Suppose that I drew an infinite number of disjoint closed curves in the plane (such as circles, squares, etc.). Suppose that I then tell you that there is one curve for each positive real number.
You would not have too much trouble believing my assertions at this point. For example, I could have drawn all circles with center at the origin. They are all disjoint, and for each positive real number x, there is a corresponding circle - namely, the circle of radius x.
But suppose that I also tell you that all the curves I drew were figure eights. Can you believe my assertions now?
(A figure eight is a curve in the plane obtained from the basic "8" shape by any combination of translation, rotation, expansion, or shrinking.)
(In reply to
re(10): I think it's... - Elaboration by David Shin)
I was thinking it was impossible for any figure 8 to be enclosed in
infinitely many right compartments, just as it is for the left
compartments.
But as I recall, that wasn't one of the things I proved, I merely
proved that there would be countable number of 8s if right compartments
are unused.
In fact, I just thought up a counterexample: there can
be a figure 8 enclosed in (countably) infinitely many right
compartments, even if there is a single largest figure 8. The
counterexample goes as follows:
The largest figure 8 has a small figure 8 in its right
compartment. We can draw a middle-sized figure 8 with its right
compartment around the smaller 8, and its left compartment half the
size of that of the small 8. We can draw another figure 8 with
its right compartment around the middle-sized 8, and its left
compartment half the size of that of the middle-sized 8. We can
do this infinitely (not infinite induction since though there will be
infinite 8s, the infinitieth step is not taken). The end result
will be that the smallest has infinite right compartments enclosing
it. The empty left compartments, while there are aleph null, take
a finite amount of space.
Perhaps I can build on this counterexample to a solution that shows that the assertions are possible, quickly invalidating my developing proof.
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Posted by Tristan
on 2005-02-19 17:28:58 |