Re: Calculus XOR Probability
- From: Tony Orlow <aeo6@xxxxxxxxxxx>
- Date: Wed, 3 May 2006 15:59:44 -0400
Virgil said:
In article <MPG.1ec29dedde64c5c198acac@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:
Virgil said:
In article <MPG.1ec1404cde84263398aca0@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:
Virgil said:
In article
<MPG.1ebfda9786fc0aa298ac93@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:
Of course IFR depends on the definition of quantity and
arithmetic formulas. Consider those already defined as usual.
Defining them "as usual" requires the mechanism of the very set
theory that TO objects to, o that TO is already assuming all
that he wants to reject.
The usual definitions for arithmetic operations have nothing to
do with transfinite set theory, though that may be artificially
woven into some axioms.
The same axioms that give us transfinite set theory are needed to
give us the naturals, and from them the other number systems on
which TO is depending.
TO can't get to where he wants to go without the same foundation
that provides the structures he is trying vainly to avoid.
What, specifically, prevents the same axioms appied to infinite or
infinitesimal units wouldn't work the same way as with finite units?
No one has to show that TOmatics doesn't work, TO has to show it does.
The position of standard mathematics is that nothing other than an axiom
system has to be accepted until it is proved, and axiom systems are only
accepted for purposes of argument.
Every mathematical theorem is really of the form:
"Given such and such a set of axioms then...."
So until TO can give his set of axioms, and the proofs, he has nothing.
If by "nothing" you mean a set of complementary concepts, then you're right.
Without a specific set as domain and another as codomain for a
given function, there is no way to tell whether it is invertible or
not.
The invertible functions of IFR map reals to reals.
Does TO mean that they are both bijections of R to itself?
No, TO means they are a bijection between the subset of the reals called the
naturals and another subset of the reals.
When measuring a
discrete set of reals, we compare to the set of naturals, which is
the "unit", or standard, discrete set.
The set of integers is also a "discrete" set
That's true, and that could also be used as a "standard" infinite set. I've
thought about which would be better, without much conclusion either way.
So, if we map the naturals in
N to reals in S through f (n),
Then the domain at issue is the set of naturals, not the set of reals.
Right, the domain of f in the bijection is the naturals, but the inverse
relationship with g covers all the reals, and the domain of g is the set of all
reals mapped from naturals by f.
then applying g(n) to the values in S
And the set S, not the reals is the codomain of f and the domain of g
and there is no reason to suppose that there is any x for which f(x) and
g(x) are both defined, and certainly no x for which
f(g(x)) = g(f(x)) = x, as TO requires.
Huh? That's gobbledygook. There is no reason to assume that f(x) and g(x) are
both naturals or are equal, but if the domain of g is f(n) for all n in N, then
g(f(n)) is defined as an element n of N. Are you saying there is no such thing
as an inverse function, and therefore no such thing as a quantitative formulaic
bijection? I can't even tell what convoluted excuse you have this time.
will result in a set of reals which all happen to be naturals. Is
that so hard to picture?
Quite impossible in the way TO describes it.
It's impossible, not because of the way I describe it, but because of whom I'm
describing it to.
If " f(g(x))=g(f(x))=x" for all relevant x, as TO claims, then
they have to be bijections from some set to itself.
No, they're mutual injections between the naturals and some other
sequential set, together defining a bijection.
if " f(g(x))=g(f(x))=x" for any x, then f(x) and g(x) must both be
defined for that x.
Yes, and if x is real and f and g are defined over all the reals, or at least
the domain and codomain in question, then they will both be defined for any
such x. That doesn't mean that both functions will produce a natural with any
real x, but if g is the inverse of a, and x=f(n) for some n in N, then g(x)
will be a natural, the original n we started with.
If g(x) and f(x) are both defined for any x, then that x must be in the
domain of both f and g.
No, that is not correct. The domain of f is N. The domain of g is the codomain
of f, that is, all f(n) for n in N. The two domains are mapped using f and g.
Similarly if for any x both f(g(x)) = x and g(f(x)) = x then x must be
in the codomain of both f and g.
No, for the same reason. f is defined for all n in N, and g is defined for all
x in S mapped from the naturals using f.
So that TO's ignorance of mathematics trips him up once more.
In your dreams. Review your objection. It's empty.
No, but the set of rationals is infinitely larger than the set of
naturals, your diagonalized bijection notwithstanding.
Larger in what sense?
In the same sense that Zuhair's intuition is infinitely larger than
Virgil's.
Considering the uselessness, and even detrimental effects, of such
"large intuitions" in mathematics, than benefits me more than Zuhair,
or TO.
Okay, I'll refrain from any further "largeness" jokes. Let's keep this PG. ;)
Can you say, "dense vs. sparse"? I knew you could. Nice start.
Let's look at the line. We'll close our eyes and pick any unit
interval, at random. No, don't ask me about the probability
distribution. Let's just pick ANY unit interval and look at it. How
many naturals does it contain? Uh....one!
None or two for the interval I have in mind, depending on its
topological properties.
(0,1) is not a full unit interval, and neither is [0,1]. The first is missing a
natural, and the second has an extra. In any case, it's a finite number. Call
it 0, 1 or 2. I don't care, for this particular argument.
Okay, that makes sense.
Now, how many rationals does it contain? Ummmm..... a lot. An
infinite number. Gee, and every unit interval, everywhere on the real
line, has this property. There is an infinite number of rationals,
for every single natural. That sound precisely like an infinite
ratio, now doesn't it?
But taking the union of all those one-natural rational intervals results
in the same "number" of naturals as rationals.
Yeah, sure, that makes sense. If I have a cup, put a drop of milk and fill it
up with water, pour it into an empty ocean, and do this an infinite number of
times, the infinite ocean of liquid will end up being half milk. I don't know
how you can say this with a straight face. Maybe you can't. Pick any two
naturals on the standard line. There are an infinite number of rationals
between them. Pick any two rationals. There are a finite number of naturals
between them. Conclusions of equality are delusions of rationality.
One finite set is larger that another if and only if there is an
injection but no surjection from the "smaller" to the "larger".
Sure, for finite sets.
Since TO wants other limiting processes to carry over from finite to
infinite, why does he object to this one carrying over?
There's no limit involved, just a primitive finger counting method for infinite
hands.
That rule rule (that one set is larger than another if and only if
there is an injection but no surjection from the "smaller" to the
"larger") need not be constrained to finite sets.
It need not be extended to infinite ones either
But it has been quite successfully, and TO has not produced anything but
his intuitions as objections.
Nothing but alternatives that actually work, and detailed objections to the
particular foundational properties that lead to the standard nonsense. If you
think the standard treatment is successful, then you have a strange concept of
success.
Perhaps you could consider it a "mapping" from the mapping
function used to used to define the bijection and the value range
to some expression of the set size. I think of it as a formulaic
relationship.
In other words, a mapping.
So, I thought you said there was NO mapping?
But TO rushes in again to say there is a mapping where mathematicians
know better than to tread.
Because mathemticians have tender feet? That's from spending too much time in
the Ivory Tower and sipping tea in the Garden. Take your shoes off and get your
feet dirty. Trod through the bramble, and you'll find things unseen for years.
Infinity itself was one of those places, once, and it still is to a large
extent. The timid discover nothing. So, take your shoes off, and get your feet
dirty. :)
--
Smiles,
Tony
.
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