Re: Cantor K.O.'d -- again!
From: Randy Poe (poespam-trap_at_yahoo.com)
Date: 01/12/05
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Date: 12 Jan 2005 11:26:27 -0800
Mark Adkins wrote:
> Our little day-hike of infinity may thus be considered illustrative
> of the adage solvitur ambulando. (Despite Randy Poe's hysterical
> invective to the contrary, this does not properly translate as "proof
> by foot-stomping", even if it does entail a kind of forced march.)
"Hysterical"? Did you mean that in the sense of
hilariously funny, or in the sense that somehow my
making that accusation indicated panic on my part.
At any rate, I now have a Latin phrase for proof by
foot-stomping: Argumentum ad pedem supposionem.
> Our tour originates at the Origin. Here, the first symbol of the
> diagonal is laid. It is also, of course, the first (and only) symbol
> of the first list member. At this point, the diagonal and the first
> list member are identical.
Cut to the chase: Each number is mapped to a set of discrete
points ending on the diagonal, being squeezed tighter and
tighter down to the bottom of the square. So element
n is a distance 1/n from the bottom, and terminates on
the diagonal at a distance 1/n from the right.
Clearly along the diagonal there are an infinite
number of entries, one for every n=1,...,inf.
> Did we ever reach a stage of the list at which the list members
> became infinite? No, because by design each list member was
> indexed to a natural number, and every natural number is finite.
> Did the diagonal whose construction we were supervising, stage
> by stage, ever become infinite, then? No, because at *every* stage
> the diagonal mirrored the list at that point. No infinite list
member,
> no infinite diagonal.
>
The pedem supposionem occurs here.
> Because the diagonal intersects and mirrors the list at every stage,
> the only way the diagonal could be infinite would be for a list
member
> to be infinite,
That's a bald assertion, but unfortunately is not true.
The diagonal is infinite, every list member (at distance
1/n from the bottom) is finite.
> and for the two to meet at the lower-right corner of the
> unit square (1,-1), where we are now.
Clearly there's no list member "at" the bottom. Also
clearly there is no end to the process of going from
1/n to 1/(n+1).
> Of course, there is no "last" list member; and each member, being
> by design indexed to a natural number, must contain a finite number
> of symbols; therefore, a member with an "infinite" number of symbols
> is not possible; and a symbol consisting of a circle with an
"infinite"
> number of dots inside it is not a possible symbol for our list.
That's right. There's no last member, and there's no
infinite member.
> And,
> of course, the diameters of the circular symbols have been
> decreasing as the limit point at (1,-1) has been approached: at the
> point (1,-1) any such circle would have to have a diameter of zero,
> and a "circle of diameter zero" is also a contradiction in terms.
No it's not. It's perfectly consistent with the definition
of a circle as the set of points at radius r from a
center C. If r=0, the set is {C}, which is a perfectly
good set.
Again, pedem supposionem.
So what is it you're claiming is impossible?
> The solution, of course, is that L* does not exist, and therefore
> cannot be mapped into a unit square. It does not exist because
> its very postulation entails logical inconsistency. L* poses as an
> entity consisting of discrete parts which is simultaneously
> "limitless" and "completed": but there is no such entity because that
> is a contradiction in terms.
Perhaps you're imposing this new condition "completed"
on the list that none of us require.
> One cannot have "all" of something that is definitionally limitless.
In
> fact, the term "natural numbers" does not even specify a set of
> individual numbers,
Huh? Are you saying that integers aren't numbers, or
aren't "individual numbers"?
> but merely references a particular *concept* of
> number; a set of rules or template through which individual,
specific,
> actual entities can be created. There is no {N}.
Of course there is an N. I'm not sure what the notation
{N} is supposed to mean. We can easily test whether any
object is a member of N or not. That makes it a set
with clearly defined membership.
> There are an
> unlimited number of {n}, each of which is finite. N is merely a
general
> criterion for classification, much like "individuals named Smith".
The set would be all objects meeting that classification.
The set isn't the criterion.
> There is no "set of all Smiths" because there is no theoretical limit
> to the number of Smiths; one can keep constructing new Smiths
> as long as the original Smith, or his progeny, continue to procreate
> and pass the name Smith through their lineage.
And every one is a member of the set of Smiths. And
those not named Smith are not. What's wrong with that?
That you don't a priori know how big the set is?
> Similarly, new
> natural numbers can always be created and named,
Well, no. We can't create something new which doesn't
fit the Peano axioms and call it a "natural number".
Here we have an advantage over the Smiths in that time
is not involved and we have defined a priori every element
of the set N.
Just as your diagonal contains a point (1-1/n, 1-1/n)
for every natural number n. You can't "create and
name" any more... they're already there. Your line
is complete. It contains every single such point.
- Randy
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