Virgil
Posts:
8,833
Registered:
1/6/11


Re: Matheology � 263
Posted:
May 14, 2013 4:43 AM


In article <fe808d300f124c9587083d6053afe0b6@oy9g2000pbb.googlegroups.com>, Graham Cooper <grahamcooper7@gmail.com> wrote:
> On May 14, 4:46 pm, Virgil <vir...@ligriv.com> wrote: > > In article > > <d6f681f1613b4e57a3365ab501a04...@wb17g2000pbc.googlegroups.com>, > > Graham Cooper <grahamcoop...@gmail.com> wrote: > > > > > > > > > > > > > > > > > > > > > On May 14, 1:36 pm, Virgil <vir...@ligriv.com> wrote: > > > > In article > > > > <d8620fe3928d4bc5bf24b16bee326...@wb17g2000pbc.googlegroups.com>, > > > > Graham Cooper <grahamcoop...@gmail.com> wrote: > > > > > > > On May 14, 11:09 am, Virgil <vir...@ligriv.com> wrote: > > > > > > In article > > > > > > <4f6cc18e90b2415e83aa963e1c083...@n5g2000pbg.googlegroups.com>, > > > > > > Graham Cooper <grahamcoop...@gmail.com> wrote: > > > > > > > > > such as Virgil's favorite number! > > > > > > > > > 0.44444454444444444445444444545544444444445444444444444... > > > > > > > > That denotes, as yet, any of a range of real numbers, not any > > > > > > specific > > > > > > one, and whichever ones in that range Graham finds his favorite, > > > > > > none of > > > > > > them are anything like my favorite. > > > > > > > Real numbers of that form are all you need to show > > > > > > I don't need to show any any such numbers. > > > > > > >  POINTS  >  INFINITE LIST  > > > > > > > between these 2 bars! > > > > > > > >< > > > > > > > Here's another one > > > > > > > 0.4444444444445444444444454444445444444444454444445444444... > > > > > > > Remember your hero CANTOR showed you how to CONSTRUCT that number! > > > > > > > You post 20 times a day the Algorithm (sic) to construct that real! > > > > > > The algorithm I regularly post, and Cantor first used, is for binary > > > > sequences not decimals. > > > > > > Neither type of "antidiagonal" is defined without an infinite list of > > > > sequences of the the appropriate type from which to build it, which > > > > lists you have not provided, so no antidiagonal need exist until you > > > > do. > > > > > Such algorithms have been posted 100 times. > > > > > Though You have no clue what Cantor's Missing Set function actually > > > does. > > > > > SET1 = { 1 , 3 , 6 } > > > SET2 = { 1 , 5 , 11 } > > > SET3 = { 2 , 4 , 6, 8 , 10 , ... } > > > SET4 = { 4 , 5, 6, 7, 8 } > > > > > [VIRGIL] > > > > > Given an arbitrary function f from N to the powerset of N (set of > > > all subsets of N), the set S = {n in N  ~ n in f(n)} is a subset of > > > N not in the image of f, and thus is a proper "Cantor's missing > > > set". > > > > > You learnt this magic formula off by heart and you have no idea how to > > > apply it! > > > > I have learnt the quadratic formula off by heart, too, though, at need I > > can derive it from the quadratic equation, a*x^2 + b*x + c = 0, and > > apply it. > > > > > > > > > and the Missing Set from the above enumeration is.... ? > > > > In order to be able to use the definition "S = {n in N  ~ n in f(n)}" > > and thus determine which sets are missing in the image of a given > > function, f: N > 2^N, one must first be able to determine all the > > values of that function, i.e., one subset of N for each member of N.. > > > > If you only give me > > > > f(1) = { 1 , 3 , 6 } > > f(2) = { 1 , 5 , 11 } > > f(3) = { 2 , 4 , 6, 8 , 10 , ... } > > f(4) = { 4 , 5, 6, 7, 8 } > > > > All I know so far is that that your f cannot be such a function > > because 1 is in f(1) and 4 is in f(4). > > > > > <BZZZT!> > > Wrong! Try again, what about 2? Is that in your missing set ?
Depends on whether f(2) = { 1 , 5 , 11 } or not.
If f(2) = { 1 , 5 , 11 } and f(3) = { 2 , 4 , 6, 8 , 10 , ... } then 2 and 3 will be in that set, S, but that leaves all infinitely many n in N with n > 4 still undetermined as to membership in S where "S = {n in N  ~ n in f(n)}" > > > Herc >  > www.BLoCKPROLOG.com 

