Re: An uncountable countable set

Tony Orlow wrote:
imaginatorium@xxxxxxxxxxxxx wrote:
Tony Orlow wrote:
David Marcus wrote:

<snip-snop: the valiant shall see>

Time is actually irrelevant.

If you are trying to determine the limit of the sequence of operations,
time does appear to be irrelevant, yes.

Individual operations are indistinguishable at noon. You must take the
limit as the number of iterations approach oo. Then what do you get? Why
do you have a conflict between looking at it in terms of iterations vs.
time? Because of the clever little Zeno machine. Nice obfuscation.

I'm not sure what you mean by "obfuscation" - just because you are
confused doesn't mean anyone is trying to confuse you.

Consider my blue sliver - if you think of sliding along it, almost
vertically, closer and closer yet never quite touching the y-axis, this
is a journey that never ends. But sliding along the x-axis under the
sliver will surely reach (0, 0), as long as you don't put the brakes
on. There is no obfuscation here - though like most of maths there is
certainly something you have to think carefully about. Anyway,...


... The sequence is measured in iterations as
n->oo, and the number of balls in the vase at iteration n is represented
by sum(x=1->n: 9). The limit of this sum as x diverges also diverges in
linear fashion.

Certainly does. I mean that sum from x=1 as x increases 2, 3, 4, ...
without limit of (10-1) diverges.

Right, and that characterizes the salient features of the gedanken.


Let me ask you another question, Tony, as I don't think you answered
the last one.

I don't see any previous question at this point, but I'm relatively sure
I answered what was asked.

Here is an argument, ending with a conclusion I don't
personally swallow. Can you tell me at what point it goes wrong?
(Or do you think it is valid?)

You don't think it's valid - good.

Consider the function step0: R -> R mapping x to 0 if x<0 and mapping x
to 1 if x>=0.

A discontinuous function at x=0.

Right. Well, slightly more precise to say "a function with a
discontinuity at x=0" I think.


FWIW, we can write this function in a C-like way (taking 'TRUE' and
'FALSE' to have the numeric values 1 and 0 respectively), so it is just
a simple expression:

step0(x) = (x>=0)

OK, for n a positive integer, now consider the sequence of values of
step0(p) for p=-1, -1/2, -1/3, ... -1/n, ... without end

For any n, -1/n < 0, therefore step0(-1/n) = 0.

So the sequence of values is simply the constant sequence 0, 0, 0, 0,
.... without end

The limit of a constant sequence of values is the single value itself.

Therefore lim(n->oo) step0(-1/n) = 0

By the Orlovian limit-swapping axiom, therefore:

step0(lim(n->oo) -1/n) = 0

But lim (n->oo) -1/n = 0.

Thus step0(0) = 0.

But by definition, step0(0) = 1

Therefore 0 = 1.

A function with such a declared discontinuity has two limits at that
point, depending on the direction of approach. So, what else is new?

Ah. Is a "declared discontinuity" somehow significantly different from
a simple discontinuity? I mean, is there such a thing as an "undeclared
discontinuity" to which different rules apply? (I've no idea: this is
not normal mathematical terminology you see.)

That proves nothing.

It illustrates that for a function f(), the value of f(0) is not
necessarily equal to lim(x->0) f(x). Which is of rather crucial
importance in the current problem.

What causes a discontinuity at noon? I'll tell you. The von Neumann
limit ordinals. That's schlock.

Um, that's foaming at the mouth. Von Neumann limit ordinals have
nothing to do with it - the very simplest notion of the natural numbers
being an unending sequence - Wolf Kirchmeir's six? eight?-year old
grandson's understanding - is absolutely all that is needed. You can't
grasp the notion of an unending sequence, which is why you are in such
a total tangle.

According to your "view" then, there is no discontinuity at noon - is
that right? The number of balls identified by natural numbers increases
without limit, and despite the fact that there is no ball not removed
before noon, at ten past an unlimited number of them are somehow still
lurking in the vase?

