An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Virgil on 27 Jun 2006 22:28 In article <1151441124.445419.327130(a)i40g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1151332059.355968.171920(a)p79g2000cwp.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > But whether or not this is accepted: If infinity exists, then also > > > infinitely many transpositions do exist, and none of them will destroy > > > the well-ordering. > > > > It remains for "mueckenh" to provide any transposition, or sequence of > > transpositions, which will convert a non-empty well ordered set into one > > not having a first element. > > My mapping is chosen, on purpose, so that it does not convert > well-order into non-well-order. That supplies the contradiction. I fail to see a contradiction of any sort, except for the contradictory claim that there is some contradiction "supplied".
From: Virgil on 27 Jun 2006 22:44 In article <1151441411.720682.246290(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > It is not "Cantor's list" but ANY list. > > You are challenged to find a list which lists all reals, then Cantor > > presented a rule for showing any list you provide is incomplete. > > I have already shown that any list of all or even some rationals and > its diagonal number is uncountable. There is no need to consider all > reals. You have not shown me that the set of all rationals, or any of its subsets, is uncountable. And as I have many times proven to myself otherwise, and seen many others' similar proofs, you never will. > > > > And each time I implement the diagonal number K(f) into the list f > > > (insert it before the first ordinary list number, for instance), you > > > will construct another diagonal number K(g) and say, look here, this > > > diagonal is not contained in your g. > > > > As that K(g) was not in M(f) , that is irrelevant. > > That the diagonal of Cantor's list is not in the list is irrelevant. That is the one thing about Cantor's anti-diagonal number that IS relevant. How you can so carefully miss the point time and again, argues that you have some skill at being unskilled.
From: Virgil on 27 Jun 2006 23:06 In article <1151441565.220298.319790(a)j72g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > But, independent of what were Cantor's actual thoughts, if infinity > > > would exist, we could order the rationals by magnitude and maintain the > > > well-order. > > > > Actually, that would require that no infinity exist, or at least that > > the set of rational be finite, so that the rationals in their natural > > order be well-ordered, as is every finite ordered set. > > > > That an infinite set of rationals in their usual order are not, and > > cannot be, well ordered is trivial: > > > > In order for a set to be well ordered, every non-empty subset > > must have a smallest member > > > > The non-empty set of positive rationals has no first element, > > since for every positive rational, 1/2 of it is strictly between > > it and zero. > > This shows that set theory is self contradictory, because the following > mapping does not destroy well-order but establishes order by magnitude. > > > For n = 1 to oo: > (q_2n-1, q_2n) --> (q_2n-1, q_2n) if q_2n-1 < q_2n, else: (q_2n-1, > q_2n) --> (q_2n, q_2n-1) > For n = 1 to oo: > (q_2n, q_2n+1) --> (q_2n, q_2n+1) if q_2n < q_2n+1, else: (q_2n, > q_2n+1) --> (q_2n+1, q_2n) > And again: > For n = 1 to oo: > (q_2n-1, q_2n) --> (q_2n-1, q_2n) if q_2n-1 < q_2n, else: (q_2n-1, > q_2n) --> (q_2n, q_2n-1) > For n = 1 to oo: > (q_2n, q_2n+1) --> (q_2n, q_2n+1) if q_2n < q_2n+1, else: (q_2n, > q_2n+1) --> (q_2n+1, q_2n) > ... > repeat aleph_0 times. Where is your proof that it takes the rationals in some given sequential order and ever produces an ordering by magnitude? As there is no such thing as a smallest (most negative) rational, how does all of your transposing produce what does not exist? As far as I can see what you have is sort of like an infinite bubble sort, with large values being bubbled upwards past smaller ones, one at a time, until they bump into an even larger value. But at no stage of this operation does there ever get to be infinitely many values between two other values. For example, in the original sequence there are at most finitely many values between 0 and 1. You are claiming that there will be some single transposition which makes the up until then finite number of values between 0 and 1 in one step into an infinite number of values between 0 and 1.
From: Virgil on 27 Jun 2006 23:09 In article <1151441681.849284.48750(a)i40g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > If you can describe the number as you have done, you have made it > > exist, even if you cannot necessarily compare it for size with another > > number. > > If numbers cannot be compared for size, they do not deserve the name > "number". If no one got anything undeservedly, the world would be a much sadder place.
From: Virgil on 27 Jun 2006 23:14
In article <1151441760.374615.105520(a)b68g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1151319403.823915.220910(a)c74g2000cwc.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb:> > > > > > Virgil. Such proofs must be wrong in general, and even the allmighty > > > > > could change this fact. > > > > > > Should read: could not change this fact. > > > > > > Regards, WM > > > > Your lack of faith in an Almighty to be almighty, is self-contradictory. > > I recently proved that there is no Almighty (stone, force). That alleged "proof" is no more valid than your alleged mathematical proofs. While as far as I know there very well may not be any "Almighty", no one yet has a valid proof of that. And considering "mueckenh"'s incompetence with mathematical proofs, he will not be the first to produce a valid proof on no almighty. |