Uncountability of the Rationals?
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From: Jesse F. Hughes on 26 Jun 2010 08:47 Tony Orlow <tony(a)lightlink.com> writes: > On Jun 25, 7:37 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> Tony Orlow <t...(a)lightlink.com> writes: >> > On Jun 25, 12:41 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> >> Tony Orlow <t...(a)lightlink.com> writes: >> >> > On Jun 25, 10:48 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> >> >> Tony Orlow <t...(a)lightlink.com> writes: >> >> >> > For the purposes of IFR, N+, the positive naturals, is the standard >> >> >> > countably infinite set. >> >> >> >> Can you tell us exactly how you define N+? This question may seem >> >> >> silly, but it really is important. >> >> >> > 1 e N+ >> >> > x e N+ -> n+1 e N+ >> >> ^n [...] >> And, of course, that typo is irrelevant to my comment. You didn't pick >> out a defining characteristic of N+, since the sets Z, Q, R, w+w, and so >> on, all satisfy the condition you wrote. > > N+ contains all and only the elements described by the rules > mentioned, plus the term "standard", or "finite", if you so desire. Yes, that's what you wanted to say, but what you said is that the definition is: 1 e N+ n e N+ -> n+1 e N+ That does *not* say that *only* such elements are in N+. In order to say that, one wants to say that N+ is the *smallest* set containing 1 and closed under successor. Hence, we define ind_1(x) <-> 1 in x & (Ay)(y in x -> y+1 in x) and let N+ be the unique set satisfying ind_1(N+) & (Ax)(ind_1(x) -> N+ c x). >> >> >> For comparison, I would define N+ to be the least inductive_1 set, where >> >> >> >> ind_1(x) <-> 1 in x & (Ay)(y in x -> y+1 in x). >> >> >> > I like mine better, but same difference. >> >> >> You're confused. I haven't yet defined N+, so it does not compare to >> >> your "definition". >> >> > If that wasn't the full definition then you have little business >> > drawing conclusions prematurely. >> >> Er, that was the full definition of ind_1. The definition of N+ is just >> below. > > That was equivalent to my full definition. What else is there? > >> >> >> >> Thus, N+ has the property >> >> >> >> ind_1(N+) & (Ax)(ind_1(x) -> N+ c x). >> >> >> > How do you draw that conclusion? > > Okay, "thus" you drew this conclusion, without having defined N+? This can be taken as the definition of N+, though one should prove that there is a unique set satisfying the above condition. I will do so on request. Perfectly standard. >> >> That's what I mean when I say that N+ is the *least* inductive_1 set[1]. (I >> >> haven't proved there *is* a least inductive set here, but I will if you >> >> wish. It's perfectly standard.) >> >> > Mmmmmm....standard. Nummies! >> >> > So, your proof is that N+ is a member of the set of inductively >> > defined and well-founded sets, and supposedly the "least" among them. >> >> "Inductive", not "inductively defined". And while it happens to be >> well-founded, that's not something that I've mentioned. It is a >> consequence of the definition, not a requirement. > > Yes, it has a first element. And a set can be defined inductively, if > you don't mind, so that becomes a vacuous correction. You are confusing my use of the predicate "inductive_1" with the notion of inductive definition. I've defined inductive_1 explicitly. >> > It's a member, but far from the least. COnsider the size of the set of >> > natural tetrations. >> >> Is that set an inductive set? That is, does it satisfy the property >> >> 1 in x & (Ay)(y in x -> y+1 in x)? > > No, that's the definition of N+, not tetration(n) for neN+. Then it is not an ind_1 set, so it's irrelevant. N+ is the smallest ind_1 set. >> >> >> Is this also your definition? >> >> >> > Apparently not. >> >> >> I guess I wasted my time with this question. You don't really >> >> understand the need for specifying N (or N+) as a *least* set satisfying >> >> some condition, because you mistakenly think that your two conditions >> >> already define N+. I hope that my examples showed you why your >> >> so-called definition does not suffice, but I honestly don't expect so. >> >> > What doesn't suffice for your