that > > define which picture will be on the screen. The set defined is not its > > own definition any more than the picture of the cowboy on the screen is > > the sequence of buttons on the remote that brought that picture on. > > Yet another meaningless analogical insult. How enlightening. TO seems impervious to all attempts to enlighten him. I think the appropriate phrase is "invincibly ignorant".
From: Virgil on 6 Sep 2006 02:14 In article <44fe1d38(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > MoeBlee wrote: > > You ignore instead of asking me for justification. The justification is > > in the incompleteness theorem, which is even STRONGER than what I > > mentioned. Aside from incompleteness, if you knew even just a little > > bit about the subject you'd understand the sense in which you can't get > > adequate mathematics (such as, say, enough to do calculus) from just > > logical axioms. > > Listen, in retrospect, I agree with the statetement, "To get an adequate > amount of mathematics, you have to adopt axioms that are not derivable > from pure logic alone." However, that does not mean they should not have > ANY justification. The axioms of set theory whether ZF, ZFC or NBG or some other variation has a long and serious history of development, and the reasons why those and not other axioms and axiom systems have survived are in that history. So that if TO wants justifications, they are available, but they are often too technical for someone of TO's level of understanding to comprehend, and even if not, tend to be buried in papers that are otherwise highly technical in precisely the ways TO object to all the time. >As I said below, IFR is justified geometrically given > the graph of a function On the contrary, there are lots of "graphs" for which it is nonsense. > very intuitively. N=S^L is based on > combinatorics. But is only valid for naturals, even when properly interpreted. And is often misinterpreted. > Somewhere in the statement of an > axiom should be included the intent The intent is discussed in detail in all those books that TO declines to read. > > Oooh boy. You don't even have a logicistic system, logical axioms, nor > > rules of inference, and yet you're telling me that you can prove your > > mathematical axioms, let alone your notion of inductive proof as > > provable by logic by forming an infinite loop shows you really have no > > idea what induction is. > > If you say so. What is it, then, in a nutshell? I mean, you can't define > "mathematics", but maybe something a little more restrictive could be a > good place to start. So, why don't you explain induction? It has been defined quite clearly any number of times for TO, but he has such a weak memory that nothing sticks. Induction says: Given a subset S of the set of all naturals,N, if the first natural is in S and the successor of every element of S is also an element of S, then S = N. And that is all it says. Is that so difficult to remember, TO?
From: Virgil on 6 Sep 2006 02:22 In article <44fe2642(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <44fd9eba(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Dik T. Winter wrote: > >>> In article <44ef3da9(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> > >>> writes: > > > >>> Your axiom uses things that are not defined. What is the *meaning* of > >>> "x<z"? > >> Geometrically it means that x is left of z on the number line. > > > > And for someone standing on the other side of the number line would x be > > on the right of z? > > > > And does the line stay horizontal as one moves around earth? Which way > > is larger if the line ever goes vertical. And how does the "larger" work > > at antipodes? > > > > Silly questions. In response to a silly definition. > > > > > > >> It means > >> A y (y<x ^ y<z) v (x<y ^ y<z) v (x<y ^ z<y). That says about all it > >> needs to, wouldn't you say? > > > > Not hardly. > > A y (y = x) v (y = z) v (y<x ^ y<z) v (x<y ^ y<z) v (x<y ^ z<y) > > is a bit better but still insufficient. > > True, I should have specified y<>x and y<>z. I guess it's usually done > using <= for this reason, eh? > > > > >>> > > That is not a definition, because it makes no sense. "The set of > >>> > > naturals > >>> > > is as large as every natural"? > >>> > > >>> > It is not larger than all naturals > >>> > >>> That is something completely different again. > >> It's not LARGER than every finite. > > > > Which natural(s) is it "not larger" than", in the sense of not being a > > proper superset of that natural or having that natural as a member? > > ....11111 binary (all bit positions finite) Unless that string has only finitely many bit positions as well as only finite bit positions, it is not a natural at all, as it is then neither the first natural nor the successor of any natural, and every natural has to be one or the other. > > > > >> If I say it's larger than all naturals, how do you read that? > > > > As its being a proper super set of that natural and containing that > > natural as a member. > > Which natural? The "all" natural? Every natural. Only TO believes in an "all" natural. > >> Nowhere has the set ever become infinite in count, as long as you only > >> count finite units. > > > > For the definition of cardinality, that is not an issue. The only issue > > is the possibility of bijection with other sets. > > Yes, cardinality ignores many issues. Only those irrelevant ones that TO homes in on, to the exclusion of all the relevant ones.
