> By definition (Aleph-null)^2 = Aleph-one. You are not scaling Aleph-null by
> a finite number, you are squaring it.
aleph-null^2 != aleph-one. A table with aleph-null rows and aleph-null columns obviously still has aleph-null elements, because they can be numbered like so:
1 3 6 10 . . .
2 5 9 14 . . .
4 8 13 19 . . .
7 12 18 . . . .
. . .
. . .
> For a set of n elements, there are n^2 subsets.
2^n subsets, you mean. 2^n is also the number of n-digit binary
numbers. There are clearly aleph-one binary numbers with aleph-null
digits, just as there are aleph-one decimal numbers with aleph-null
digits, based on Cantor's diagonalization proof.
> The number of subsets of
> Aleph-null is Aleph-one (the 'continuum' infinity (used for counting all
> line segments, partitions, etc of a line - or counting polygons, surfaces,
> etc in a real plane) as opposed to the 'enumerable' infinity (counts real
> numbers/rationals, etc).
>
> From another point of view, the number of steps to produce Cantor dust is
> enumerable, but the number of points in the dust is Aleph-one (not
> one-to-one with the rationals).
Hmm. Well, this makes sense, since the number of points is indeed
2^(number of subdivisions).
Suppose I try to make Cantor dust by a different process: instead of
subdividing *every* interval on each step, suppose I only subdivide the
*largest* interval, by removing a subinterval from the middle of it,
dividing the remainder into two subintervals. Then, the number of
endpoints becomes only 2*n, if n is the number of times I subdivide
(plus one), since I'm adding 2 endpoints at every step.
It seems that repeating this process aleph-null times should produce
the same structure that the usual Cantor-dust process produces, but if
so, it clearly only has aleph-null endpoints. Am I completely mad
here? I must have missed something in my assumptions.
Another interesting thing: What is the topological dimension of Cantor
dust?
The sci.fractals FAQ says:
A set has topological dimension 0 if every point has arbitrarily small
neighborhoods whose boundaries do not intersect the set.
This is true of Cantor dust -- for any point in the set, you can find a
neighborhood whose boundaries are outside the set, and for any such
neighborhood, you can find a smaller one. (Right? This *is* what the
definition above is calling for, right?)
It would seem that Cantor dust is a bona fide fractal, then. If we do
the usual thing of taking out the middle third of the interval at each
subdivision step, then by increasing the scale by three, we increase
the size by two. (Right?) So the fractal dimension should be log 2 /
log 3, or about 0.6309.
Kragen
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