In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 9 Apr., 20:15, William Hughes <wpihug...@gmail.com> wrote: > > > > > > You said that the Binary Tree, when constructed from its FISONs, does > > > not contain their suprema. > > > > nope. > > So it contains the suprema? Uncountably many?
Of the same cardinality as the power set of |N. > > > > > > > > > You said that the sequence > > > 0.1 > > > 0.11 > > > 0.111 > > > ... > > > does not contain 1/9. > > > > yep > > > > > But this list contains an infinite sequence of > > > finite unions. > > > > Nope. The list only contains lines. > > Every line is the union of itself and all its predecessors. Unless you > 'd like to consider indices, consider the sequences of FISONs > > 1 > 1, 2 > 1, 2, 3 > ... > > There it is clear. > > > No line is an infinite > > sequence of finite unions. > > But if the list contains infinitely many (more than any finite number > of) FISONs, then it contains infinitely many (more than any finite > number of) unions, doesn't it?
Nope. Its members are not unions. > > > So now we have: > > > > D is the collection of all finite > > lines. > > If you remove the collection of all finite > > lines from D > > i.e., if you remove, according to induction, with FIS n also FIS n+1 > > > nothing is left > > If you remove any one line (and all its predecessors) > > every natural number is left. > > > > This is exactly what you keep saying > > is a contradiction.- > > > A contradiction is your claim: > > 0.1 > 0.11 > 0.111 > ... > > does not contain 1/9
Is WM is now claiming that a strictly increasing sequence will contain its limit value? Note that every member of that sequence is clearly strictly less than that limit.
While that may work in his WMytheology, it does not work anywhere else.
> infinitely many appended 1's are not sufficient > to yield the decimal fraction of 1/9
Sure they are sufficient, because appending infinitely many digits results in the limit.
> but *there is* a decimal > fraction of 1/9. And this is accomplished by the union of all lines
But the "union of all lines" is not a member of the sequence
> > This is the contradiction: In my list
Your 'lists' are always too ambiguous.
You deliberately obscure the essential difference between membership in a set and subsets of a subset, and between a sequence and its limit. --