Uncountability of the Rationals?
in [Math]
|
Prev: Collatz conjecture
Next: Beginner-ish question
From: David R Tribble on 16 Jun 2010 18:08 David R Tribble wrote: >> I'll ask the question again, since you didn't bother answering >> it before: Are there square roots r in S such that r*r is not a >> natural in N+? If so, can you provide an example r? If not, does >> this mean that there are more members in S than in N+?- Hide quoted text - > Tony Orlow wrote: > Imanswered this already. I do not disagree that there exists a > bijection, but within any segment of R greater than measure 2 exist > more square roots of naturals than naturals. Sure, you can find a > member in each set corresponding to a unique member of the other. They > are equicardinal. They are not equibigulous. Okay, finally, perhaps, some progress. Bigulosity of a set is equivalent to the relative "density" (or perhaps "frequency") of its members when they are mapped onto the real number line. I.e., how frequently the numbers occur *when considered as points on the real number line*, as compared to the frequency of the naturals (what you call "units") on the real line. So it's probably fair to say that Bigulosity is more about the numeric values of members of sets than it is about the sets themselves.
From: Virgil on 16 Jun 2010 18:32 In article <027111c0-77c9-423a-ac49-93194059545b(a)b5g2000vbl.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On Jun 16, 2:52�pm, Virgil <Vir...(a)home.esc> wrote: > > In article > > <f6d161b0-150a-4c47-bea4-bc898e5a0...(a)i31g2000yqm.googlegroups.com>, > > �Tony Orlow <t...(a)lightlink.com> wrote: > > > > > Sometimes there are logical arguments themselves which don't appeal to > > > everyone equally. What I have been told is that the size of omega must > > > be larger than every natural since no natural is large enough to > > > express it, and so it is some infinite number, aleph_0. However, that > > > logical argument, as I pointed out, really just proves that aleph_0 > > > cannot be finite. Aleph_0 is certainly infinite, and if we chose to call it a number, that does not contradict anything > Along with the argument that any initial segment of N > > > + of size x contains an xth element whose value is x, But with the right N it doesn't. For vN's N it is false. > > > which would > > > imply that aleph_0 or omega is a member of N+, we arrive at a > > > contradiction implying that aleph_0 cannot actually exist. Which axiom > > > contradicts this logic? > > > > Using the von Neumann naturals, or any other set of naturals which > > starts with 0, as is becoming fairly standard in mathematics, no natural > > need be anything like a member of itself, which is a consummation > > devoutly to be wished. > > Did you just state an axiom? Why don't you answer this simple > question? You know why, don't you? > > Tony If the question is "Which axiom contradicts this logic?", I do not find it meaningful.
From: MoeBlee on 16 Jun 2010 18:53 On Jun 16, 4:52 pm, Tony Orlow <t...(a)lightlink.com> wrote: > On Jun 16, 1:33 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > On Jun 16, 8:15 am, Tony Orlow <t...(a)lightlink.com> wrote: > > > > On Jun 15, 8:22 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > > > If TO accepts all of the axioms of ZFC, but rejects the > > > > theorem "there exists a cardinal (or ordinal) number," > > > > then I'll agree to call TO "wrong." Still, I believe that > > > > if we can show him a theory which does satisfy his > > > > intuitions, he'll have less of a reason to criticize the > > > > adherents of ZFC. > > > What do you mean Transfer Principle "adherent of ZFC"? > > > > Okay, perhaps I am "wrong" about this. I am going over the axioms of > > > ZFC, and I simply don't see any reference to any primitive referring > > > to ordinality or cardinality > > This is egregious: INDEED your ignorance as you yet again post it is egregious. BIG NOTE: I'm replying line by line to you again, but I might not followup to further of your foolishness, just as I did not followup recently. At a certain point, such exchanges with you are just not productive. I need not be a hen continually cleaning up your confusions as when I do that, and get in yet another back and