Virgil
Posts:
9,012
Registered:
1/6/11


Re: Matheology � 285
Posted:
Jun 13, 2013 7:50 PM


In article <d63675360d8241c3a4b42d6f576dc485@googlegroups.com>, mueckenh@rz.fhaugsburg.de wrote:
> On Wednesday, 12 June 2013 22:59:24 UTC+2, Zeit Geist wrote: > > > > > > Every finite one can, but not the entire set of Q. > > > > I prove that all elements of Q can be wellordered by size. You cannot > > > disprove it, because you cannot find any element that stays outside, can > > > you? > > > > > > > > You proved nothing! > > Your nded unfouopinion is not of interest in this connection.
But WM's opinion is provably of immensely negative value on this and most other mathematical issues. > > > > You proposed an algorithm, > > > > But never showed that it conclusively does what you say. > > You can take whatever algorith you like to wellorder by size any finite set > of rationals.
And it will fail to wellorder any subset of the rationals that is not normally wellordered.
A FINITE union of well orderedsets with compatible orderings will also be wellordered.
An INFINITE union of wellordered sets with compatible orderings need not be wellordered.
Since infinite unions, like other infiniteness, is exiled from the wild weird world of WMytheology, the set of rationals in WMytheology, being necessarily finite in WMytheology, may well be wellordered to begin with, but outside WMytheology, it is not and cannot be made so by WM's wand waving.
The set of rationals with its standard ordering is never wellordered outside the wild weird world of WMytheology, no matter what WM tries to do to it.
WM's false claim is that an INFINITE union of compatibly ordered wellordered sets must be wellordered. It is this provably false assumption on which WM's provably false argument depends.
WM claims that unioning any sequence of wellordered sets , for example, {1}, {2,1}, {3, 2, 1}, ... will force the resulting infinite union, such as {...,3, 2, 1}, to be wellordered (i.e., every nonempty subset of {...,3, 2, 1} including {...,3, 2, 1} itself has, at least according to WM's claim, a most negative member). To repeat, WM claims that every nonempty subset of {...,3, 2, 1}, including the set itself, must have a first/mostnegative member.
I challenge WM to provide us with any first/mostnegative member of that set of all negative integers.
Or, for that matter, of any of its supersets, including Q.
> Since every rational number q_i belongs to a finite set of i > rationals, this proves that all rationals can be ordered by size 
Only in the wild weird world of WMytheology, where the set of rationals is finite. > > > > > > > > > Any finite set of real numbers can be well according to magnitude. > > > > Infinite ones, not necessarily. > > I stated that all rationals that belong to FIS of the enumeration can be > wellordered by size.
Which may well be the case WITHIN the wild weird world of WMytheology, where all sets are finite, since l finite ordered sets are all automatically wellordered, but is not true anywhere that allows the rationals to be densely ordered. > > > > > > > > > > > Either find a q that stays outside of the wellorder by size. If such a q > > > exists and is enumerated as q_n, then there is a first natural number n > > > that cannot be put in a permutation such that all q's are in order by > > > size. If such a q exists and cannot be enumerated, then countability is > > > nonsense. > > > Sure thing. > > > > If the enumeration has f(j) = 1/2, > > > > what is f(j+1), f(j+2) and f(j+3)? > > That depends on the enumeration you choose at the outset. Take Cantor's > original one. > > > > > > > > If f keeps the order we must have > > > > > > > >  f(m)  <=  f(m+1)  > > f does not keep the order. Always when you extend the set of rationals, you > order them by size. This is possible for every finite set, i.e., for every > FIS of the enumerated set. > > > > > > > > And no rational r, such that the magnitude > > > > Of r is in between that of f(m) and f(m+1). > > You are lacking understanding. > > Take the first i rationals, order them by size. Add the rational with number > i+1put them in the correct order by size with the others. And so on without > ever arriving at an infinite i and without leaving out any rational q_i or > q_i+1 or any desired rational. When inside WMs wild weird world of WMytheology, where all sets are finite, all ordered sets are also wellordered, but as soon as any infinite ordered sets are allowed, WM's arguments fail. 

