Uncountability of the Rationals?
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From: Brian Chandler on 17 Jun 2010 10:17 Tony Orlow wrote: > On Jun 16, 7:45 pm, David R Tribble <da...(a)tribble.com> wrote: > > Tony Orlow wrote: > > >> Sometimes there are logical arguments themselves which don't appeal to > > >> everyone equally. How can that be? If the "logical argument" is actually invalid or nonsensical it wouldn't appeal to me, but then I wouldn't call it a logical argument. If the logical argument is valid why should anyone care whether you like it? > > >> 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? Invalid? Nonsense? Hard to decide. Anyway this is not what you have been told, it is your mangled misunderstanding of what some misguided soul whiled away a few hours trying to drum into your head. > If the set is an initial segment of N+ with no largest member it has > no size. I have said this about a zillion times (or at least tav). I think (hope) "N+" means something recognisable like the naturals. These go on and on ("to the right") without ever ending. To me "initial segment" means a proper subset of the naturals that "stops somewhere", which is equivalent to saying "has a largest element". Can you explain what an initial segment with no largest element would look like? Hereabouts, you seem to have been saying that if a set doesn't have a last member to "be its size" this means the set "has no size"? Please confirm: is this the same as "has no bigulosity"? Do you have a definition of "has a size" for a set? Brian Chandler
From: David R Tribble on 17 Jun 2010 11:42 Tony Orlow wrote: >> [...] the tree of values, the foundation. Additionally, if one wants to >> include all reals positive and negative then the H-riffics may be >> redefined as: >> E(0) >> E(x) -> E(-2^x) ^ E(2^x) > David R Tribble wrote: >> This still omits vast (uncountable) subsets of of the reals. >> 3, for example, is not a node in the binary tree that is E. > Tony Orlow wrote: > It is an uncountable number of iterations down the tree. No, it is an countably infinite path of the tree. However, you don't define the tree as having infinite paths, but as each H-riffic being a node in the tree, the end of some finite path, each being at the end of a countable number of left/right bifurcations. You appear to have a basic misunderstanding of "countable" and "uncountable". Tony Orlow wrote: > Unless, of > course, you think there are only countably infinite segments oof this > tree, in which case it would apprear to be some kind of well order, > not havbing any infinite descending sequences. Yes, as long as the H-riffics are nodes along finite-length paths (which is what your definition above produces), then the set of H-riffics is countable and well orderable. But it is composed of only a countable subset of the reals, and omits a rather large uncountable subset of the reals. So I wouldn't get too excited about it if I were you. David R Tribble wrote: >> As I pointed out to you several times, any real of the form >> r = f*k^m, for integers f, k, and m, and k /= 2, is not a node >> in the E tree. You've never addressed this. > Tony Orlow wrote: > I have as I just did. Can you think of any sequence or finitely- > branched structure that could include all the reals and yet not have > two that were infinitely distant from each other within that > structure? I can't. I can't either. If I could, I could probably imagine a structure that could well order the reals (or some other uncountable set). David R Tribble wrote: >> Likewise, your definition is inherently described by a countable >> binary tree, where each node is an H-riffic. Therefore, E can't >> possibly contain all the uncountable reals, nor any uncountable >> subset of them. > Tony Orlow wrote: > Not by a countable tree, no, but by an uncountable one, yes. Right. As longs as the each H-riffic is a node along a finite path in the tree, the H-riffics are only a countable set. Your definition above is exactly that. David R Tribble wrote: >> Now (as Walker pointed out previously) if you want to amend the >> H-riffics to include all infinite paths in the tree, that's fine, as >> there are an uncountable number of such paths. However, once >> you do that, you lose the well-ordering of the tree. The countable >> nodes are still well-orderable, of course, but the infinite-length >> paths are not. (Choose any infinite path in the tree, then ask >> what the "next" path is; there is none.) > Tony Orlow wrote: > If there is an infinite path from 2 to 3, then there is a next one > after that, as Transfer pointed out. Not within any tree within infinite paths. The infinite path that is (the Cauchy limit to) 3 does not have a sibling path. Each finite path (H-riffic node) has left and right sibling nodes, sure, but no infinite path does. It's the same problem with any infinite binary tree containing infinite paths. Choose any infinite path (which has no end node) and you can't get to its "next" right or left neighboring path. That's essentially the continuum problem, and the whole problem with trying to well-order the reals (or any other uncountable set) using countable constructions. David R Tribble wrote: >> So you can't have both an uncountable number of elements >> and a well-ordering on them. > Tony Orlow wrote: > Then you believe there is no well ordering on the reals, and you must > therefore reject the Axiom of Choice? What on god's green earth gave you that idea? When I say "you can't have" I mean "you", Tony Orlow. Specifically, "your" H-riffics can't be both an uncountable set and a well-ordered set. Perhaps I should have said: | So you can't have both an uncountable number of elements | and a well-ordering on them with a structure like yours.
