| > Could someone tell me what the Peano Function is? I know that
> it is a function whose value passes through every point of the unit
> square, but what is the function itself?
I'll try to reconstruct my memories of it.
We're looking for a transformation which is applied on
-----------
to give some other line, eg:
---+ +---
| |
+---+
and is then applied iteratively.
For the above example, we can say:
5X = (3^D)X, where X is the surface covered by the limit line, and
D is the fractal dimension of the surface. (This is, roughly,
just the quantification of the notion of self-similarity.)
5 is the number, N, of line segments, and (1/3) is the ratio,(1/R), of
the subline length to the original length.
So we want to have:
N >= R^2
in order to cover the whole plane.
So, something like
*
/ \
/ \
/ \
/ \
where the angle at the top is right
will be large enough, but no way will it cover a square, or
(aha!)
+---+
| |
---+---+---
| |
+---+
This is going to cover a square of which the start and end-points
are opposite diagonals. (d'you see?: I can't draw fractals
very well in ASCII chars).
Moreover, the graph is Eulerian (only 2 vertices of odd order) so
that we can define a function of time which navigates the shape.
The infinite iteration of this function is going to be a function
of the form your looking for.
Is there anyone knows a good book to learn about this kind of thing?
It's all fascinating, but I'm pretty ignorant about the details.
Andrew.
|
| There is an easy way to define a Peano function. It is
in Rudin's book on real analysis, in one of the problems
(with an attribution to the source where Rudin found it).
I can't include that since the book is at home.
This process gives a way to define a continuous function
on I = [0,1] (the unit interval) which maps the Cantor
set onto I^k.
The Cantor set is a compact, uncountable perfect set obtained
by the following process. Let C[0] = I. Given C[n], define
C[n+1] by removing the (open) "middle thirds" of the segments
of C[n]. So for example, C[1] = [0,1/3] union [2/3,1]. Then
C[2] is [0,1/9] union [2/9,1/3] union [2/3,7/9] union [8/9,1].
Then C[0], C[1], ... is a nested sequence of closed subsets
of the unit interval. The Cantor set C is the intersection of
all of the C[n]. (It has measure zero if you are into that
kind of stuff.)
A number like 3/10 = 0.3 has a finite decimal representation,
I just gave it, but it also has an infinite decimal representation,
3/10 = 0.3 = 0.299999.... Likewise, each element of [0,1] has a
unique infinite ternary (base three) representation of the form
0.abcd.... where the "digits" are either 0, 1, or 2. You can
verify that the Cantor set C is the set of all x in [0,1] such
that the infinite ternary representation of x consists of all
0's and 2's, i.e., no 1's. Example, 1/3 = 0.0222222....
The Peano function constructed will take any x in I = [0,1]
into I^k, and will be continuous. If furthermore x is in the
Cantor set C, then the function will take that infinite stream
of 0's and 2's in its ternary representation and split it apart
into k infinite streams of 0's and 1's which will be the binary
representation of the k coordinates of f(x). (The 0's stay 0 and
the 2's become 1's). This is an extremely clever idea and I wish
I had the reference to its author.
So define g:R -> [0,1] so that it is continuous, periodic with
period 2, is 0 on [0,1/3], and is 1 on [2/3,1]. For concreteness,
let g(y) be 0 on [0,1/3], linear from 0 to 1 on [1/3,2/3], 1 on
[2/3,1], linear from 1 down to 0 on [1,2], and periodic with
period 2. g as defined is continuous.
Now consider g(3^n x) for some x in the Cantor set and some nonnegative
integer n. Take x in its ternary representation. Then 3^n x is
the same stream of base three digits, but with the "decimal" point
moved n places to the right. All of the digits are 0's and 2's,
so the integral part of this number (to the left of the "decimal"
point) is even. Since g has period 2, we can therefore ignore
this integral part, and g(3^n x) will be g of the part of the
representation to the right of the shifted "decimal" point. If the
first base three digit there is a 0, then g of that number will
be 0. If the first base three digit there is a 2, then g of that
number will be 1.
So g(3^n x) for x in the Cantor set is 0 or 1 depending on whether
the n+1st digit (indexing those starting at one) of the infinite
ternary representation of x is a 0 or 2. Also g(3^n x) is continuous.
Now consider the function f(x) = sum(n=0 to oo) g(3^n x)/2^(n+1).
This infinite series is uniformly convergent and so its limit, f,
is also continuous. Each term is between 0 and 1/2^(n+1) so f takes
on values between 0 and 1/2 + 1/4 + ... = 1. Finally, each number
in I is the value of f(x) for some x in the Cantor set ... take the
binary representation of the number in I, change the 1's to 2's and
that gives the ternary representation of x in the Cantor set such that
f(x) is that number. So f:R -> I, and f[C] is *all* of I.
You can cover I^k for some finite k in the same manner. Let
fi for i = 0, ..., k-1 be given by
fi(x) = sum(n=0 to oo) g(3^(kn+i) x)/2^(n+1)
Let f(x) = (f0(x),...,fk-1(x)). Then just as above each fi is
continuous, fi:R->I, the image of fi on C is all of I, f:R->I^k,
and the image of f on C is all of I^k. (watch it, that last one
does not follow from merely the fact that each fi covers I, but
from the fact that each k-tuple of infinite binary sequences comes
from an x in the Cantor set). Restrict the domain of f from R to
I (or even to C) and you have a continuous function from I (or C)
which covers I^k.
I don't know what the function will look like, and I wonder if it
will also be a function which is nowhere differentiabe, which is
another nice example. But I haven't really considered that problem.
Dan
|
| re .3
In Rudin's book Principles of Mathematical Analysis
it is on page 168, problem 14 in the problems at the
end of chapter 7. He states
(This simple example of a so-called
"space-filling curve" is due to I. J.
Schoenberg, Bull. A.M.S., vol. 44,
1938, pp. 519.)
Furthermore in Munkres's Topology a first course (one
of my favorites) in Chap. 7-2 exercise 5, page 274 it
states
(a) Let X be a Hausdorff space. Show that if there
is a continuous surjective mapping f:I->X, then X
is compact, connected, locally connected, and
metrizable.
(b) The converse also holds; it is a famous theorem
of point set topology called the Hahn-Mazurkiewicz
theorem (see [H-Y], p. 129). Assuming this theorem,
show there is a continuous surjective map f:I->I^omega.
A Hausdorff space that is the continuous image of the
closed unit interval is often called a Peano space.
Of course he gives no hint as to a proof of (b) [don't you
just love when they do that? (-: ]. The reference [H-Y] is
to Hocking and Young, Topology, Addison-Wesley.
Dan
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