Questions about Hausdorff and lower packing dimension
Dave L. Renfro
By "lower packing dimension", I mean the version obtained via the
lower box dimension (= lower Minkowski dimension), and similarly
for "upper packing dimension". ["lower" means a lim inf is utilized;
"upper" means a lim sup is utilized.]
1. Does anyone know of a not too obscure (if possible) reference
for the existence of a CLOSED Lebesgue measure zero set in
the reals that has lower packing dimension 1? Even more, what
about a CLOSED Hausdorff h-measure zero set with lower packing
dimension 1, for any preassigned Hausdorff measure function h
such that the limit as t --> 0 of t/h(t) equals 0?
If "lower" is replaced with "upper", giving what is usually meant
by "packing dimension", there is no problem. In fact, the
Baire-typical compact subset of the reals has this property.
2. If the lower packing dimension of a set E is zero, then we
know that for any epsilon > 0 there is a covering of E by
countably many sets each having lower box dimension less than
epsilon. [See the bottom of p. 81 of Pertti Mattila's 1995
book.] Is it always possible to find a covering of E, under
the assumption that E has lower packing dimension zero, by
countably many sets each having lower box dimension equal
to zero? In other words, is every set with lower packing
dimension equal to 0 a countable union of sets having
lower box dimension 0? What if "0" is replaced with "1", or
with some other number between 0 and 1?
3. It is not difficult to show the following property is
equivalent, when E is bounded, to the closure of E having
Lebesgue measure zero:
For each epsilon > 0 there exists a FINITE covering of E by
open intervals whose lengths have a sum less than epsilon.
Suppose that in addition we now require the finite covering
of E to be with open intervals all having the same length. Is
this new requirement equivalent to some type of Minkowski
content-zero-in-dimension-1 statement?
Dave L. Renfro
Dave L. Renfro
[sci.math 11 Jan 01 00:23:42 -0500 (EST)]
<http://forum.swarthmore.edu/epigone/sci.math/jashanpeh>
wrote late last night (in part):
> 3. It is not difficult to show the following property is
> equivalent, when E is bounded, to the closure of E having
> Lebesgue measure zero:
>
> For each epsilon > 0 there exists a FINITE covering of E by
> open intervals whose lengths have a sum less than epsilon.
>
> Suppose that in addition we now require the finite covering
> of E to be with open intervals all having the same length. Is
> this new requirement equivalent to some type of Minkowski
> content-zero-in-dimension-1 statement?
Looking at this the next day I see that requiring the open
intervals to have the same length (for each value of epsilon > 0)
doesn't give anything new. In fact, given any finite collection
of open intervals and given any delta > 0, there is a covering of
these intervals by another finite collection of open intervals
that all have the same length and such that the sum of the lengths
of the intervals in the latter collection exceeds the sum of the
lengths of the intervals in the former collection by no more than
delta. [Thus, I can not only require that delta --> 0 as
epsilon --> 0 (all that's needed to prove my initial weaker claim),
but I can require that delta/epsilon stay bounded as epsilon -->0,
and even that delta/epsilon --> 0 (arbitrarily quickly, relative
to epsilon) as epsilon --> 0.]
Suppose there are M open intervals. Then replace each
of these M intervals with an interval concentric with
the original interval so that it's length is a rational
number at most delta/M more than the length of
the original interval. The lengths of these new intervals
are commensurable, and so we can cover them *exactly* by
open intervals all having the same length, and the sum
of these equal-length intervals will be at most delta
more than the sum of the lengths of the original M open
intervals.
Of course, we could also consider the minimum number, call it
M(epsilon), of open intervals needed to cover a compact Lebesgue
measure zero set for a given value of epsilon (epsilon being an
upper bound on the sum of the lengths of these intervals), and
then ask how M(epsilon) behaves as epsilon -->0, but I believe
this leads directly to a consideration of upper and lower
Minkowski dimensions. [The Minkowski dimensions involve looking
at the minimum number, call it N(epsilon), of open intervals having
length at most epsilon that are needed to cover the set, and then
seeing how N(epsilon) behaves as epsilon --> 0.]