From: Jesse F. Hughes on 17 Jun 2010 12:17 David R Tribble <david(a)tribble.com> writes: > David R Tribble wrote: >>> This still omits vast (uncountable) subsets of of the reals. >>> 3, for example, is not a node in the binary tree that is E. >> > > Tony Orlow wrote: >> It is an uncountable number of iterations down the tree. > > No, it is an countably infinite path of the tree. However, you > don't define the tree as having infinite paths, but as each > H-riffic being a node in the tree, the end of some finite path, > each being at the end of a countable number of left/right > bifurcations. It's not even obvious to me that every real is represented by a (finite or infinite) path in the tree. Tony's condition is pretty strange to me. The tree is built something like this: 0 1 2^(-1) 2 2^(-1/2) 2^(1/2) 2^(-2) 2^2 2^-(2^-(1/2)) 2^(2^(-1/2)) 2^-(2^(1/2)) 2^(2^(1/2)) 2^-(2^(-2)) 2^(2^(-2)) and so on. That last row should have two more entries: 2^-4 and 2^4. Approximations would be (apologies for any errors): 0 1 0.5 2 0.707 1.414 0.25 4 0.613 1.633 0.375 2.666 0.841 1.189 0.0625 16 Every entry in Tony's tree (aside from 0 and 1) has the form 2 ^ (+/- (2^ (+/- (2^ ... (+/- (2 ^ (+/- 1))) ...)))) I don't really see that we can guarantee to get as near to any arbitrary real as we want by traversing this tree. The Stern-Brocot tree this ain't! Of course, there might well be a proof that any real can be approached by paths in this tree, but I wouldn't want to bet on it. -- Jesse F. Hughes "There are VERY FEW real mathematicians and I am one of them. Few of you can handle the pressure of real mathematics, like being wrong, while I demonstrably can." -- James S. Harris
From: MoeBlee on 17 Jun 2010 12:26 On Jun 17, 7:47 am, Tony Orlow <t...(a)lightlink.com> wrote: > On Jun 16, 6:53 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > 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. Damn, you went line by line again anyway, and, against all my better judgment, here I go again. But PLEASE don't just come back with another line-line post. The proper response by you to this post now is one of two: (1) Oh, I understand how it works now. I see how mathematical definitions work now. And I see what you mean about definining 'ordinal' in terms of primitives (though I'd have to see the full details in a book or work out the rest of the exercise myself in order to know to a moral certainty). Thank you very much. or (2) I still don't get it. It seems I would need a book to follow along with more specifics. Because as it stands now you are FLAUNTING that you are VERY BASICALLY ignorant on this matter. And without even all the technical details spelled out, this really boils down to VERY BASIC common sense in mathmatics that virtually anyone who does mathematics understands (even as it goes back (in possibly imperfect) form to Euclid and his method for geometry): We define terms with other terms. Those other terms are in turn defined by other terms. But after going back far enough in a finite sequence, every term ultimately reduces to the primitives which are not defined. PLEASE just be quiet for a moment to understand the above. > > > > Also not among the primitives are mentions of > > > > > subsets > > > > Wrong. > > > In the PRIMITIVES, there is NO subset symbol. > > Oh. What do I see in the Axiom of the Power Set? Just as I said, 'subset' is not a PRIMITIVE. That does not contradict we may write certain axioms using defined symbols. You're not LISTENING at all. We can write the power set axiom at least two ways (I'm leaving off the initial universal quantifiers, as I will do generally): (1) ExAy(yex <-> y subset of z) (2) ExAy(yex <-> Av(vey -> vez)) With (1) the axiom is NOT written in only the primitives, since 'subset of' is NOT a primitive. With (2) the axiom IS written in only the primitives. > Of course, these can be written without the subset operator, as > "S c T" is equivalent to "Ax (xeS -> xeT)". EXACTLY! And 'is an ordinal' can be written with just the primitives. > Is 'e' a primitive > operator in set theory? It's a primitive RELATION symbol. It's a primitive 2-place predicate symbol. > > > "X is a subset of Y" can be expressed as "aeX -> aeY", > > > Actually as Aa(aeX -> aeY), just to be exact. > > Yeah, "forall" is a little