purposes? What can't you prove? >> >> That there is a unique set satisfying the so-called definition. > > There is one set which satisfies all and only the conditions > specified. There's no such thing as satisfying *only* the conditions specified. This is why logicians specify N as the *smallest* inductive set. >> There are just oodles of sets containing 1 and closed under successor. >> Lots of them. Z contains 1 and is closed under successor. > > Not unit increment, aka, +1. Yes, Z contains 1 and is closed under "unit increment" (which others call successor). That is, 1 e Z and if n e Z, then n+1 e Z. This is getting tedious. If you still don't see *why* we define N+ as the smallest inductive set, then there's no reason to continue going point by point here. One last correction however: >> >> Footnotes: >> >> [1] That is, "least" modifies "inductive_1 set", not "inductive_1". N+ >> >> is the smallest set which is inductive_1. This is the definition of N+. >> >> > Consider any subset of N+, with a somewhat naive perspective, if only >> > for a moment. >> >> Any proper subset S of N+ either does not contain 1, or contains an n such >> that n+1 is not in S. Thus, no proper subset of N+ is inductive_1. > > Piffle beside the point. An inductive set is one where each member can > be generated from the last, not necessarily by adding 1. No. I've told you what ind_1 *means*. A set is ind_1 iff it contains 1 and is closed under +1. I will not use ind_1 in any other way in this conversation. In this conversation, the meaning of inductive is fixed. It is *not* something that can be generated from the last. It means it contains 1 and if n is in it, then so is n+1. That is what inductive means here. Note: I've always disliked calling this property "inductive", because I've found that the term is misleading. Nonetheless, the terminology is standard. -- "This sucks," said a Pennsylvania State University student [...] " Why can't the college let me do what I want to do with my computer? The school computer security guys are being way more annoying than the spyware was." -- A student pines for his disabled spyware
From: Jesse F. Hughes on 26 Jun 2010 08:57 Tony Orlow <tony(a)lightlink.com> writes: > On Jun 26, 12:01 am, Transfer Principle <lwal...(a)lausd.net> wrote: >> On Jun 24, 10:59 pm, Tim Little <t...(a)little-possums.net> wrote: >> >> > On 2010-06-24, Transfer Principle <lwal...(a)lausd.net> wrote: >> > > So tav appears to contain all the pofnats. But it also contains >> > > elements which aren't pofnats. >> > And so we have a contradiction. Gee, that didn't take long. >> >> But now TO is posting again, so we can fix this error based >> on TO's latest comments. > > There is no error. The contradiction concerning omega or tav is > deliberate. It doesn't exist as a number, as evidenced by its own > self- contradiction. Well, *that*'s encouraging! Your system is purposely, not accidentally, inconsistent. -- Jesse F. Hughes "I think the Iraqi people owe the American people a huge debt of gratitude." -- G.W. Bush in January, 2007
From: David R Tribble on 26 Jun 2010 15:49 David R Tribble wrote: >> R+ is the uncountable set of positive reals. We'll order it >> "linearly" with the usual ordering ('<' over the reals). So it >> meets the two requirements you state above. >> >> So then, according to you, R+ "must contain an element >> which is infinitely distant from the foundation of the set". >> So: >> 1. Which element of R+ is that? >> 2. What is the foundation of set R+? > Tony Orlow wrote: > In the H-riffics the foundation is 0, and in base 2, 3 is infinitely > distant from the foundation (and vice versa). You may require proof of > this fact, and I'll work on that, because it's getting pretty > interesting again, but for now, I'm just pretty convinced, personally. > If you're not, I understand. You said nothing about your H-riffics in the requirements for an uncountable "linearly ordered" set above. Are you saying that in addition to the set being a) uncountable and b) linearly ordered, it must also be c) derived from your H-riffics, in order to answer the question, which member is infinitely distant from the foundation of the set?