Look, I know my "sliver" corresponds to a slightly different sequence,
but it's simpler. Consider the sliver between y=-2/x and y=-1/x, for
x<0. Consider it "hatched" with horizontal lines on integral values of
y. Think of every one of the horizontal bars as representing a time
some ball spends in a vase. You seem to agree that the sliver goes ever
upward, ever closer but not actually reaching the y-axis. If we were to
travel upward, we would see each line corresponding to a ball's stay in
the vase - always in then out halfway towards the y-axis; and
importantly, this viewing journey would never end.

But if we were to travel along the x-axis towards the origin, looking
upwards (this is maths, not physics; we pretend we could view the
sliver however far away), we would notice that the number of balls was
increasing without limit. Then we would reach the origin. Looking up we
would see the sliver to the left of the y-axis. You agreed at one point
that the sliver is entirely in the neg-x/pos-y quadrant, so obviously
there is no sliver to the right of the y-axis.

But in your view (do I understand?) somehow there would just be some
more sliver that had crept around the "top"? Or what? Do enlighten
us...

Meanwhile, it would greatly help if you also considered some different
slivers.

How about the one between

y=-2/(x+1) and y=-1/x ... (sliver-2: height diverges)

This is exactly the same as the first, except that the upper hyperbola
lobe has been shifted left by one. This sliver does not become
indefinitely narrow - the width as we go up tends to exactly 1. This is
the case where every ball is inserted one minute (are we in minutes?
makes no difference...) earlier, so by 11:59 all the balls are in the
vase. In this case how much of the sliver leaks into the x>0 quadrant?

Or:

y=-1/x + 1 and y = -1/x ... (sliver-3: height constant = 1)

Or:

y = -1/x - x and y = -1/x ... (sliver-4: height tends to 0)

For sliver-3 there is one ball in the vase, and at each -1/n it is
removed and replaced by the next ball. So the function is close (in
some sense) to a step function, which mathematics says has a
discontinuity at x=0, since there is no numbered ball put in at or
after noon.

For sliver-4, the number of vertical lines of hatching (averaged out)
tends to 0. In other words as you get closer and closer to 0 there is a
ball in the vase less and less of the time.

I wonder if we can agree (probably can, amazingly!) on the limits as x
-> 0 of the number of hatching lines in each of the slivers. (For
sliver-4, of course, there is no proper limit, because the function is
oscillating more and more rapidly between 0 and 1.)

You're concluding that a linear increase
results in nothing, when it's clearly a series of operations which diverges.

Of course it's a series of operations which diverges.

Corollary: set theory is inconsistent.
That's not my conclusion, especially since that's not my logic.

Good.

Brian Chandler
http://imaginatorium.org

.

Relevant Pages

  • Re: An uncountable countable set
    ... If you are trying to determine the limit of the sequence of operations, ... Consider my blue sliver - if you think of sliding along it, ... A function with such a declared discontinuity has two limits at that ... some ball spends in a vase. ...
    (sci.math)
  • Re: An uncountable countable set
    ... discontinuity" to which different rules apply? ... and despite the fact that there is no ball not removed ... You seem to agree that the sliver goes ever ... ever closer but not actually reaching the y-axis. ...
    (sci.math)
  • Re: An uncountable countable set
    ... why people think it has anything to do with set theory. ... Each ball is involved ... and each iteration of the sequence is involved with 10-1 balls. ... There is no discontinuity at noon, ...
    (sci.math)
  • Re: An uncountable countable set
    ... 'sliver height', not 'sleight of hand'...) ... By way of a different example, consider the circle radius 1, centre (0, ... same function, possibly some including a "declared discontinuity", ... TO claims I pretend to have a "specific number" of naturals. ...
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  • Re: An uncountable countable set
    ... discontinuity" to which different rules apply? ... supposedly the infinite process completes. ... Well then, nothing can change at noon that was true at every time before noon, when there is a growing positive number of balls in the vase. ... You seem to agree that the sliver goes ever ...
    (sci.math)
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