Re: Cantor Confusion

Tony Orlow wrote:
David Marcus wrote:
Tony Orlow wrote:
David Marcus wrote:
Tony Orlow wrote:
Now, sequences may be said to derive from ordered sets, but sets are
said to be determined solely by membership, with order unimportant. So,
the notion of a sequence derives really from an inductive definition
such as Peano's, and not from the one primitive in set theory,
membership, alone. The notion of order is not captured by "is an element
of". Do you disagree?
Of course I don't agree. You seem to be saying that infinite sequences
can't be handled in ZFC. Since ZFC has no trouble modeling the natural
numbers and defining functions, it clearly has no trouble acting as a
foundation for all of calculus and analysis.
Is there not a single primitive in set theory, namely, e (element of)?

Sure. But that just says that there is only one relation that is built
into the language of ZFC. We are perfectly free to define new stuff,
just as we do in any math class or book.

How is order derived from that,

In the usual way. If we model the natural numbers as 0 = {}, 1 = {0}, 2
= {0,1}, then we can define n < m to mean n in m. We then define Z, Q, R
in the usual way from N and define addition, multiplication, and order
for all of them. Haven you seen the constructions?


Ugh, yes, I guess I have. The von Neumann ordinals appear to be the
vehicle connecting set membership and order this way. Okay. I don't like
it, but it works in its way.

It is just supposed to work. No one is saying zero really is the empty
set (whatever "really is" means).

It seems like it would be better to have
another primitive, such as Peano's successor, than to use this strange
definition of the naturals, but I'll have to think about that.

The idea is to be parsimonious. Since you can define the relation you
want, there is no reason to be redundant by including another primitive.
Doing so just makes things more complicated without any benefit.

and why is it not applicable in the case
of the staircase?

For any given n, the number of steps, the staircase is defined as the
sequence of segment offset pairs:
(x=1->n: {(0,1/n),(1/n,0)}

What do you mean "segment offset"? If n = 2, then you wrote something
like

(0,1),(1,0),(0,0.5),(0.5,0)

You are reading it as if it said (x=1->n: {(0,1/x),(1/x,0)}, but 1/n is
constant for each iterated value of x. It would be
(0,0.5),(0.5,0),(0,0.5),(0.5,0). That is, up 1/2, right 1/2, up 1/2,
right 1/2.

Oh. Simpler to just write the coordinates of the points as functions of
x.

In other words, the pair of numbers denote the x and y
offsets from the start of the segment to its end.

Do you mean the staircase is the path connecting the points

(0,0),(0,0.5),(0.5,0.5),(1,0.5),(1,1)

?

Yes, that's what I meant, and that's what I said. We start at the
origin, (0,0). Then we add 1/n to the y value to get the next point, 1/n
to the x value to get the next, and repeat that n times, until we get to
(1,1). See? Each segment in the curve is defined by a single pair, the
starting point being already defined as the last ending point, and the
overall starting point being the origin.

The diagonal may be divided into corresponding pairs of segments with
the same overall segment offset:
(x=1->n: {(sqrt(2)/2n,sqrt(2)/2n),(sqrt(2)/2n,sqrt(2)/2n)}

The segments in the first are always vertical or horizontal, while those
in the second are all diagonal. The point set interpretation does not
catch this. Why?

What "point set interpretation"? What doesn't it catch?

I refer to the point set interpretation of the limit of the staircase,
put forth some time back by Chas as a counterexample to infinite-case
inductive proof. By his argument, since the points in the staircase
become arbitrarily close to the corresponding points in the diagonal as
n grows without bound, the staircase "in the limit" can be considered to
be the same object as the diagonal line. His conclusion is that, since
one can prove inductively that the length of the staircase is 2 for
every value of n, that proof only applies in the finite case, since his
version of the infinite case obviously has a length of sqrt(2). My
response to that was that the staircase in the limit is clearly a
different object than the diagonal line, preserving its right angles on
the infinitesimal scale, and therefore not straight, but a kind of
fractal object. The fact that the segments of the diagonal in the limit
are always at a 45 degree angle to the diagonal accounts for the
discrepancy in measure of sqrt(2), the inverse of the cosine of the
angle. In other words, approximating the diagonal using the staircase
segments cannot provide accurate measure, because the segments are not
parallel to the object they are approximating.

This doesn't have anything to do with whether set theory can be used as
a foundation for mathematics. You need to give a precise definition of
what sort of limit you are doing that would allow the staircase to
approach your "kind of fractal object". If you want such a thing, you
have to invent it. Developing new mathematics isn't easy.

So, the point set approach has that the same object has two different
measures, because it cannot distinguish between two objects which are
locationally the same, and directionally different.

Then come up with an approach that does what you want. By "approach", I
mean definitions (of objects, convergence, etc.) that let you prove the
theorems you want. That's what everyone does. For example, Einstein came
up with Brownian motion, but it wasn't clear how to mathematically model
it. Weiner figured out how.

I'm sure that cleared things up for you, eh?

Pretty much.

By the way, the only other counterexample to infinite-case induction
suggested made obvious use of a discontinuity based on a buried
difference with a limit of 0 in the infinite case, and thus violated the
rules as I put forth. So, if you happen to have a good counterexample to
infinite-case induction that doesn't rely on such differences, I'd be
happy to decompose them for you. :)

--
David Marcus
.

Relevant Pages

  • Re: Cantor Confusion
    ... For any given n, the number of steps, the staircase is defined as the sequence of segment offset pairs: ... the pair of numbers denote the x and y offsets from the start of the segment to its end. ... His conclusion is that, since one can prove inductively that the length of the staircase is 2 for every value of n, that proof only applies in the finite case, since his version of the infinite case obviously has a length of sqrt. ... By the way, the only other counterexample to infinite-case induction suggested made obvious use of a discontinuity based on a buried difference with a limit of 0 in the infinite case, and thus violated the rules as I put forth. ...
    (sci.math)
  • Re: Cantor Confusion
    ... the limit of the staircase from to, as the number of steps ... the notion of a sequence derives really from an inductive definition ... But ZFC does have considerable difficulty dealing with infinite ... proof by induction that these "infinite quantities" were also simply ...
    (sci.math)
  • Re: Calculus XOR Probability
    ... of thing as the members of the sequence. ... infinite number of infinitesimal stairs, the length IS 2, because that's ... Then you have no business talking about the identity between the staircase ... because the segment definition preserves the notion of direction IN THE ...
    (sci.math)
  • Re: Cantor Confusion
    ... So, the notion of a sequence derives really from an inductive definition such as Peano's, and not from the one primitive in set theory, membership, alone. ... For any given n, the number of steps, the staircase is defined as the sequence of segment offset pairs: ... No, that is no simpler, and does not capture the direction or magnitude of any segment in a single pair. ... If this is a valid formulation of the two objects, and an explanation for Chas' counterexample to infinite-case induction, where does this fit with set theory? ...
    (sci.math)
  • Re: Two results of set geometry
    ... So M is not the nom-completed matrix. ... Then the whole staircase with infinitely many ... An infinite union of finite steps is ... Every finite initial segment of 111... ...
    (sci.math)