Re: infinitely many nn's = infinite nn's?

In article <45F46F84.60607@xxxxxxxxxxxxx>,
Phil <toob-headman@xxxxxxxxxxxxx> wrote:

G. Frege wrote:

On Sun, 11 Mar 2007 16:00:59 +0100, G. Frege <nomail@invalid> wrote:


Your "argument" from above doesn't work for R and Q. It's true: between any
two (different) numbers e Q there is a number e R. And between any two
(different) numbers e R there is a number e Q. BUT (in contrast to the
example you mentioned above) there are _infinitely many_ numbers e Q
between any two numbers e R and there are _infinitely many_ numbers e R
between two numbers e Q. Hence the nice "picture" you considered above
"breaks down."

( In contrast to your picket fence ...-picket-space-picket-...,
h e r e there are no immediate "successors" of the considered
objects. )

Actually, it's completely misleading since there are _uncountable many_
numbers e R between any two numbers e Q (but only countable many numbers e
Q between any two numbers e R).


Note though (especially the last sentence)...

"The irrational numbers may just be defined as the set of gaps in Q. Since
the set of the irrational numbers has the cardinal aleph, in our case the
set of the gaps has a greater cardinal than Q itself; this phenomenon is
not surprising since the cuts are essentially subsets and not members of
the set. By mere spatial intuition or classical geometrical attitudes
without set-theoretical methods, the phenomenon can neither be stated nor
comprehended." (Fraenkel, Levy)


F.

Actually, you cannot show that one argument's conclusion is false by
providing another argument with a different conclusion. That merely
shows that one of the arguments is probably flawed (it could also be a
premise problem, but that's uncommon). You have to find a flaw in one of
the arguments. In other words, although you keep stating that "since
proof B is true, therefore proof A is false," that doesn't necessarily
follow, since it could be proof B that is flawed. Since the guy is
attacking the beliefs of current set theory, simply REPEATING the
current beliefs of set theory is not a valid defense against those
attacks. Mind you, in this case I'm on your side, since I also think
there are infinitely many more irrationals than rationals, but I hate to
see sloppy reasoning being used to defend my beliefs.

Phil

The set of rationals has been shown to be countable by construction of
explicit injections from the rationals to the naturals and surjections
from the naturals to the rationals.

To claim that the reals ,or even the irrationals are countable would
require either an injection from the irrationals to the naturals or a
surjection the other way round.

Such constructions have been proved impossible, by proofs that have not
been successfully attacked.

So until someone has overcome the proof of that impossibility by
actually constructing either the needed injection or the needed
surjection, the proof of their impossibility stands.
.

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