
Re: Uncountable Diagonal Problem
Posted:
Dec 31, 2012 10:21 PM


On Dec 31, 1:18 pm, Virgil <vir...@ligriv.com> wrote: > In article > <6eaa4cc3ff1d4cf0b27100155cca9...@oi3g2000pbb.googlegroups.com>, > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > > > > > > > > > > On Dec 30, 9:58 pm, Virgil <vir...@ligriv.com> wrote: > > > In article > > > <166aaacd16b647b8925bbb5b42023...@vb8g2000pbb.googlegroups.com>, > > > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > > > > This is from that, the constructed sequences (here ordinallyvalued) > > > > of nesting endpoints, of an interval, either meet or don't. > > > > In comprehensible! > > > > Why is it that Ross' attempts at expression himself about things > > > mathematical are always more obfuscating than clarifying? > > > > The rest is silence! > > >  > > >http://en.wikipedia.org/wiki/The_Oak_and_the_Reed > > But Ross is, as yet, no more than an acorn. > 
Ah, that's rich. It's a rather poor metaphor, but how low you'll stoop for it is notable.
Unless you've discovered some way to blank the memory of others, there are quite a few who could recount the salient points from memory, at least of the definition of what the E.F. or EF is, whether or not they would care to defend it or discount it. And, even though you've displayed of yourself a lack of memory, I wouldn't so ascribe that lack thereof to others.
I see you won't quite comprehend, a scene of a battle of wits: http://princessbride.8m.com/script.htm#19 .
Good luck with that.
So, in the wellordering of the reals, for any initial segment, as defined by an ordinal, where the wellordering is a function from the ordinals onto the reals, in the courseofpassage there is a concomitant interval defined by the elements of the reals in their wellorder. These intervals, nested in the previous, see that the endpoints converge. Do they meet? If they don't, the interval is non empty, and there's not an element of the reals there, eventually in order. If they do meet, where there are no points within the interval, they are consecutive, or duplicate.
You could well note that the interval endpoints converge to each other, in the countably infinite. The endpoints meet: or don't. Cantor's first has their intersection: nonempty.
http://en.wikipedia.org/wiki/Cantor%27s_first_uncountability_proof
So, do they: do, or don't?
Maybe it's as simple as that any wellordering of the reals, contains duplicates. Would that not be a surprising concomitant fact? Yet, then only note to nest the unique ones.
Wellorder the reals. While you're at it: count the integers.
Regards,
Ross Finlayson