superfluous. Ultimately it suffices to say > "if a is an element of X then a is an element of Y". But not if you write it this way: X subset Y <-> (aeX -> aeY) That is not correct and does need a universal quantifier, so that correct is: X subset Y <-> Aa(aeX -> aeY) > > And, yes! subset is DEFINABLE from 'e'. > > > Just as 'is an ordinal' is DEFINABLE from the mere primitives. > > Is it? Demonstrate as I did for you just now. Why should I do your homework for YOU?! Damn, I'll at least get you started: x is an ordinal <-> Er(the membership relation on x is a well ordering of x and the converse of membership on x is a well ordering of x) then we only need to define each of the following in terms of 'e': membership relation on well ordering converse And each of those may be defined in terms of other defined terms, and soon enough (usually it takes a bout a chapter or two of a book in set theory), we get all terms used defined back to 'e' and '='' (from now on, I'll say just 'e', since even '=' can be defined by 'e' if we want to). > > > 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. > > Or, you just say X=the empty set <-> Ax (~xeX) That uses the "the" definite description operator. I guess it's okay. But the point is that FIRST we had to PROVE that there exists a unique object X such that Ax ~xeX. > > 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. > > Please demonstrate. Get a book in set theory! I shouldn't be asked to do your homework for YOU. (Anyway, I gave the start up above in this post). > > 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. > > You only have 7 or so axioms. How many steps can it take? As many or as few as we like. We can do it in one rather complicated step. Or we can break it up into as many smaller steps (successive defined terminology). You REALLY know virtually NOTHIHNG about mathematical definitions! And you're not LISTENING to what I've said. And I can tell that AGAIN you're just replying to me line by line instead of reading my whole post first so that you can get the context. Here, you missed my later point that for PREDICATE symbols the axioms DON'T MATTER. It has nothing to do with what axioms we have whether we can define 'is an ordinal' from 'e'. For OPERATION (function) symbols we do need axioms to prove the existence and uniqueness clauses. (And another note about that regarding the Fregean method I also mentioned). > > 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.) > > You "mind the drudgery" of defining ordinals? How do you expect me to > be convinced that you have any idea what you're talking about? NO, NO, NO. I did NOT say "mind the drudgery of defining ordinals". I said, "mind the drudgery of [writing the definens in the pure primitive language]". You're understanding virtually nothing of what I've said, because (for reasons I can surmise) you don't WANT to understand this subject. And my comment about drudgery doesn't in the least reflect that I wouldn't know what I'm talking about. > > > > union > > > > Axiom > > > The union axiom has NO union symbol in it. The union axiom is in the > > PRIMITIVES of ZFC. > > No, union is defined by 'e': > X U Y = {a: aeX v aeY} Listen, FOOL, the union AXIOM pertains to UNARY union not to BINARY union. It is from the combination of union AND pairing axioms that we get BINARY union. The definition is: XuY = U{X Y} (Note: I'm speaking in terms of Z set theory. In ZF, we don't need a pairing axiom, but rather derive it from the other axioms. But still, the definition I gave above is ordinary in ZF too). And IN ANY CASE, my point is made: Neither the unary union symbol nor the binary union symbol are PRIMITIVES and we may state (and usually do) the union axiom WITHOUT the the union symbol or the binary union symbol. > > > 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. > > Wouldn't it be easier to simply demonstrate how ordinals are logically > implied by the axioms than go through all this verbiage? I begin to > think you cannot do it. You're question is ILL PREMISED. You're understanding virtually NOTHING of this. 