On 11 Jan., 15:44, Zuhair <zaljo...@gmail.com> wrote: > On Jan 11, 3:52 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > >
> > > > They cannot spring off Cantor's argument. > > > > They do of course, they are a consequence of his arguments. > > > No, you misunderstand again. Cantor's opinion was (and did not change > > until he died) that undefined items are nonsense. And of course he was > > absolutely right.
No answer. You are babbling "of course", of course, but you are unable to support your babblings, of course, because you don't know Cantor's writings. You exlude yourself from scholarly discussion.
> It is easy to prove that ALL reals are discernible.
It is easy to prove that two given reals are discernible. But undefinable reals cannot be given. > > Now the question is: Is all reals finitely definable? > or Equivalently: is it the case that for each real r if r is > discernible then r must be finitely definable. > or Equivalently: Can there exist a real r that is discernible but not > finitely definable.
Can it exist in mathematics, i.e., in the discourse, i.e., can we talk about it? If you believe in some remote existence outside of mathematics, you may do so, but I am only interested in mathematical discourse. > > YOUR answer to that question is in the negative, i.e. you think that > there cannot exist a real that is discernible and not finitely > definable. That's your stance. > > While Cantor's arguments PROVES the existence of at least one real > that is discernible but not finitely definable.
No, it does not. If the list is undefined, then there is no discernability. If the list is defined, then the diagonal is defined too. >
> Take the list of ALL finitely definable reals.
How would you do that? These reals cannot be "listed", i.e., written line by line, and the list cannot be defined. Therefore in mathematics this list does not exist. (You can talk about it like about the set of all sets. But that does not make it appear in mathematics.)
> You hold that this list > is countable, i.e. there is a function f such that f is bijective from > N to the set of all finitely definable reals.
No. I *can prove* that there cannot be more than countably many finitely defined reals. > > Ok it is easy to prove that there can be infinitely many such > bijective functions.
You see what a rubbish your formal "proofs" are. If a list cannot be defined, then there is no function.
> Actually the standard is that MOST of reals are > not finitely definable! And yet ALL of them are discernible! i.e.
"Actually accepted" is not tantamount to "generally true". And the heads of "leading mathematicians", i.e., matheologians who spread such nonsense are obviously predominantly filled with mud.
> each > one of them do differ from all other reals at some finite position of > its decimal expansion.
You cannot understand that this is just a definition of uncountably many rationals?
If you were right, Cantor's argument would still remain true if you cut every entry of the Cantor list after its n-th digit. Then you see that the anti-diagonal differs from every entry q_n at a finite place q_nn. But if the list contains every rational q_n (which is possible), then this differing is impossible because, as I mentioned already, the anti-diagonal has at least one double up to every finite n. - And to know more than all finite digits of the anti-diagonal is not possible. Further it has no chance to differ at digits that have no finite index.
> Anyhow I don't think that further discussion with you would be > fruitful.
In particular since you would be forced again and again to answer this question, at least towards yourself. But you cannot as long as you believe in matheology.