Uncountability of the Rationals?
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From: David R Tribble on 8 Jun 2010 23:21 Tony Orlow wrote: >> By the axiom of extensionality, as sets they are identical. > David R Tribble wrote: >> No, they are not. Each sequence is a *different* set. >> >> I'll use a more expressive notation to depict these sequences >> as actual sets: >> S1 = {<0,0>, <1,1>, <2,2>, <3,3>, <4,4>, <5,5>, <6,6>, ...} >> S2 = {<0,0>, <1,2>, <2,1>, <3,4>, <4,3>, <5,6>, <6,5>, ...} >> >> Different sequences, thus different sets. > Tony Orlow wrote: > Yeah I get all that. What is your point? By the axiom of > extensionality they are the same set with different order. No, it's painfully obvious that, no, you don't get it. Not at all. Did you even bother to look at S1 and S2? The Axiom of Extensionality says that two sets are identical if they have exactly the same members. S1 and S2 do not have the same members. So they are not the same set. The Axiom says nothing about order. More to the point, sets don't have order. Even when you do impose an ordering on S1 and on S2, they still are different sets, because they still have different members. Really, Tony, you're not helping your cause any if you can't be bothered to learn the basics. -drt
From: David R Tribble on 8 Jun 2010 23:43 David R Tribble wrote: >> But then what about other ordered sets containing members >> other than reals, hyperreals, naturals, or numbers at all? Does >> your "bigulosity" still apply to them, in general? >> >> For example, consider the sets constructed from the "primitive" >> element 'o'. Some examples: >> T1 = { o } >> T2 = { o, {o} } >> T3 = { o, {o}, {{o}} } >> >> Likewise consider some infinite sets built in this same way: >> R1 = { o, {o}, {{o}}, {{{o}}}, ... } >> R2 = { o, {{o}}, {{{{o}}}}, {{{{{{o}}}}}}, ... } >> >> Now we'll define an order relation '<' for these particular sets. >> We define o < {o}. So in general, >> o < {o} < {{o}} < {{{o}}} < {{{{o}}}} < ... >> >> Now we can see that all of the sets above can be ordered >> sets, using our new '<' relation. >> >> So the question is, given these ordered sets, can you calculate >> the relative bigulosity for them? Specifically, does R1 have a >> larger or smaller bigulosity than R2? > Tony Orlow wrote: > That depends how the strings are interpreted. Standard set notation (obviously). > If each bracket in one > is equal to one bracket in the other, R2 is a subset of R1, including > all the "even" members of the set. So far, so good. > When dealing with the countably > infinite, they can only be compared parametrically, and the > correspondences need to be defined. Bigulosity is not as simple as > cardinality, for sure. Not all situations are easily handled, and some > may be viewed from a couple different perspectives, giving different > results. But, since omega doesn't really exist as a number in my > theory, that's not really a problem. Oh, there you dropped the ball. You completely avoided answering my question, which was meant as a concrete example to answer the bigger question: "is bigulosity general enough for to be useful for all sets?" Because, you see, cardinality applies to every set. If bigulosity only applies to a tiny fraction of sets, well, its usefulness is equally tiny. More to the point, it can't be a replacement for cardinality, since there are so many sets it can't be used for, and cardinality works for every set. But since you'd like it to be just such a replacement, and it just plain can't, that makes it, well, uninteresting. Your bigulosity kind of resembles Natural Density (or Asymptotic Density) (for sets) or perhaps Schnirelmann density (for sequences), both of which have been around for some time: http://en.wikipedia.org/wiki/Natural_density http://en.wikipedia.org/wiki/Schnirelmann_density Perhaps you should try expanding on those ideas. -drt
From: Jesse F. Hughes on 8 Jun 2010 23:51 Tony Orlow <tony(a)lightlink.com> writes: > Bigulosity is not as simple as cardinality, for sure. Not all > situations are easily handled, and some may be viewed from a couple > different perspectives, giving different results. In other words, Bigulosity is not a well-defined mathematical concept at all. -- Jesse F. Hughes "Dead men can't talk. Especially when they've been cremated." --- From the 1944 radio program "Adventures By Morse"
