Uncountability of the Rationals?
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From: Brian Chandler on 25 Jun 2010 14:30 Tony Orlow wrote: > On Jun 22, 12:03 am, Brian Chandler <imaginator...(a)despammed.com> > wrote: > > Incidentally, you [Transfer] managed to write quite a long screed about "strong > > bijections", without as far as I can see capturing in any sense at all > > the distinction from normal bijections. (In practice "non-strong" just > > seems to mean "Tony doesn't like it"!) > > It really means the relationship can't be formulaically quantified in > any way that leads to quantitiative ordering using IFR. In order to > state that one diverging formula is greater than another in the limit, > one needs to extend inductive proof in a way that orders such > formulas, much like Big O notation in computer and information > science. Yes, as I said, it seems to mean "Tony doesn't like it". > > 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". > > Now, you are bringing N=S^L into the picture. I am? In what way am I doing that? I merely mentioned some sets, and it looks awfully as though this triggers a particular response in the oracle. (That's you) > What you don't > understand is that the set must be bijected with N+ in some manner. It must? Any particular manner? Your next sentence appears to give me the option of doing things just so I get a particular answer for the bigulosity of A, which is odd, since I'm asking you to tell me what the bigulosity of A is. > If you are going to allow for tav elements in set A, then you are > talking about log2(tav) bit positionss (still countably infinite, so > that shouldn't bother you). If you want to say there are tav bit > positions possible, then you have 2^tav bit strings in your list > (still countably infinite, as far as Bigluosity is concerned, which > might bother you). So I can choose at least two different bigulosities for A. Hmm. I don't recall anything *remotely* like that in mathematics at university. And why would I expect *tav* to be the bigulosity of the set of strings that represent precisely(?!) *tav+1* non-negative integers in binary. Is there an odd one on the end somewhere? > Set B *might* be represented in digital format and thus become strings > of characaters rather than quantities, and if so, then we parameterize > according to string length and digital number base. As pure > quantities, N is N having Bigulosity tav+1, and not some list of > strings. So B, only has this bigulosity of tav+1 when it seems to be convenient? > When it comes to set C, it *looks* like set A, however, if there are > tav+1 naturals (starting at 0) in decimal, there are log_10(tav+1) bit > positions. Given N=S^L, with S=2 and L=log_10(tav+1), we have > 2^log_10(tav+1)=log_5(tav+1) elements. This doesn't seem to be right. Just suppose tav=999. Then 2^(log10(tav +1)) = 2^3 = 8. And 5^8 does not equal 1000. Not that details like this seem very important, somehow. > Yes, different rules work with languages than with pure quantities. > That doesn't make the system inconsistent. It appears to make the "system" entirely incoherent, since the answer to any question of the form "What is the bigulosity of set Q?" starts with a bit of negotiation, deciding what sort of answer is appropriate. I think I give up. Brian Chandler
From: Tony Orlow on 25 Jun 2010 17:55 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 > > That "definition" does not specify a single set. There are many sets > that satisfy that definition. The set R of real numbers, the set Q of > rationals, the set Z of integers *all* satisfy the definition you've given. Uh, what? I don't know where the "^n" came from, but it wasn't mine. Please don't insert deliberate obfuscation. 1 is an element of N+ and if x is an element of N+ then so is x+1. > > >> 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. > > > >> Thus, N+ has the property > > >> ind_1(N+) & (Ax)(ind_1(x) -> N+ c x). > > > How do you draw that conclusion? > > 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. It's a member, but far from the least. COnsider the size of the set of natural tetrations. > > >> 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? > > 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. TOny > > -- > Jesse F. Hughes > "That's what's brutal about mathematics! When you're wrong, you can > have spent years, and lots of effort, and come out at the end with > nothing." -- James S. Harris on the path of self-discovery (?)
