Uncountability of the Rationals?
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From: MoeBlee on 17 Jun 2010 12:54 On Jun 17, 8:51 am, Tony Orlow <t...(a)lightlink.com> wrote: > On Jun 16, 7:59 pm, "K_h" <KHol...(a)SX729.com> wrote: > > 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. > > Well, that defines the von Neumann ordinals, as described by the Axiom > of Infinity. No, there's no need for the axiom of infinity in this regard and the axiom of infinity does not "describe" the von Neumann ordinals. Whether anyone even THOUGHT of the axiom of infinity, from the 'e' symbol we define 'is an ordinal'. And whether anyone even THOUGHT of the axiom of infinity, we can prove that there are ordinals. Then the axiom of infinity is that there exists a successor-inductive set. Then, using the axiom of infinity, we prove that there exists a limit ordinal. This has been explained to about a HUNDRED times already. Really, literally, getting close to a HUNDRED times. > For N+ it is certainly true that it's well ordered, N+ is the set of positive natural numbers here? But you use 'N+' also for some set of hypernaturals. It would be better if you use some other symbol then for whatever set of hypernaturals. > but > to say that 2 is exactly a subset of 3 doesn't sound particularly > necessary. "necessary" in what way? It comes out that way with the particular definitions we have. Also, it is convenient in certain ways. If you don't want it that way, then get your primitives and axioms and define and prove as you like. > I understand that that provides the set-theoretic basis for > N, but to view quantities as sets might not capture all the qualities > of quantities. That may sound like gibberish, but my point is this. > While one MAY define something this way and call it an ordinal, it > doesn't mean this definition is ENTAILED by the axioms, DAMN, DAMN, DAMN, did anyone say the definition is entailed by axioms? NO definition is entailed by previous axioms! What we said is that 'ordinal' is definable back to the primitive SYMBOLS (or primitive concepts if we're only informal). Axioms have NOTHING to do with it. Once the language is specified to have a 2- place predicate symbol (we call it 'e') then we can make a formula with just the primitives that is our definition of 'is and ordinal'. > it simply > means it does not CONTRDICT the axioms, and so is CONSISTENT with the > axioms. Otherwise, the definition of the ordinal could be deduce > directly from the axioms, no? NO, NO, NO. Please LISTEN. Not only are definitions consistent with the axioms, but definitions have two additional special properties: elminability and non- creativity. And I've explained what that means about a hundred times to you already over YEARS and in about as many different ways too. And it doesn't even MAKE SENSE to talk about deducing definitions from axioms. All your posting reveals how TOTALLY clueless you are about such a basic matter as definitions in mathematics. > > 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. Personally, I wouldn't define cardinals in terms of alephs, but rather vice-versa. > Right, that definition is based on the previous one, which is what is > in question, not in terms of consistency with ZFC, but in terms of > whether the axioms imply the definition. Are you REALLY so obtuse? REALLY? Definitions go back in finite sequence to primitives. NO MATTER WHAT. Else they are NOT definitions (in a formal sense). And it doesn't even make SENSE to talk about axioms implying definitions. For OPERATION symbols ('is an ordinal' and 'is a cardinal are NOT operations but rather PREDICATES) we rely on axioms to prove the existence and uniqueness clauses that ensure the definition is proper. But still, axioms do NOT "imply" definitions (excpet in the extremely pedantic sense that definitional axioms imply themselves as any formula implies itself). > > 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. > > One can define operations consistent with the axioms which the axioms > don't directly imply. Axioms don't imply definitions. > Thanks for your explanation. That's exactly what I was looking for. And you tenaciously MISunderstood it. MoeBlee
From: MoeBlee on 17 Jun 2010 12:57 Tony Orlow: Please stop to think about this subject for at least a day or two. You just keep posting more and more replies that pile up to more and more that needs to be corrected in them. You're completely confused and in the dark about even what a mathematical definition IS. Please read back over the posts, or, if they don't clear it up for you, then get a book. MoeBlee
From: Rotwang on 17 Jun 2010 13:09 Jesse F. Hughes wrote: > Tony Orlow <tony(a)lightlink.com> writes: > >>> 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. > > x is an ordinal iff (Ay,z in x)(y in z or z in y) & > (Aw,y,z in x)((w in y & y in z) -> w in z) & > (A z)( ((Ay)( y in z -> y in x ) & (Ey)( y in z )) -> > (Ey in z)(Aw in z)( y in w ) ) & > (A y in x)(A z in y)( z in x ). ITYM (Ay,z in x)(y in z or z in y) & (Aw,y,z in x)((w in y & y in z) -> w in z) & (A z)(((Ay)( y in z -> y in x ) & (Ey)( y in z )) -> (Ey in z)(Aw in z)(y in w or y = w) ) & (A y in x)(A z in y)( z in x ) (bearing in mind that "in" is a strict order).
