Re: Calculus XOR Probability

imaginatorium@xxxxxxxxxxxxx said:
Tony Orlow wrote:
Matt Gutting said:
Tony Orlow wrote:
MoeBlee said:
Tony Orlow wrote:
MoeBlee said:
Tony Orlow wrote:
To say
all elements are finite but the set is infinite implies that there is some
lartgest finite with an infinite successor.
In what theory? In set theory? No, what you said is not true of set
theory. So in some other theory? If in some other theory, then what are
its axioms and primitives and what are its definitions of 'infinite',
'largest', 'finite', and 'successor'?
I was speaking of the logic behind the limit ordinals, and how it is flawed.
You said "To say all elements are finite but the set is infinite
implies that there is some lartgest finite with an infinite successor."
That statement contradicts set theory. So my question, which you did
not answer, is: In what theory do you cliam that your statement holds?

And there is no flaw iin the logic behind limit ordinals. The only
logic involved in set theory is first order predicate logic.

I suppose what it boils down to here is infinite induction. Since the size of
the set is the successor to the maximal element in all finite sets, this
relationship should hold, being an equality, in the infinite case. If that is
so, then omega is successor to the largest finite, and the notion is self-
contradictory. Infinite induction appears to be discredited, but the only
counterexample in this thread is easily explained otherwise, and "infinity did
it" doesn't fly when it comes to explaining an error of sqrt(2). The limit
ordinal omega is in direct contradiction with infinite induction, so one of
them is wrong. Hint: it's not infinite induction.

The problem I see with this is that the finite cases, as you yourself
point out, deal with sets having maximal elements. Since this is not true
for "the infinite case", I don't see how you can apply the same reasoning
there.

Matt

*** Posted via a free Usenet account from http://www.teranews.com ***


What does that have to do with anything? This is an inductive proof that for
every finite natural, the number of naturals up to that point is finite. Does
that not mean that there is no point at which the set of finite naturals is
infinite? If it does not become infinite within the range of elements that it
includes, how is it infinite? By including things outside the set?

I'm not prepared to discuss anything with you involving the i-word,
because I have no idea what you mean by it (and I wonder if you do
really).

There really is a problem with this stuff you keep saying about sets
"growing", or in this case talking about the "point" at which a set
"becomes" [Tinfinite]. Well, do you remember saying this:

"Yes, the set of bits is bounded for every specific pofnat, but there
is no
bound on the length of the string required for all pofnats." Tony Orlow

That's how it comes about that for any particular pofnat P given at the
beginning of the argument, it is the case that the set of pofnats up to
P is a finite set. But there is no Q you can give at the beginning of
the argument such that Q is a pofnat, and the entire set of pofnats
does not include anything exceeding Q. There is no "point" in the
pofnats at which the set "becomes [Tinfinite]", but neither is there
any point at which the pofnats stop. Thus those of us constrained not
to use the i-word say that the set of pofnats is unending.

Brian Chandler
http://imaginatorium.org



I agree it's unending, and in that sense may be considered infinite, as that is
compatible with the definition regarding injection into a proper subset. Within
set theory, that works without contradiciton. However, I am speaking in more
quantitative terms, and relying on a set of rules which determine whether an
arithmetic operation produces a finite or infinite value depending on its
arguments. For instance, if A and B are finite and nonzero, A+B, A-B, A*B, A/B,
A^B and logA(B) are all finite, and nonzero except for subtraction if A=B.
Given those rules, and N=S^L, we can say that if S and L are both finite, that
is, we have only finite length strings and a finite alphabet, then N, the size
of the language so defined, is also finite. This relates directly to the
naturals as well, through the digital number systems, which are languages,
since any finite natural only requires a finite string, and if all strings are
finite in the set, the set is also finite, given any finite number base. There
is no bound on L, but if L cannot be infinite, then neither can N.
--
Smiles,

Tony
.

Relevant Pages

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