Re: Corners in metric spaces
- From: rsheskey@xxxxxxxxxx (Robert Sheskey)
- Date: Sat, 29 Apr 2006 15:53:48 -0500
In article <1146211357.797570.323720@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, we_pretty@xxxxxxxxx says...
I thought about combining solids in R^3. For example an ellipsoid
and a ball are both smooth in an intuitive sense. Their union (if
they have common points) is also smooth except at the points that
belong to the boundary of both objects. I tried to think what
characterizes this "smoothness" - property.
What is true is that when "corner point" is defined properly then
a (finite) ellipsoid does not have corner points and a (finite)
smooth torus does not have corner points. Neither has their union
corner points except at the points that belong to the boundary of
both objects. Also the set of the corner points of their union
is cornerless.
I tried to define "corner" in metric space so that the definition
would match the intuitive picture of a corner.
Let M = (X, d) be a metric space.
Definition of corner point
Let S be subset of X. p is a corner point of S in X iff
there exists a sequence B_1, B_2, B_3,... of open balls such that:
1. the elements in the sequence are pairwise disjoint
3. for each B_i:
3.1 there exists (at least) two distinct points such that
both are boundary points of the closure of S and
both are boundary points of the closure of B_i
3.2 the intersection of B_i and boundary of S is empty
3.3 the intersection of the closure of B_i and the closure of
B_i+1 is a singleton and not in the closure of S
2. each sequence p_1, p_2, p_3, ... where each p_i is in B_i, has
p as a limit point
This definition works in many cases, for instance in R^2, with the
usual metric, the solution set of y = |X| has a corner point at
the origin while y = x^2 does not have corner points. In R^3 the
set of corner points of the intersection of any two ellipsoids
(degenerate or not) is cornerless.
Definition of function C: C(S) is the set of corner points of S.
Definition of cornerless set: S is cornerless <-> C(S) is empty.
Definition of n-cornerless set:
S is 0-cornerless <-> S is cornerless,
S is n+1-cornerless <-> C(S) is n-cornerless
Since the empty set is cornerless,
S is n-cornerless -> S is n+1-cornerless
I would like to have the following sentence true:
Sentence 1:
Let S_1 and S_2 be any two n-cornerless sets.
Then their union is n+1-cornerless
and their intersection is n+1-cornerless.
[snip]
I think that the following sentence should be true at least in
some spaces, for instance R^2 with usual metric:
Sentence 2: For any set S in M,
for the functions CSeq_0(S) = S, CSeq_n+1(S) = C(CSeq_n(S)),
there exists a n such that CSeq_n+1(S) = CSeq_n(S)
(this is trivially true for n-cornerless (with finite n) sets)
[snip]
I thought that with the proper definition, Sentence 1 would have
been easy to prove even for me. If the definition does not work,
I would appreciate references or working definitions. Then I
would like to learn a proof or counterexample of Sentence 1 and
to get scetch of proof or counterexample of Sentence 2
I have sets in R^2 that provide counterexamples (I think) to your
sentences S1 and S2. We need the following two lemmas.
Lemma A. Let x_0 < x_1 < ... be real numbers, with (x_n) -> p. Then
(in R^2) the set S ={ (x_n, 0) : n=0,1,...} has exactly one corner,
namely the point (p,0).
I'm not going to attempt a rigorous proof. The idea is that you can
find an increasing sequence (n_k) and circles C_k such that
1. n_{k+1} > n_k + 1
2. the center of C_k is above the x-axis.
3. C_k meets the x-axis in the two points x_{n_k} and x_{n_k + 1}
4. C_k and C_{k+1} are tangent.
5. the radius of the C_k's tends to zero.
Then if B_k is the ball with boundary C_k the B_k's and (p,0) satisfy
the definition.
Lemma B. Let alpha be a countable ordinal, s:alpha -> R a bounded
order-preserving injection, S = { (s(eta),0) : eta < alpha }. Then the
corners of S are exactly the points s(beta+omega), for every beta
such that beta+omega <= alpha.
The idea here is that for every such beta we can find an epsilon>0
so that the interval ( s(beta+omega)-epsilon, s(beta+omega) ) intersects
S in a sequence of points as in Lemma A, thus making s(beta+omega) a
corner. That such points are the only corners I leave to the reader.
Since this result does not depend on any specific embedding we can
speak of the corners of an ordinal. So we get that CSeq_1(alpha)
consists of ordinals of the form beta + omega, CSeq_2(alpha) ordinals
of the form beta + omega^2, in general, CSeq_n(alpha) ordinals of
the form beta + omega^n. So taking alpha = omega^omega gives
us a counterexample to your sentence S2, since the iterated
corner-taking sequence for such an alpha never stabilizes.
To get counterexamples for your sentence S1 let S be the same set,
S = { (s(eta),0) : eta < alpha } for say alpha = omega^omega.
Let A be the set of vertical lines through S, A = s(alpha) x R.
And let B be the x-axis. Then both A and B are cornerless, but their
intersection is S, which as we've seen has infinite 'corner rank',
or in other words, is not n-cornerless for any n.
To get a counterexample for the other part of your S1, let A be as
above, A = s(alpha) x R, and let B be the corresponding set of
horizonal lines, B = R x s(alpha). Then A and B are cornerless but
their union also has infinite corner rank.
Robert Sheskey
.
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- Corners in metric spaces
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