
Re: Uncountably Nested Intervals
Posted:
Jan 4, 2013 11:52 AM


On Jan 3, 9:32 pm, Virgil <vir...@ligriv.com> wrote: > In article > <b0302dd56a044af09ae4690cffd26...@pd8g2000pbc.googlegroups.com>, > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > > > > > > > > > > On Jan 3, 7:02 pm, Virgil <vir...@ligriv.com> wrote: > > > In article > > > <b7e06477b83641b1be03c4d0fe3c2...@q16g2000pbt.googlegroups.com>, > > > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > > > > On Jan 3, 9:07 am, "Ross A. Finlayson" <ross.finlay...@gmail.com> > > > > wrote: > > > > > On Jan 2, 12:48 am, Virgil <vir...@ligriv.com> wrote: > > > > > > > In article > > > > > > <de9ee3af08234a9982167b6033235...@po6g2000pbb.googlegroups.com>, > > > > > > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > > > > > > > On Jan 1, 11:22 pm, Virgil <vir...@ligriv.com> wrote: > > > > > > > > In article > > > > > > > > <ef09c567163746b8932abcb856e41...@r10g2000pbd.googlegroups.com > > > > > > > > >, > > > > > > > > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > > > > > > > > > On Jan 1, 8:59 pm, Virgil <vir...@ligriv.com> wrote: > > > > > > > > > > In article > > > > > > > > > > <5e016173aa1b48349d700c6b08f19...@jl13g2000pbb.googlegroup > > > > > > > > > > s. > > > > > > > > > > com>, "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > > > > > > > > > > > On Jan 1, 7:29 pm, Virgil <vir...@ligriv.com> wrote: > > > > > > > > > > > > In article But in that proof Cantor does not require a > > > > > > > > > > > > well > > > > > > > > > > > > ordering of the reals, only an arbitrary sequence of > > > > > > > > > > > > reals > > > > > > > > > > > > which he shown cannot to be all of them, thus no such > > > > > > > > > > > > "counting" or sequence of some reals can be a count or > > > > > > > > > > > > sequnce of all of them.  > > > > > > > > > > > > Basically > > > > > > > > > > > Nonsense deleted!  > > > > > > > > > > Nonsense deleted, yours? > > > > > > > > > Nope!  > > > > > > > > Great: from demurral to denial. > > > > > > Seems clear enough: in ZFC, there are uncountably many irrationals, > > > > > each of which is an endpoint of a closed interval with zero. And, > > > > > they nest. Yet, there aren't uncountably many nested intervals, as > > > > > each would contain a rational. > > > > > To whit: in ZFC there are and there aren't uncountably many > > > > > intervals. > > > > > Then, with regards to Cantor's first for the wellordering of the > > > > > reals instead of mapping to a countable ordinal, there are only > > > > > countably many nestings in as to where then, the gap is plugged (or > > > > > there'd be uncountably many nestings). Then, due properties of a well > > > > > ordering and of sets defined by their elements and not at all by their > > > > > order in ZFC, the plug can be thrown to the end of the ordering, the > > > > > resulting ordering is a wellordering. Ah, then the nesting would > > > > > still only be countable, until the plug was eventually reached, but, > > > > > then that gets into why the plug couldn't be arrived at at a countable > > > > > ordinal. Where it could be, then the countable intersection would be > > > > > empty, but, that doesn't uphold Cantor's first proper, only as to the > > > > > finite, not the countable. So, the plug is always at an uncountable > > > > > ordinal, in a wellordering of the reals. (Because otherwise it would > > > > > plug the gap in the countable and Cantor's first wouldn't hold.) > > > > > > Then, that's to strike this: > > > > > "So, there couldn't be uncountably many nestings of the interval, it > > > > > must be countable as there would be rationals between each of those. > > > > > Yet, then the gap is plugged in the countable: for any possible value > > > > > that it could be. This is where, there aren't uncountably many limits > > > > > that could be reached, that each could be tossed to the end of the > > > > > wellordering that the nestings would be uncountable. Then there are > > > > > only countably many limit points as converging nested intervals, but, > > > > > that doesn't correspond that there would be uncountably many limit > > > > > points in the reals. " > > > > > Basically that the the gap _isn't_ plugged in the countable. > > > > > > Then, there are uncountably many nested intervals bounded by > > > > > irrationals, and there aren't. > > > > Yes there are, as I pointed out in a posting that Ross has carefully > > > snipped entirely. > > > > The set of intervals { [x,x] : x is a positive irrational} is one > > > such set of uncountably many nested intervals bounded by irrationals. > > > > A simple, and obvious, example of what Ross claims does not exist. > > > > > Point being there are uncountably many disjoint intervals defined by > > > > the irrationals of [0,1]: each nonempty disjoint interval contains a > > > > distinct rational. Thus, a function injects the irrationals into a > > > > subset of the rationals. > > > > This too is false. > > > { [x,1x] : x is an irrational between 0 and 1/2} being an explicit > > > counterexample. And as there are way more such intervals than rationals > > > in their union, no such injection from intervals as Ross claims to > > > rationals can exist. > > > > And Ross is totally wrong again!!! > > > > And Ross will, no doubt, snip all of this proof of his errors too, just > > > as he did the last one, if he repies at all. > > >  > > > That example contains zero, a rational, no? > > No! For z between 0 and 1/2, no interval from x to 1x will contain 0. > > > > > What, that is news? Once again your plain arguments against the man > > instead of for the argument show your lack of argumentative ability, > > and responsibility, and poor form. > > Since my ARGUMENT was entirely a mathematical example refuting your own > mathematical claim, it is ad mathematics not ad hominem. > > Though I did enjoy being able to show your mathematics to be totally > wrong! > > > But, for me to note that, is it ad > > hominem, to note ad hominem? > > It is certainly an ad hominem to claim it when it did not exist, as you > did. > > > See, for that I would refrain: because > > it's less than perfectly ethical to argue ad hominem. > > Particularly when you are in the wrong and trying to cover your ass. > > > Also quit > > bullying me, I'm bigger than you. A suitable change of topic for the > > thread, to respect the time of readers, is more along the lines of > > "Uncountably Nested Intervals". > > > Uncountably many nested intervals, each pairwise disjoint contains two > > rationals, or rather as nested their disjoint contains a rational. > > "Uncountably many nested intervals, each pairwise disjoint"? > > Nested intervals are not pairwise disjoint, at least in any real world. > > > > > The rationals are dense in the reals. Deal with it. > > What misleads Ross onto thinking I don't already? > > Re the original issue: > > There are countably infinite sequences of nested intervals with rational > endpoints. For example { (1/n,1/n) : n in N }, but obviously no > uncountable nested set of such sequences. > > Ross then claimed that there could not be any uncountable set of nested > intervals with irrational endpoints, which is trivially false: > { (x, x) : x is a positive irrational} is just such an uncountable but > nested set of intervals with irrational endpoints as Ross had claimed > did not exist. > > So Ross was wrong, and too chicken to own up. > 
No, what I said was there are and aren't. I simply constructed examples where there are and examples where there could not be. That's a poor and ungenerous representation. Your argument is simply fallacious, where generally your mathematical content can be replaced with a text reader. You should speak well of yourself, I don't care if you do or don't. Obviously enough I was thinking of the irrationals in [1/2, 1/2] containing zero.
Here, the conundrum is that in ZFC there are uncountably many irrationals, that there be uncountably many disjoint intervals, that each contains a rational, which are countable in ZFC, which would be a contradiction.
Regards,
Ross Finlayson