From: Brian Chandler on 9 Jun 2010 03:02 Jesse F. Hughes wrote: > Tony Orlow <tony(a)lightlink.com> writes: > > > Bigulosity is not as simple as cardinality, for sure. Not all > > situations are easily handled, and some may be viewed from a couple > > different perspectives, giving different results. > > In other words, Bigulosity is not a well-defined mathematical concept > at all. Just about sums it up, I think. Reminds me of the bright child: "OK, Johnny, what's 7 times 12?" "Right, Dad, are we buying or selling?" Brian Chandler
From: Tony Orlow on 9 Jun 2010 09:51
On Jun 8, 11:43 pm, David R Tribble <da...(a)tribble.com> wrote: > David R Tribble wrote: > >> But then what about other ordered sets containing members > >> other than reals, hyperreals, naturals, or numbers at all? Does > >> your "bigulosity" still apply to them, in general? > > >> For example, consider the sets constructed from the "primitive" > >> element 'o'. Some examples: > >> T1 = { o } > >> T2 = { o, {o} } > >> T3 = { o, {o}, {{o}} } > > >> Likewise consider some infinite sets built in this same way: > >> R1 = { o, {o}, {{o}}, {{{o}}}, ... } > >> R2 = { o, {{o}}, {{{{o}}}}, {{{{{{o}}}}}}, ... } > > >> Now we'll define an order relation '<' for these particular sets. > >> We define o < {o}. So in general, > >> o < {o} < {{o}} < {{{o}}} < {{{{o}}}} < ... > > >> Now we can see that all of the sets above can be ordered > >> sets, using our new '<' relation. > > >> So the question is, given these ordered sets, can you calculate > >> the relative bigulosity for them? Specifically, does R1 have a > >> larger or smaller bigulosity than R2? > > Tony Orlow wrote: > > That depends how the strings are interpreted. > > Standard set notation (obviously). > > > If each bracket in one > > is equal to one bracket in the other, R2 is a subset of R1, including > > all the "even" members of the set. > > So far, so good. > > > When dealing with the countably > > infinite, they can only be compared parametrically, and the > > correspondences need to be defined. Bigulosity is not as simple as > > cardinality, for sure. Not all situations are easily handled, and some > > may be viewed from a couple different perspectives, giving different > > results. But, since omega doesn't really exist as a number in my > > theory, that's not really a problem. > > Oh, there you dropped the ball. > > You completely avoided answering my question, which was > meant as a concrete example to answer the bigger question: > "is bigulosity general enough for to be useful for all sets?" In some cases Bigulosity cannot make a greater distinction than cardinality, and may have to simply classify a set as countably infinite. It is not as simple as cardinality, and therefore not as universal, but includes a collection of notions which lead to solutions to a great many situations which better staisfy most intuitions. > > Because, you see, cardinality applies to every set. If bigulosity > only applies to a tiny fraction of sets, well, its usefulness is > equally tiny. More to the point, it can't be a replacement for > cardinality, since there are so many sets it can't be used for, > and cardinality works for every set. But since you'd like it > to be just such a replacement, and it just plain can't, that > makes it, well, uninteresting. Well, when you come up with a tool that replaces the screwdriver, hammer and wrench, I will admire your rock with great disinterest as well. > > Your bigulosity kind of resembles Natural Density (or Asymptotic > Density) (for sets) or perhaps Schnirelmann density (for sequences), > both of which have been around for some time: > http://en.wikipedia.org/wiki/Natural_density Only works for finite fractions of N, and not for any higher power. For instances it works for any set of naturals mapped from N by y=ax+b but not, for instance, for the set of squares of naturals, which has a density of 0. Not general enough. > http://en.wikipedia.org/wiki/Schnirelmann_density This looks a little heady to wade through right now but, "the Schnirelmann densities of the even numbers and the odd numbers, which one might expect to agree, are 0 and 1/2 respectively". That's nothing like Bigulosity. I'm rather amazed that folks may be willing to accept simpler, less general ideas such as these, but object to infinite-case induction and the inverse function rule as being too restrictive because it's not as general as cardinality. It does not compute. > > Perhaps you should try expanding on those ideas. > > -drt- Hide quoted text - > > - Show quoted text - IFR goes far beyond either of those already. (sigh) TOny |