forth with you, even MORE confusions come issuing from you. Here, I'll reply point-to-point, but I suggest you save yourself the trouble of doing the same in return thus to set off yet another round of unproductive exchange. All you have to do, in the meantime, is get any kind of decent book that discusses the subject of mathematical definition. > > Also not among the primitives are mentions of > > > subsets > > Wrong. In the PRIMITIVES, there is NO subset symbol. > "X is a subset of Y" can be expressed as "aeX -> aeY", Actually as Aa(aeX -> aeY), just to be exact. And, yes! subset is DEFINABLE from 'e'. Just as 'is an ordinal' is DEFINABLE from the mere primitives. > which is > used explicitly in the axiom of the power set, if not implied elswhere > in the axiom set. > > > the empty set > > Implied by the other axioms. Yes, exactly! The axioms IMPLY that there exist a set x such that Ay ~yex, as either that is itself an axiom (with some authors) or entailed by the axiom schema of separation (as with other authors). Then from the axiom of extensionality we get that there exists a UNIQUE such set. THEN we DEFINE 0 to stand for that unique such set. There is no primitive of the language of set theory that is "the empty set" symbol. '0' is DEFINED from the primitives and as enabled by axioms that prove that there exists a unique such set. And as 'is an ordinal' is DEFINABLE from the mere primitives. It's a long trail going through previous definitions that go back to 'e', but still a definiens of 'is an ordinal' can be stated in terms of the mere primitives. There is a formula in the PRIMITIVE language of ZFC that defines 'is an ordinal'. (We don't usually SHOW it that way, because it's pedantic and unnecessarily laborious to do so, but still the formula is there and we could present if we didn't mind the drudgery of doing it.) > > union > > Axiom The union axiom has NO union symbol in it. The union axiom is in the PRIMITIVES of ZFC. THEN we DEFINE the unary union symbol. Again, a operation symbol defined in terms of primitives. And 'is an ordinal' is a locution for a predicate (technically, a predicate symbol) defined in terms of previous defined symbols that themselves all can be traced back to a definiens in the PRIMITIVE language. > > intersection > > Implied by separation using xeG as condition phi on set F. Extend the > size of the intersection to the infintie case... (1) We don't "extend" to infinite case, but rather, we prove all at once (with no mention of infinte or infinite) Ax(x is non-empty -> E!zAy(yez <-> Av(vex -> yev))) Then we DEFINE If x nonempty -> /\x = the unique z such Ay(yez <-> Av(vex -> yev)). All in the primitives ('unique' and 'nonempty) definable back to primitives, or we could dispense with 'unique' and do it this say: If x nonempty -> (/\x = z <-> Ay(yez <-> Av(vex -> yev))) > > pairs > > ordered pairs > > Not axioms? Pairing axiom, yes. Same remarks, mutatis mutandis as above. But no ordered pair axiom. Rather, ordered pairs defined in terms of pairs which are themselves defined in terms of the PRIMITIVES. The axioms state IN PRIMITIVES that there exist sets having certain properties (properties in terms of the PRIMITIVES), and THEN the DEFINITION of the { } operation is given, and, as with ALL definitions in ZFC, including 'is an ordinal' ultimately traced back to a formula in the PRIMITIVE language. > > natural numbers > > Axiom of infinity, SOME authors wait for the axiom of infinity. But that's not required or essential. We can give an equivalent definition of 'is a natural number' without the axiom of infinity and in the PRIMITIVE language. Indeed NO predicate needs ANY axioms at all to support a definition. Operations (operation symbols, pedantically) require axioms to prove the existence and uniqueness clauses, but if we have just one defined object (such as 0) then we may use the Fregean method so that even operations do not require appeal to axioms (other than those required to construct at least one unique object). > but without reference to its size, only its > existence. Correct. > > prime numbers (thank you, Aatu) > > No, that comes later, and has nothing to do with cardinality, except > peripherally, as a tough example for Bigulosity. I