Re: Visualization of unmeasurable sets and computability

mike3 a écrit :
On Jun 19, 5:14 am, Denis Feldmann <denis.feldmann.asuppri...@club-
internet.fr> wrote:
mike3a écrit :> Hi.

Is it possible to visualize, even in a rough way (ie. an
approximation, it does not have to be exact! No visualizations are
exact anyway!), non-Lebesgue-measurable sets like those produced in
the ball-cutting of the Banach-Tarski Paradox?
No. And remember you are probably unable to wisualize perfectly good
easy to construct sets, like Cantor sets with positive measure, or even
just the set of all transcendant numbers (or their cartesian square, if
you are more atease with subsets of R^2 than with subsets of R


They cannot be visualized *exactly*, but this picture shows a
rough approximation of the postivie-measure Cantor set:

http://en.wikipedia.org/wiki/Image:Smith-Volterra_set.png
(white bars represent points in the set)

You know too much for your own good, so I will assume you are trolling. But see below.



and this type of approximate visualization is what I mean by
"visualization". Of course it's "infinitely inexact", but
since you could not see the finer lines even if you were able
to display an hypothetical (and physically impossible)
infinitely-precise one simply because your eyes are finite in
their resolving power, it's good enough and gives you at least
some idea of what's going on. We can estimate a few elements
that are members of the set, and spit them out on the computer
screen, yielding a rough visualization. It. Does. Not. Have.
To. Be. Exact. Heck, to throw more fuel on this fire, the
Mandelbrot set cannot be drawn exactly, yet it is possible to
draw a rough, approximate visualization of it that gives you
some idea of what it looks like.

So then the question comes out to: can we construct an algorithm
that spits out approximations of an arbitrary number of points to
any given degree of accuracy, that are inside a non-measurable
set?

Yes (see below)



If not, does this mean that non-measurable sets are also non-
computable (not sure if that is the right term, but what I mean by
"computable" is that you can construct an algorithm for a Turing
machine that will generate approximations of an arbitrary number of
points in the set to arbitrary accuracy?)? As if they are computable
then one could spit the estimated points out on a screen and generate
an approximatevisualization.
Try it already for one of the sets mentioned above, and good luck...
Anyway, your dreams of "seeing" the Banach-Tarski paradox with your own
eyes will never be fulfilled, I am afraid


Even in a rough, approximate way? I didn't say it needed to
be exact.


But in fact, you *hint* at it, expecting the approximation to somehow let you understand what is happening...



Therefore I direct you to the more general question, which
covers what I am trying to say even if you do not get my
drift:

***
Does non-measurability of a set imply that no algorithm can
exist that computes an arbitrary number of points of that
set to an arbitrary degree of accuracy?
***

As this question is incredibly stupid, given what has already be said to you, just one easy answer : take any non-measureble set. Union it with the rationals (or *any* dense resursive set of measure 0). plot the points in the added set. Is not that a perfectly good algorithm answering your question? If not, why?



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