Re: Interval in a continuum is a fractal
- From: mike3 <mike4ty4@xxxxxxxxx>
- Date: Mon, 3 Dec 2007 12:33:45 -0800 (PST)
On Dec 2, 1:25 am, Venkat Reddy <vred...@xxxxxxxxx> wrote:
On Dec 2, 1:20 am, mike3 <mike4...@xxxxxxxxx> wrote:
On Nov 29, 8:07 pm, Venkat Reddy <vred...@xxxxxxxxx> wrote:
This is another perspective to explain things when I say you can't
break the conituum interval into points.
An interval in continuum such as time span is bounded by two time
instants, which are modeled as real numbers in mathematics. The
interval also has a non-zero extent or magnitude or span which is
modeled by the "measure" in mathematics.
Any statement that is even remotely related to "such an interval is
made up of certain things", has to deal with division process of such
interval. Simply defining that such interval is made of time instants
is like saying I just said, and you can't question. We need to attempt
a division process to see if it is right.
Such a division process on an interval results in smaller intervals
which has the same properties of the original interval - having bounds
of same type, and having a non-zero measure of extent which is also
same type. This can be continued for ever but the intervals or parts
do not cease to hold the properties of the whole, which qualifies the
partitions as fractals.
Fractals do not collapse to zero extent. It is just forced by our
arrogance at not recognizing the nature and it symmetry. Nature has no
jumps. If the measure has to collapse to zero, there needs to be a
reason such as hitting the "largest number" of divisions.
This is the crux of what I meant in all my recent posts and threads.
OK, then, how large is a point? According to you, it should not
have zero extent. Does it have extent "b"? Then if you want it to
build up and extent, b + b should equal 2b. Concede that point.
There is nothing to concede here. I have already stated that a finite
number of b would not build to a finite extent. However it is a non-
zero limiting measure of finite extent reached when divided by
unlimited number of splits. Actually thats how we arrived at b.
So then why should not a finite number of b build a finite number
of b, but exactly one b, no matter how big that finite number is?
Ie, why doesn't b + b = 2b, b + b + b = 3b, etc.? Furthermore,
do you admit that some seemingly-okay arithmetic operations in
your system are undefined, for example, b/2 is not defined as you
should not be able to split b if b is the "smallest" possible amount
but not zero, like how you can't split 1 in the integers?
Contrast this with reals. Can you split the real line into zero-width
point and rebuild it back? You can't do any.
Yes, I can. Splitting up the real line into zero-width points, ie.
*single real numbers*, leaves behind a huge infinity of singletons.
If I union all those singletons into a set, why is it not R that
I get back?
- venkat
Let's *actually* attempt the division process, instead of just
talking about it, to attempt to prove our claims, and see what
really happens. We will use the mathematical continuum, since we
are discussing pure mathematics.
We begin with a partition of the segment [0, 1] into two intervals
of equal length:
[0, 1/2) and [1/2, 1].
Now we split those into two intervals of equal length:
[0, 1/4), [1/4, 1/2), [1/2, 3/4), [3/4, 1].
Continuing this process n times, we have 2^n intervals:
I_i = [i/(2^n), (i+1)/(2^n)), i <> 2^n - 1,
[i/(2^n), (i+1)/(2^n)], otherwise.
for i = 0...2^n - 1.
Then
U_{i=0...2^n - 1} I_i = [0, 1].
(where "U" is union.). The measure of each interval is, of course,
(i+1)/(2^n) - i/(2^n) = 1/(2^n). Continuing this process an infinite
amount of times, we find the measure of each interval becomes
1/(2^infinity) = 1/infinity = 0.
Where did you get this equation (1/infinity = 0)? Any proof?
Are you stating that the sequence 1/2, 1/4, 1/8, .... hits zero?
Yes, but not on the real number line. On the "extended" real number
line (with a special "point at infinity"), it indeed "hits" zero,
at that point at infinity.
Also, on the surreal numbers, a different system, they do not
reach zero, but instead some infinitesimal, and this infinitesimal
gets smaller as one gets further into the infinite regime. Why
don't you like the surreal numbers?
This contradicts our idea that we
can break something down and still always have finite extent left.
WRONG! Only for a *finite* number of breakdowns! It turns out the
claim was wrong.
A fractal does not "collapse to zero extent", no, but that's only
because we are considering the WHOLE fractal, just as you are
considering the WHOLE line.
Oh, are you not considering WHOLE line when you assert the real
numbers fill the line?
Yes, but I'm also addressing the pieces that result from breaking
it apart (partitioning) it into smaller and smaller subsets.
However, the subdivision process used to generate many fractals
results in similar pieces whose size, when examined individually,
in order of their construction, goes to zero as we get to infinity.
Are you sure? Is that the limitation of the fractal rendering program,
or is it the limitation of our imagination?
It's the limit of mathematics. The pieces get smaller and smaller,
so "at infinity", they shrink to zero.
.
So then yes, it does make sense for points to have extent zero.
True Mathematics is rescued, Venkat Reddy's Mathematics is sunk.
Just like the Titanic.
Not so fast mike :) Seems I'm still batting at 1000 (is that cricket,
WH?)
- References:
- Re: Interval in a continuum is a fractal
- From: mike3
- Re: Interval in a continuum is a fractal
- From: Venkat Reddy
- Re: Interval in a continuum is a fractal
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