'is an ordinal' is a PREDICATE. We are not concerned to proof EXISTENCE just for the mere purpose of DEFINING a PREDICATE. THEN, we may turn to proving that indeed there do exist ordinals. And we DO that. It's trivial: 0 is an ordinal. (Your exercise, if you doubt.) Thus Ex x is an ordinal. And we may prove the existence of many more ordinals. Then from the axiomn of infinity, we prove that there exists a limit ordinal. And from the axiom schema of replacement we prove that there are more limit ordinals too. > > > > 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))) > > If it takes you that long to define intersection then deriving > ordinals from ZFC must be drudgery indeed. Whether it is drudgery is a psychological question for each person. It does not bear on my MATHEMATICAL point that 'is an ordinal' is definable from the primitives. And also that we can prove that there exist ordinals (as we defined from the primitives) from the axioms. And, as a matter of fact, it is an interesting process to a lot of people. What I said would be drudgery is actually writing the definition in the primitive language. We don't need to do that just to show that we CAN do that. Instead, we prove that definitions are elminable and, so then, by induction on any finite sequence of definitions, it follows that any defined term at the end of a finite sequence of definitions reduces to primitives. > XnY = {a: aeX v aeY} > > It's that easy. You FOOL, that is BINARY intersection. Again, that is defined from unary intersection and pairing: XnY - /\{X Y} Anyway, my point is sustained that we define back to primitives: /\ is defined back to primitives { } is defined back to prmitives so n (the binary intersection symbol) is defined back to primitives > > > > 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. Damn, IF you're going to post back to me line-line, then don't skip stuff as the above. Tell me either you understand it or not, IF you're going to line-line me in other places. > > > > 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). You went right past this. And that you did that shows among your responses. Predicate symbols (relation symbols) don't rely on axioms. > > > 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. You understand or not? > > 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'. You understand or not? > > > > 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. You understand or not? I didn't write all this stuff just for you to AVOID understanding it. > > > > 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. > > The equality symbol doesn't even require 'e'. a=b <-> (a->b ^ b->a). > It's basic logic. You're a TOTALLY confused FOOL. Above you used '=' in the sense of a SENTENTIAL CONNECTIVE. Your 'a' and 'b' are PROPOSITIONAL symbols (or meta-variables ranging over formulas) and not TERMS. You used '=' in the sense of '<->'. And, since '<->' is there being defined, we can't use it as the connective in the definition itself. So it would be: "a <-> b" stands for "(a->b) & (b->a". Now, getting back to '=' as a RELATION SYMBOL that is used with TERMS, what I said is correct, we may take '=' as a primitive 2-place predicate symbol (and add the axioms of identity theory) or we may take '=' as defined: x=y <-> Az(zex <-> zey) & (xez <-> xey)) > > > > 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. > Okay, so, rather than simply explain how ordinals are implied by the > axioms you'd rather go through all of this, No, FOOL, they are TWO DIFFERNT matters: (1) Defining 'is an ordinal' in terms of the primitives (or in terms of previous things that are themselves definable back to the primitives, so that 'ordinal' too is reducible to the primitives). (2) Once (1) has been done, the proving Ex x is ordinal. > which is equally as > tedious, and much less fruitful. Too bad. Let me know if you change > your mind. Let me know if you EVER grasp the most basic concept of definition in mathematics. MoeBlee
From: MoeBlee on 17 Jun 2010 12:32
On Jun 17, 11:26 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > x is an ordinal <-> Er(the membership relation on x is a well ordering > of x and the converse of membership on x is a well ordering of x) CORRECTION. I was thinking of something else (and didn't even quantify correctly). I should have said: x is an ordinal <-> (x is well ordered by membership on x & x is membership-transitive). Anyway, that all can be reduced to primitives. (See Jesse's post earlier.) MoeBle |