From: Virgil on 6 Sep 2006 02:28 In article <44fe27fe(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > But is it a property of z, necessary to the identification of z, that > > z e y? > > Not in ZF, ZFC, or NBG. > > if zey but ~xey, can it be possible that x=z? No. But the axiom of extension says that is irrelevant to the "identity" of z. > > > > >> No, but we can take as axiomatic that a=b = A P P(a)=P(b), > > > > Not in ZF, ZFC or NBG. > > > > Come out of the cave. The sun won't hurt you. That's only a shadow of a > caterpillar. > We are in the sunlight of ZF, ZFC and NBG, having a variety of axiom systems that all insist that TO is wrong and it their members which identify sets, and not what they are members of. > > > >> and that P=Q = A a P(a)=Q(a). > > > > Not in in ZF, ZFC or NBG. > > Come out and wallow in the Pearl Pond with the swine, O Albino Cave > Being. You need a bath. :) That is the response of someone who willfully rejects reason. > > > > >> Thus properties are defined by the objects to which > >> they pertain, and objects are defined by the properties which pertain to > >> them. Despite the fact that this statement is not first-order, I see no > >> problem arising from it. > > > > Certain problems arise from the "property" of a set being, or > > especially not being, a member of itself. > > Yes, Russell's Paradox indicates a need to define what kinds of > properties are allowed and which are not. That kind of recursion causes > problems. Not in ZF, ZFC or NBG. > > > > > So unless allowable "properties" are somehow constrained to avoid those > > problems, TO's set theory will be self contradictory. > > Agreed. > > > > > In ZF, et all, such constraints are built in. > > Oh? Yup!
From: Mike Kelly on 6 Sep 2006 04:31
Tony Orlow wrote: > Virgil wrote: > > In article <44fd9eba(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Dik T. Winter wrote: > >>> In article <44ef3da9(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> > >>> writes: > > > >>> Your axiom uses things that are not defined. What is the *meaning* of > >>> "x<z"? > >> Geometrically it means that x is left of z on the number line. > > > > And for someone standing on the other side of the number line would x be > > on the right of z? > > > > And does the line stay horizontal as one moves around earth? Which way > > is larger if the line ever goes vertical. And how does the "larger" work > > at antipodes? > > > > Silly questions. > > > > > > >> It means > >> A y (y<x ^ y<z) v (x<y ^ y<z) v (x<y ^ z<y). That says about all it > >> needs to, wouldn't you say? > > > > Not hardly. > > A y (y = x) v (y = z) v (y<x ^ y<z) v (x<y ^ y<z) v (x<y ^ z<y) > > is a bit better but still insufficient. > > True, I should have specified y<>x and y<>z. I guess it's usually done > using <= for this reason, eh? > > > > >>> > > That is not a definition, because it makes no sense. "The set of > >>> > > naturals > >>> > > is as large as every natural"? > >>> > > >>> > It is not larger than all naturals > >>> > >>> That is something completely different again. > >> It's not LARGER than every finite. > > > > Which natural(s) is it "not larger" than", in the sense of not being a > > proper superset of that natural or having that natural as a member? > > ....11111 binary (all bit positions finite) That isn't a natural number, Tony. -- mike. |