Re: Cantor K.O.'d -- again!
From: Randy Poe (poespam-trap_at_yahoo.com)
Date: 01/13/05
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Date: 13 Jan 2005 09:28:13 -0800
msadkins04@yahoo.com wrote:
> No, you don't. So far as I can tell, "supposionem" is
> not Latin and your phrase as a whole is wrong. The Latin
> for stamp (with the foot) is pedem supplodere; the Latin
> for to stamp underfoot is conculcare.
You're right. I trusted too much to my own memory.
The term is "pedum supplosionem". "Pedes" is an
adjective meaning by foot, and "pedes supplosionem"
is the term for a foot stomp. The term is spelled
correctly in my thread in alt.language.latin, and in
a thread I started here with the term in the subject line.
> > > 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.
>
> Instead of "cutting to the chase" why don't you try doing
> the math. Your formulae are wrong. For example, given
>
> y = -1 + 1/2^(n-1)
>
> the third member (n=3) is 1/4 from the bottom, not 1/3.
I thought the difference between n and n+1 was
unimportant when we are talking about limits as n
goes to infinity, but if you like, then fine. The n-th
member is 1/(n+1) from the bottom. What possible difference
does that make in the argument?
How about if we redefine m to be n+1 and then substitute
m for n in what I said. Does that help?
> > The pedem supposionem occurs here.
supplosionem
> Since you're using a meaningless pseudo-Latin expression,
> you've said nothing. I suppose you're trying to say that
> my argument (which, in your hysteria, you refer to as
> "foot-stomping") is erroneous: but you haven't demonstrated
> that, you've merely asserted it.
Yes, I'm saying you made a statement ("because... the only
way...") which claims a causal connection between precedent
and conclusion which does not actually exist.
So let's see. So far your critiques of my argument are
that I used n instead of (n+1), and that I omitted the
letter "l" in the latin phrase. You are aware these are
not substantive critiques, right?
If you prefer your phrasing to "ad p.s." I'll use it,
as in...
> > > 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,
You haven't demonstrated that, you've merely asserted it.
Because in fact the diagonal is infinite, and no list
member is infinite.
"Infinite" is defined as "not finite". If the diagonal
were finite, it would have a length M and no element
at M+1. Yet for any M, the corresponding element is
of length M, and there is an element M+1. Therefore
the diagonal is not finite, since no such M exists.
Also, every list element is of course finite.
> > That's a bald assertion, but unfortunately is not true.
>
> It's not a bald assertion: it's supported by an examination
> of the list design and construction, as explained in patient
> detail in my post. Why "unfortunately"?
No, it's not. The conclusion you are drawing from the
"list design and construction" does not follow from
the list design and construction. There is a leap
happening here.
> > The diagonal is infinite, every list member (at distance
> > 1/n from the bottom) is finite.
>
> Again, this expression of yours, 1/n, is incorrect:
1/m, where m = n+1
> and I've
> already carefully demonstrated why the diagonal cannot in fact
> be "infinite".
Is the diagonal finite?
- Randy
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