From: MoeBlee on 26 Jun 2010 17:22 On Jun 25, 9:01 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > On Jun 24, 10:59 pm, Tim Little <t...(a)little-possums.net> wrote: > > > On 2010-06-24, Transfer Principle <lwal...(a)lausd.net> wrote: > > > So tav appears to contain all the pofnats. But it also contains > > > elements which aren't pofnats. > > And so we have a contradiction. Gee, that didn't take long. > > But now TO is posting again, so we can fix this error based > on TO's latest comments. > > Let's summarize what TO has stated. > > 1. The set omega is exactly what we expect it to be. I thought Orlow doesn't accept limit ordinals. Are there new updates released on that matter this week? > 2. N+ is the set of all pofnats, i.e., omega\{0}. It is a > countably infinite set. > > 3. *N contains the nonstandard naturals (which Chandler calls > "tnats," but TO calls T-riffics). There are many different sets that are a universe for a nonstandard model of (first order) PA. Indeed, PA (I'll omit saying 'first order' in this context) has models in ALL infinite cardinalities. So item 3 above is not a definition (I don't know whether it is supposed to be), even if it said "let *N be the set whose members are all and only the members of the universe of the nonstandard model of PA", because there is no "THE" nonstandard model, not even to within isomorphism. > we > can just go back to ZF and assign Bigulosities to the sets > of ZF, as I originally intended to do. This means that we can > go back to determining which bijections are "strong." I haven't been reading all the ruminations here. Please, what is the definition of a 'strong bijection'. Then what is the definition of 'Bigulosity of x' such that every x ('x' ranging over all objects, I presume?) has a bigulosity assigned to it. > But in ZF, _everything_ is a set (i.e., > there are no urelements) No, everything is a set in ZF not just because there are no urelements but also because every object in ZF is a MEMBER of some object, so there are no urelements and no proper classes. MoeBlee
From: Tony Orlow on 26 Jun 2010 18:24
On Jun 25, 1:37 pm, Brian Chandler <imaginator...(a)despammed.com> wrote: > Tony Orlow wrote: > > On Jun 24, 3:55 am, Transfer Principle <lwal...(a)lausd.net> wrote: > > > On Jun 21, 9:03 pm, Brian Chandler <imaginator...(a)despammed.com> > > > wrote: > > > > Transfer Principle wrote: > > > > > On Jun 20, 9:29 am, Brian Chandler <imaginator...(a)despammed.com> > > > > > wrote: > > Right, well, dipping down into this lengthy wossname... > > > > > > I wrote: > > > > Here's a start for you: consider the sets > > > > A = { "0", "10", "11", "100", ... } of (two-ended!) strings over > > > > alphabet {0,1} starting with 1 > > > > B = N ... the set of naturals (including 0), which we might represent > > > > in binary > > > > C = { 0, 10, 11, 100, ... } of integers whose decimal representation > > > > only includes digits 0 and 1 (no sign) > > > > I hope you can immediately see canonical bijections A <-> B and A <-> > > > > C. Tony claims that B and C have different bigulosities, so your job > > > > is to say which of the bijections (or both!) is not "strong". > > > > As I mentioned before, any alleged bijection between a set and one > > > of its proper subsets indicates that the set contains extra elements > > > which aren't mapped to unique elements. > > > No it doesn't. There are no elements unmapped. There is a difference > > in frequency of elements as the range increases. That is all IFR > > detects, but it does it accurately. > > Ah, yes, the range. I remember that... Brain still has a working neuron. > > > > In particular, we know that B contains extra elements. There is no > > > set N of pofnats in this theory, but instead the set tav of tnats: > > This seems to be Transfer's attempt to address my question... Which specifically was? > > > No, tav is the Bigulosity of N+, not a set. > > ... but your response seems to indicate he has missed the point. So > let's skip again... Sip to my loo, my daling. > > > If the strings of B are bijected string-wise with those of C, they may > > be considered equibigulous, > > OK, well, here's another copy of the definitions of B and C. Please > explain what it means to "biject string-wise" the set of naturals and > the set of integers whose decimal representation only includes digits > 0 and 1. I mean that, according to N=S^L, the number of naturals in decimal and binary differ, depending on S. > > > > > B = N ... the set of naturals (including 0), which we might represent > > > > in binary > > > > C = { 0, 10, 11, 100, ... } of integers whose decimal representation > > > > only includes digits 0 and 1 (no sign) > > Next: > > > ... but when the strings of C are specified to > > be binary naturals, then one needs to decide how one is comparing the > > two sets, and decimal notation comes into play with N=S^L. > > Well, the elements of C (sets normally have *elements* Tony, not > "strings") are integers. A particular proper subset of the non- > negative integers. What would it mean to "specify" that they are > something else. Bull. Elements may be anything, including strings, which may be mapped to other sets of strings through non-quantitative formulas. Tony |