Dave L. Renfro
Lars Olsen
L. Renfro) wrote:
> I have some questions about Hausdorff and lower packing dimensions.
>
> By "lower packing dimension", I mean the version obtained via the
> lower box dimension (= lower Minkowski dimension), and similarly
> for "upper packing dimension". ["lower" means a lim inf is utilized;
> "upper" means a lim sup is utilized.]
Asking for some clarification:
By the lower packing dimension of E, do you mean the following,
\inf_{E\subseteq\cup_{i\in N} E_i} \sup_i (lower box dimension of E_i)
? (*)
Similarly, by the upper packing dimension of E, do you mean the following,
\inf_{E\subseteq\cup_{i\in N} E_i} \sup_i (upper box dimension of E_i)
? (**)
>
> 1. Does anyone know of a not too obscure (if possible) reference
> for the existence of a CLOSED Lebesgue measure zero set in
> the reals that has lower packing dimension 1? Even more, what
> about a CLOSED Hausdorff h-measure zero set with lower packing
> dimension 1, for any preassigned Hausdorff measure function h
> such that the limit as t --> 0 of t/h(t) equals 0?
>
> If "lower" is replaced with "upper", giving what is usually meant
> by "packing dimension", there is no problem.
This remark seems to indicate that you define the upper packing dimension
by (**).
Dave L. Renfro
[sci.math Fri, 12 Jan 2001 18:02:02 +0000]
<http://forum.swarthmore.edu/epigone/sci.math/jashanpeh>
wrote (in part):
> Asking for some clarification:
>
> By the lower packing dimension of E, do you mean the following,
>
>
> \inf_{E\subseteq\cup_{i\in N} E_i}
> \sup_i (lower box dimension of E_i)
> ? (*)
>
> Similarly, by the upper packing dimension of E, do you mean
> the following,
>
> \inf_{E\subseteq\cup_{i\in N} E_i}
> \sup_i (upper box dimension of E_i)
> ? (**)
Yes, this is what I mean. These appear on pp. 81-86 of Pertti
Mattila's 1995 book "Geometry of Sets and Measures in Euclidean
Spaces" and in Falconer's 1990 book "Fractal Geometry". I don't
have a copy of Falconer's book with me, so I can't give you a
specific page reference in it. However, I do know that Falconer
uses the terms "lower modified box dimension" and "upper modified
box dimension" for what Mattila calls the lower packing dimension
and upper packing dimension, respectively.
One of my questions was:
Does anyone know of a not too obscure (if possible) reference
for the existence of a CLOSED Lebesgue measure zero set in
the reals that has lower packing dimension 1? Even more, what
about a CLOSED Hausdorff h-measure zero set with lower packing
dimension 1, for any preassigned Hausdorff measure function h
such that the limit as t --> 0 of t/h(t) equals 0?
I just realized that part of this question is fairly trivial. It
is well known that there are compact nowhere dense subsets of the
reals having Lebesgue measure zero and Hausdorff dimension 1, which
is stronger than what I asked in the first part of my question.
Here is what I was really after ---->>>>
Does there exist a CLOSED Lebesgue measure zero set in the
reals that is not a countable union of sets having lower
Minkowski 1-content zero? If not (possible or known), does
such an example exist when "lower Minkowski 1-content zero"
is strengthened to (thus weakening the example) "upper
Minkowski 1-content zero"? More generally, do such examples
exist having arbitrarily small generalized Hausdorff dimension?
Def: Let E be a nonempty bounded subset of the reals and let
E(epsilon) be the set of all points whose distance from E
is less than or equal to epsilon. The lower [upper] Minkowski
s-content of E is the lim-inf [lim-sup] as epsilon --> 0 of
(2*epsilon)^(s-1) times the Lebesgue measure of E(epsilon).
(See p. 79 of Mattila's book.)
Dave L. Renfro