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Jan 11, 2001, 8:55:08 AM1/11/01

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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.]

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

Jan 11, 2001, 4:10:20 PM1/11/01

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Dave L. Renfro <ren...@central.edu>

[sci.math 11 Jan 01 00:23:42 -0500 (EST)]

<http://forum.swarthmore.edu/epigone/sci.math/jashanpeh>

[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

Jan 12, 2001, 1:02:02 PM1/12/01

to

In article <ef0o7s...@forum.mathforum.com>, ren...@central.edu (Dave

L. Renfro) wrote:

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 (**).

Jan 13, 2001, 3:11:38 PM1/13/01

to

Lars Olsen <l...@st-and.ac.uk>

[sci.math Fri, 12 Jan 2001 18:02:02 +0000]

<http://forum.swarthmore.edu/epigone/sci.math/jashanpeh>

[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

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