From: Jesse F. Hughes on 25 Jun 2010 17:51 Tony Orlow <tony(a)lightlink.com> writes: > Where you have a minimum positive difference between successive > elements (like 1), then any infinite number of them means an infinite > difference between the first and last. Is it your opinion that N+ has a last element? -- "And God Himself won't help you if this goes bad as despite your beliefs I can assure you that angry people will against all law if necessary tear their rage out of your hides if it goes badly." -- James S. Harris, on the dangers of criticizing his mathematics
From: Tony Orlow on 25 Jun 2010 17:58 On Jun 25, 12:54 pm, David R Tribble <da...(a)tribble.com> wrote: > Tony Orlow wrote: > > Okay, I should clear something up, here, as a misconception seems to > > be spreading. Bigulosity does not just measure subsets of N+. It > > measures any set formulaically bijected with N+, which may include > > elements outside of N+. > > So by your definition, bigulosity is a relative measure of countable > sets only. Uncountable sets (such as R) can't have bigulosities.\ Wrong. They are handled by different methods, as there is no finite formulaic relationship between the two. Then, there is language using N=S^L, whether countable or uncountable. > > > Secondly, any uncountable set, when ordered linearly (or with a finite > > number of child nodes each), must contain an element which is > > infinitely distant from the foundation of the set. A countably > > infinite set never does, which is why it's not "actually" infinite. > > Since R+ (the set of positive reals) is an uncountable set, which > element of it is "infinitely distant" from its foundation? And what > is the "foundation" of set R+? How far is H-riffic 3 from 2 (in base 2 H-riffic) ? We cannot determine a rational pattern there, or can we? TOny
From: Tony Orlow on 25 Jun 2010 18:04
On Jun 25, 2:30 pm, Brian Chandler <imaginator...(a)despammed.com> wrote: > Tony Orlow wrote: > > On Jun 22, 12:03 am, Brian Chandler <imaginator...(a)despammed.com> > > wrote: > > > Incidentally, you [Transfer] managed to write quite a long screed about "strong > > > bijections", without as far as I can see capturing in any sense at all > > > the distinction from normal bijections. (In practice "non-strong" just > > > seems to mean "Tony doesn't like it"!) > > > It really means the relationship can't be formulaically quantified in > > any way that leads to quantitiative ordering using IFR. In order to > > state that one diverging formula is greater than another in the limit, > > one needs to extend inductive proof in a way that orders such > > formulas, much like Big O notation in computer and information > > science. > > Yes, as I said, it seems to mean "Tony doesn't like it". > > > > > > > > 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". > > > Now, you are bringing N=S^L into the picture. > > I am? In what way am I doing that? I merely mentioned some sets, and > it looks awfully as though this triggers a particular response in the > oracle. (That's you) > > > What you don't > > understand is that the set must be bijected with N+ in some manner. > > It must? Any particular manner? Your next sentence appears to give me > the option of doing things just so I get a particular answer for the > bigulosity of A, which is odd, since I'm asking you to tell me what > the bigulosity of A is. > > > If you are going to allow for tav elements in set A, then you are > > talking about log2(tav) bit positionss (still countably infinite, so > > that shouldn't bother you). If you want to say there are tav bit > > positions possible, then you have 2^tav bit strings in your list > > (still countably infinite, as far as Bigluosity is concerned, which > > might bother you). > > So I can choose at least two different bigulosities for A. Hmm. I > don't recall anything *remotely* like that in mathematics at > university. > > And why would I expect *tav* to be the bigulosity of the set of > strings that represent precisely(?!) *tav+1* non-negative integers in > binary. Is there an odd one on the end somewhere? > > > Set B *might* be represented in digital format and thus become strings > > of characaters rather than quantities, and if so, then we parameterize > > according to string length and digital number base. As pure > > quantities, N is N having Bigulosity tav+1, and not some list of > > strings. > > So B, only has this bigulosity of tav+1 when it seems to be > convenient? > > > When it comes to set C, it *looks* like set A, however, if there are > > tav+1 naturals (starting at 0) in decimal, there are log_10(tav+1) bit > > positions. Given N=S^L, with S=2 and L=log_10(tav+1), we have > > 2^log_10(tav+1)=log_5(tav+1) elements. > > This doesn't seem to be right. Just suppose tav=999. Then 2^(log10(tav > +1)) = 2^3 = 8. And 5^8 does not equal 1000. Not that details like > this seem very important, somehow. > > > Yes, different rules work with languages than with pure quantities. > > That doesn't make the system inconsistent. > > It appears to make the "system" entirely incoherent, since the answer > to any question of the form "What is the bigulosity of set Q?" starts > with a bit of negotiation, deciding what sort of answer is > appropriate. > > I think I give up. > Brian Chandler- Hide quoted text - > > - Show quoted text - You should have done that a long time ago, Brain. T |