didn't say it depends on cardinality. I said that there is no mention in the axioms of "prime number". Rather, we DEFINE 'is a prime number' using previously defined terms that are in a sequence of defined terms that ultimately reach to the primitive. The definition of 'is a prime number' can be given in the PRIMITIVE language, if we wish to present it that way and with such a definition EQUIVALENT in ZFC to one given with previously defined terminology. EVERY definition in ZFC (from 'subset' through 'Banach space' and beyone) has a definiens taht may be stated in the PRIMITIVE language. ordinal' is no different in this respect from 'subset', '0', 'prime number', 'Banach space'. > > metric spaces > > Not mentioned in ZFC, but rather, handled by topology, wherein which > measure is somewhat at odds with transfinite set theory. No, 'metric space' is DEFINABLE in ZFC. 'is a topology', 'is a topological space', 'is open in the topology', 'is a metric space' and all the other common terms of topology have set theoretic definitions. Thery all have definitions in the language of ZFC such that the definiens is RIMITIVE in the language of ZFC. > > > Banach spaces > > > or ANYTHING other than > > > equality (if taken among the primitives) > > and > > elementhood > > Where do the axioms mention "equality" except among identical sets ala > the axiom of extensionality? The equality symbol is in the axiom of extensionality. It's also in ordinary versions of the axiom schema of replacement. I didn't say equality symbols occurs in all axioms, but rather that it does occur among the axioms. Moreover, the equality symbol can be defined in the language of ZFC without the equality symbol. > > and the variables, the sentential connectives, and the quantifiers > > (and left and right parentheses, if we don't go Polish). > > That's all the logical foundation, about which there are only a few > questions. Whatever your questions about the logic, you need to understand that 'ordinal' is definable from the primitives of the language of ZFC, and that 'subset' and all the rest that are NOT primitives are also definable from the primitives. MoeBlee
From: David R Tribble on 16 Jun 2010 19:45 Tony Orlow wrote: >> Sometimes there are logical arguments themselves which don't appeal to >> everyone equally. What I have been told is that the size of omega must >> be larger than every natural since no natural is large enough to >> express it, and so it is some infinite number, aleph_0. However, that >> logical argument, as I pointed out, really just proves that aleph_0 >> cannot be finite. Along with the argument that any initial segment of N >> + of size x contains an xth element whose value is x, which would >> imply that aleph_0 or omega is a member of N+, we arrive at a >> contradiction implying that aleph_0 cannot actually exist. Which axiom >> contradicts this logic? > David R Tribble wrote: >> The problem is that there is no logic there. >> You consider finite sets of the form {1, 2, 3, ..., k}, all of which >> have a least and a greatest member, to have size k and to also >> have k as a member. Then you take an unjustified leap and >> claim that this implies that sets of the form {1, 2, 3, ...}, >> which do not have a largest member, to also have a size equal >> to one of their members. > Tony Orlow wrote: > No, they all have a size, and if that size is n, then there exists an > nth. If there does not exist an nth, then there are not n elements in > the sequence, which this set also is. But what if a set has an nth member, for every n in N+, but has no largest member? Does the set still have a size? Is that size a member of the set? If so, is it the largest member or some other member? David R Tribble wrote: >> So the question becomes, what axiom justifies your logical >> leap, applying a property of finite sets to infinite sets without >> largest members? > Tony Orlow wrote: > There is no leap. By your logic, you state that all sets of the form {1,2,3,...,k} have a largest member, therefore sets of the form {1,2,3,...} also have a largest member. I assume you have an axiom or theorem in mind to help you reach this conclusion. > What axiom justifies the existence of aleph_0? Axiom of Infinity. Duh.
