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Re: Uncountability of the Real Numbers Without Decimals
Posted:
Dec 5, 2013 12:39 PM


WM <wolfgang.mueckenheim@hsaugsburg.de> writes:
> Am Donnerstag, 5. Dezember 2013 13:57:41 UTC+1 schrieb Ben Bacarisse: >> WM <wolfgang.mueckenheim@hsaugsburg.de> writes: >> >> > Am Donnerstag, 5. Dezember 2013 13:06:27 UTC+1 schrieb Ben Bacarisse: >> >> WM <wolfgang.mueckenheim@hsaugsburg.de> writes: >> >> >> >> > Am Montag, 2. Dezember 2013 21:41:33 UTC+1 schrieb Zeit Geist: >> >> > >> >> > >> >> >> > The set of positive rational numbers that is less than the natural >> >> >> > number n and has not been enumerated by the first n natural numbers >> >> >> > grows with n. It is impossible eneumerate all rational numbers, >> >> >> > i.e., to remove all rationals from the state of being not >> >> >> > enumerated to the state of being enumerated. >> >> > >> >> >> Impossible? How about a proof. >> >> > >> >> > Proof (1): In order enumerate a rational, you have to take (identify) >> >> > it and map it on a natural number. Cantor tries so, but fails, since >> >> > each of his naturals belongs to a finite initial segment of N whereas >> >> > infinitely many cannot be taken (identified). >> >> >> >> How silly. Your book gives examples of bijections between infinite sets >> >> (potential or otherwise) specified by a closed formula. >> > Potential infinity! Never "all". >> >> A closed >> >> formula for a bijection between N and Q has been given here numerous >> >> times by Virgil. A formula identifies and specifies every number in the >> >> pairing at once. >> > >> > Have you ever observed a natural number other than of the initial >> > 0%set? No. And would you accept my book as the authority to falsify >> > that fact (if it said so)? >> >> I would not accept your book as an authority on anything. I am assuming >> that you, however, agree with it's contents. Why do your formulas >> define a bijection but Virgil's do not?
> My approach to mathematics is the classical one, including potential > infinity. There is a limit of a sequence like > 0.1 > 0.11 > 0.111 > ... > > namely 1/9, but there are not "all" terms. So one cannot count them. > A bijection like n <> 2n exists up to every desired or "taken" > natural number. Same is with Virgil's enumeration of the > rationals.
That's OK then. You agree that there is a bijections between N and Q just like this ones you define in your book. Is your point a quibble about words then? Does a bijection not define and enumeration? It does for the rest of us.
> But to obtain equinumerousity, you have to reach *all* > terms (or to exclude d from the rationalscomplete Cantorlist). And > that is impossible, as you could easily understanbd, if you really > tried to use a natural number that does not belong to the first 0 > percent. Why don't you try it?
No need. In mathematics the bijection defines an enumeration, and with the convenience of actual infinite sets, all elements are mapped. To top it off, we also define equinumerousity that way.
 Ben.