From: Jesse F. Hughes on 17 Jun 2010 13:26 Rotwang <sg552(a)hotmail.co.uk> writes: > Jesse F. Hughes wrote: >> Tony Orlow <tony(a)lightlink.com> writes: >> >>>> 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. >> >> x is an ordinal iff (Ay,z in x)(y in z or z in y) & >> (Aw,y,z in x)((w in y & y in z) -> w in z) & >> (A z)( ((Ay)( y in z -> y in x ) & (Ey)( y in z )) -> >> (Ey in z)(Aw in z)( y in w ) ) & >> (A y in x)(A z in y)( z in x ). > > ITYM (Ay,z in x)(y in z or z in y) & > (Aw,y,z in x)((w in y & y in z) -> w in z) & > (A z)(((Ay)( y in z -> y in x ) & (Ey)( y in z )) -> > (Ey in z)(Aw in z)(y in w or y = w) ) & > (A y in x)(A z in y)( z in x ) > > (bearing in mind that "in" is a strict order). Right. Thanks for that correction. In case others had trouble (like me) finding the correction, here it is: > ITYM (Ay,z in x)(y in z or z in y) & > (Aw,y,z in x)((w in y & y in z) -> w in z) & > (A z)(((Ay)( y in z -> y in x ) & (Ey)( y in z )) -> > (Ey in z)(Aw in z)(y in w or y = w) ) & ^^^^^^^^ > (A y in x)(A z in y)( z in x ) My expression of the well-order condition omitted "or y = w". -- Jesse F. Hughes "I'm ruler", said Yertle, "of all that I see. But I don't see enough. That's the trouble with me." -- Yertle the Turtle, by Dr. Suess
From: Virgil on 17 Jun 2010 15:23
In article <f39f6930-e0ea-450c-b418-07551788fb4a(a)u26g2000yqu.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On Jun 16, 5:51�pm, David R Tribble <da...(a)tribble.com> wrote: > > Tony Orlow wrote: > > > The H-riffics are a bit of a sidebar. Besides, it occurs to me that > > > within R there exists one element which cannot be a child node of an > > > element in R, namely, 0. So, 0 can be taken to be the root case for > > > 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) > > > > This still omits vast (uncountable) subsets of of the reals. > > 3, for example, is not a node in the binary tree that is E. > > It is an uncountable number of iterations down the tree. I was unaware that binary trees, such as the one created recursively by (�E(0), �E(x) -> E(-2^x) ^ E(2^x)) were allowed to have more than countably many iterations. How does one get those necessary infinitely many non-child elements into such a tree having only one non-child, 0? > 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. It looks to me to be similar to a standard infinite-binary-tree structure with those numbers as nodes, and it is well know that there are only countably many nodes in such a tree, but no native well ordering. > > > > > 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. > > 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. How is this irrelevancy relevant? The relevant issue is whether a set S such that 0 \e S and x \e S -> 9-2^x \e S) /\ (2^x \e S) contains all reals, which it does not seem to do. > > > > > 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. > > Not by a countable tree, no, but by an uncountable one, yes. Show us how you construct a binary tree with uncountably many nodes, TO! since every node in the standard tree every node aready has all its children, where do all those extra children get their parent nodes? > > > > > 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.) > > If there is an infinite path from 2 to 3, then there is a next one > after that, as Transfer pointed out. Nonsense. No one has yet provided an explicit well ordering of the reals, though many have tried, so that TO's casual handwaving doesn't cut it. > > > > > So you can't have both an uncountable number of elements > > and a well-ordering on them. > > Then you believe there is no well ordering on the reals, and you must > therefore reject the Axiom of Choice? We will only believe TO has found such a well ordering of the reals, or any other uncountable set, when he has presented here a proof he has found one with a proof that none of us can fault. Considering TO's past attempts at proofs, we can be reasonably confident that that will never happen. |