From: K_h on 16 Jun 2010 19:59
"Tony Orlow" <tony(a)lightlink.com> wrote in message news:f6d161b0-150a-4c47-bea4-bc898e5a0f87(a)i31g2000yqm.googlegroups.com... On Jun 15, 8:22 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > On Jun 15, 3:15 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > On Jun 15, 4:40 pm, Transfer Principle > > <lwal...(a)lausd.net> wrote: > > > According to Tribble, there are many questions which > > > TO > > > can't and won't answer about his theory. One of these > > > questions (asked IIRC by MoeBlee) is to which of the > > > axioms > > > of ZFC does TO object? But to me, the answer to this > > > question is obvious. If a poster disagrees with how > > > the > > > infinite sets work under ZFC, then they reject the > > > axiom > > > which guarantees their existence -- and that axiom is, > > > of > > > course, the Axiom of Infinity. > > You must have only skimmed the relevant exchanges in > > this thread on > > the subject of axioms. > > In that case, let me go back to the references to axioms > in > order to see TO's opinion on this subject: > > TO: > That it doesn't necessarily exist according to those > elementally > logical statements, and that I am not obliged by the > axioms or that > argument to believe that any such absolute size exists for > such a set. > > TO: > It does not follow from the axioms, as they mention > nothing about set > size or cardinality. > > Aatu's response to TO: > Just as PA proves nothing about primes since they aren't > mentioned in > any of its axioms? > > MoeBlee: > > > The formulas you mentioned don't entail that there is > > such an ordinal > > number, true. But we do prove from our AXIOMS that there > > exists such > > an ordinal number. > > So we can see what's going on here. TO doesn't accept that > ZFC proves the existence of (standard) cardinality or > ordinality because "card" isn't a primitive symbol that's > mentioned in any of the axioms. But MoeBlee points out how > that ZFC proves the existence of ordinals despite not > being mentioned in the axioms, and Aatu adds that PA does > prove the existence of primes even though the word "prime" > doesn't appear in the language or axioms of PA. > Actually, I don't see any explanation of how ZFC > proves their existence. I will probably be told to > just go read a book... //section removed > While the von Neumann ordinals are consistent > with ZFC, are they really logically entailed by the > axioms? I must be missing something. You are missing the fact that the definitions are created using the axioms. For example, using the axiom of extensionality, ZF1, the empty set axiom, ZF2, and the axiom of pairs, ZF3, you can then define an ordered pair as follows: Definition: Given sets A and B, let <A, B> denote the set {{A},{A,B}}and then <A,B> is the ordered pair with A the first member and B the second member. This is but one example. The theory ZFC, and other theories like ZF+(V=L), allow you to define the ordinals and the cardinals, for example, totally in terms of the axioms of the theory. Note, it is bad to use a definition that is not ultimately expressed from the axioms. For example, one can define a measureable cardinal but that definition doesn't follow from the axioms of the theories mentioned above. If you wanted a theory with measurable cardinals you would have to first define an axiom like MC="measurable cardinals exist" and then add it to get a new theory ZFC+MC. You can't add it to ZF+(V=L) because ZF+(V=L)+MC is an inconsistent theory. Another example, in addition to ZF1, ZF2, and ZF3 mentioned above, if you also add ZF4, axiom of separation, ZF5, powerset axiom, and ZF6, union axiom, then you can define the Cartesian product. Let P(S) denote the powerset of S. Definition: The Cartesian product of sets A and B, written C=AxB, is the set: {c in P(P(AUB)):EaEb((a in A ^ b in B) ^ c=<a,b>)} Note, this definition exploits the definition of an ordered pair which itself is defined in terms of the axioms. Informally, the bottom line is this: the definitions are just shortcuts for things that groups of the axioms provide. If you didn't want to use any definitions then you could still prove every theorem but it would be a lot more busy work -- kind of like writing a computer program in binary instead of a programming language. As for your question about ordinals and cardinals, here they are: Definition: The ordinals can be defined as follows: A set S is an ordinal if it is well-ordered by membership and if a set U is a member of S then U is also a subset of S. Cardinals can then be defined as follows: Definition: For an ordinal S, the cardinal of S, notated |S|, is the least ALEPH_K equinumerous to S (if S is infinite) or is S itself (if S is finite) where K is an ordinal. Equinumerosity can be defined in terms of bijections and bijections are defined in terms of functions and functions are defined by the axioms. Again, it can all be reduced back to the axioms. The cardinality of an arbitrary set can then be defined in terms of bijections, and so on. _ |