Re: Real Discontinuity in Cantor Diagonal

On Aug 27, 8:32 am, Balthasar <nomail@invalid> wrote:
Am Tue, 26 Aug 2008 13:06:23 -0700 (PDT) schrieb slartibartfast:



The [proof] considers the "entire infinite list of all
reals" all in one chunk and [...].

Maybe you have problems to comprehend the method of indirect proof. Where
"impossible things" are formulated as "facts" in the course of the proof
[which itself is starting with a FALSE claim, the assumption] till we
arrive at a (obvious) contradiction.

It's right that THERE is no "entire infinite list of all reals". But *that*
is what we want to prove in the first place. Hence we may start with the
ASSUMPTION that such a list exists:

        Assume that there is an infinite list of all reals.

Then we proceed with the proof, i.e. we derive conclusions from this
assumption (by applying the diagonal argument, btw.). Finally we get a
contradiction. And by RAA we conclude that the negation of our assumption
holds, i.e.

        There is no infinite list of all reals.


The issue is more subtle than that:
I do not have a problem with indirect proofs, but the reduction ad
absurdum proof that the diagonal argument appears to be may be only
that: an appearance.
for the contradiction drawn via it, is contingent upon the
diagonalizability of something called a list (or ordering) of "ALL"
the reals, it is this contingent assumption that is at issue which is
different from the fact that such an entity would produce a
contradiction IF diagonalized.
a valid reduction argument has all its steps applicable and THEN draws
a contradiction. if the "entity" (the list of all reals - whatever
that could mean) could be shown to be diagonalizable THEN (an not
before) the diagonal argument could validly be used to show that such
a list would be in fact an imposibibity.
or more precisely "that the set of reals would be UNcountably
infinite".

To put it another way;
Assume that there is an infinite list of all reals
Assume that such a list canNOT be diagonalized.
now what?
what do you do now? the argument cannot proceed!
you cannot in this case construct any number, contradictory or
otherwise.
and you cannot now via this route conclude anything about the list of
reals.



the diagonal argument rests on TWO assumptions not one:
viz 1. that "there is an infinite list of all reals " and 2. that "
such a list WOULD be diagonalizable if it existed"

the diagonal argument, as a reductio ad absurdum argument, can only
prove by reductio (ie by assuming true... then a contradiction) that
the first of these is false by IMPLICITLY ASSUMING the second of them
to be true. (and so proceed)


and it is this second "assumption" that is at issue.
one needs to have diagonalizability shown to be possible for whatever
the "infinie list of all reals" in the first assumption, is defined to
be


But actually, *I* would tend to avoid the indirect proof here. It clouds
the fact that the diagonal argument is eminently _constructive_ (if
formulated in a direct way).

Just consider an ARBITRARY (but fixed) list of real numbers. Now apply the
diagonal argument. This way we get a real number which is NOT in the
(considered) list. Since the list was arbitrary this shows that for _every_
list of real numbers there is (at least) one real number that is not in the
list.

HOWEVER THAT is not the diagonal argument:for in none of these cases
is there a contradiction. its not a contradiction to say some real
isnt in a list.

you could equally well say for any arbitrary (but finite) list of
NATURAL numbers there is a natural number that is not in the list.

the diagonalization only BECOMES the diagonal ARGUMENT of cantor, when
the list is NOT any arbitrary list of reals .... but ONLY when what is
being consitered is A list LR, of the "one" the "only" infinite set R,
of ALL reals.
(i.e. there is no element of R that is NOT in LR)
for only then would the ability to construct a real which wasn't in
the list of "ALL" of them be a self-contradiction.

THAT is the RAA. That should clarify the diagonal argument for you .

however it is contingent upon assumption 2 being true. and it is this
that I would like proven.






(Note that it's not necessary to mention if the considered list is finite
or infinite, the diagonal argument works for both cases.)

B.

.

Relevant Pages

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