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


Re: Matheology � 255
Posted:
Apr 24, 2013 1:47 PM


In article <89451078986746419ec0e21c343a2434@c9g2000vbr.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 23 Apr., 23:03, Virgil <vir...@ligriv.com> wrote: > > > > > > The question is whether there is a line with all nunbers of the > > > columns. > > > > That is a different question to which my answer is > > "No! There is no line with all the number of the columns." > > If all numbers of the columns are in lines, but not all are in one > line, then there must exist at least two lines (or more) containing > all naturals, but containing not all in one line.
I think he;s got it' > > > > The union of the set of all lines is a line > > > > If that were so, then WM should be able to give us the number of that > > line. > > No, there is no last natural number. Then, equally, no last line, since every line has a number.
> Nevertheless every line is the > union of itself and all its predecessors. Unioning again and again > will not increase the number of elements.
I am surprized that WM is actually aware of this/ > > {a, b, c} = U {a, b, c} = UU {a, b, c}.
FALSE!!!
Unless a, b and c are sets, "U {a, b, c}" is nonsense, and "UU {a,b,c}" is nonsense anyway. > > Same holds for the lines.
Lines are all nonsense? Only in Wolkenmuekenheim. > > > Actually, NO line is equal to the union of all preceding lines. > > Every line is the union of itself and of all preceding lines. > > > > > > All that is contained in all lines collectively, is also contained in > > > one single line s_i.
Unless it is not a last line, and if a last line then one can alwasy create another later line. > > > > Which line would than be that contains the largest member of the next > > line ? > > The next line.
How does your "last line" contains the last number of the line which follows it? > > > Which line, s_i, would it be that can contain i+1 ? > > Is i+1 in the column? Can it be there without being in a line?
Avoiding the question is not a legitiamete answer to it. > > > > > exist j, k, m, n : m e s_j & ~(m e s_k) & ~(n e s_j) & n e s_k. > > > > If, for each natural number i, s_i = { n in N: n <= i}, > > then for every h in N, (h in s_i) <==> (h <= i) > > > > Thus WM's "m e s_j & ~(m e s_k) & ~(n e s_j) & n e s_k" > > says (m <= j) & ( m > k) & (n > j) & (n <= k) > > or (m <= j) & (k < m) & (j < n) & (n <= k) > > but (m <= j) & (j <n) => (m < n) > > and (k < m) & (n <= k) => (n < m) > > so in WOLKENMUEKENHEIM one has n < n and m < m. > > That proves that there is no aleph! Only if there are more natural > number of numbers in the first column than in every line, we need the > contradictory condition: > exist j, k, m, n : m e s_j & ~(m e s_k) & ~(n e s_j) & n e s_k.
WM may need it in Wolkenmuekenheim, but no one else needs it at all, and it does not hold anywhere other then in Wolkenmuekenheim.
Wm has several times claimed it but never proved it, so that , as a claim, it need only hold where WM rules, i.e., in Wolkenmuekenheim, and thus need not, and does, not hold anywhere else.
Only in Wolkenmuekenheim need anyone have both m < n and n < m simultaneously. 

