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"If M is a metric space, and d > 0 is a real number, then the d-dimensional Hausdorff measure Hd(M) is defined to be the infimum of all m > 0 such that for all r > 0, M can be covered by countably many closed sets of diameter < r and the sum of the d-th powers of these diameters is less than or equal to m."
I read infimium, so I'm clear on that, but this definition is still opaque to me. Please clarify. --BlackGriffen
The box-counting dimension is not the same thing as the Housdorff dimension! -- Miguel
Hausdorff dimension isn't the only fractal dimension...it should have its own entry, and fractal dimension to talk about fractal dimensions collectively, it seems.
The banner is ridiculous and I am going to remove it. CSTAR 17:43, 25 Nov 2004 (UTC)
I put in a condition at the beginning which is basically the doubling property (which is of course much weaker). I'm not even sure I stated it correctly. Will the experts object? I perhaps should do my homework and consult my copy of Gromov-Semmes et-al. But hey, I'll let Tosha do that. CSTAR 17:49, 26 Nov 2004 (UTC)
Haha, ironic for the page-doubling bug to strike here. X) --[[User:Eequor| η υωρ]] 21:24, 26 Nov 2004 (UTC)
It's called Hausdorff dimension. Policy is that usual names should be used. This page should be moved back. Charles Matthews 19:31, 26 Nov 2004 (UTC)
It's User:Eequor on a stampede. See Besicovitch. Charles Matthews 20:08, 26 Nov 2004 (UTC)
If you really want to be accurate, you should say something like
ALso Hausdorff dimension is not a measurement and it also would be misleading to call it a measure. It's just a number assigned to a set. CSTAR 22:07, 26 Nov 2004 (UTC)
Why the name change? Virtually every authoritative work on the subject refers to the concept as Hausdorff dimension. For example
CSTAR 19:56, 26 Nov 2004 (UTC)
Firstly, Mandelbrot is much more of a publicist than a serious mathematician. I'm not prepared to take that book as an authority on mathematics. We use fractal the way he does (for better or for worse), because it's his term.
Second point: Wikipedia naming policy is (generally) to use the most usual name. The article can discuss credit where due; the title should be the normal way of naming anything. Charles Matthews 20:19, 26 Nov 2004 (UTC)
I think to split it. Tosha 06:09, 28 Dec 2004 (UTC)
They are not tied, one is defined using the other Tosha 21:39, 28 Dec 2004 (UTC)
I think it is a good idea to split them. The simplest definition of Hausdorff dimension does not involve the Hausdorff measure, but involves the Hausdorff content. Anyway, the Hausdorff content should be discussed somewhere. I think there is a lot to say about Hausdorff measure and it would not be good to put all this stuff here. Oded ( talk) 19:52, 9 May 2008 (UTC)
I think this does not work correctly! Take a look at the image at right, there you can see three examples of Hausdorff dimensions. The first is a 2D Cantor set, four main attractors and scale factor , no rotation. This gives a dimension of 2.0 and the result is also a two dimensional, 45º tilted square. The second image are the same but here a rotation of 45º is also used. From the Hausdorff method point of view this is still 2.0 dimensions but it's far from a two dimensional surface we see. Now if one choose to use a scale factor of then we vill get 4.0 dimensions and a perfec two dimensional surface. Is the Hausdorff method of counting dimensions wrong? Or did I miss something? ;-) // Solkoll 12:35, 1 Mar 2005 (UTC)
Well, you are computing the similarity dimension (for an iterated function system), not the Hausdorff dimension. There is a theorem that says you get the Hausdorff dimension when you do this, provided the Open Set Condition (OSC) holds. Basically, this means there is not too much overlap of the images. In the first case, the OSC is true. You get a solid square, and it does have dimension 2. In the other two cases, the OSC is not true. In the second case, because of the overlap, the Hausdorff dimension could be less than 2. From the picture, it looks like it is. In the third case, the image is a solid square, but there is overlap again (can you see it in the picture?)... so the Hausdorff dimension (2) could be less than the similarity dimension (4).
-- G A Edgar 15:19, 17 Mar 2005 (UTC)
I don't want to get into an discussion of the merits of pictures, but I don't particularly like the new SIrepinski image. CSTAR 12:49, 1 Mar 2005 (UTC)
"The Hausdorff outer measure Hs is defined for all subsets of X. However, we can in general assert additivity properties, that is
for disjoint A, B, only when A and B are both Borel sets. From the perspective of assigning measure and dimension to sets with unusual metric properties such as fractals, however, this is not a restriction."
This is wrong, it is for measurable A, B, not only Borel. The set of all measurable sets is strictly larger than the set of all Borel sets. This is true for at least Hausdorff and Lebesgue measure (proof can be by construction using inverse images of the Cantor ternary function, or by cardinality). 220.245.178.131 05:37, 28 February 2006 (UTC)
Why are these recent edits an improvement? Any subset of a metric space is a metric space. Moreover, it's thge behavior of the measure which is unintuitive, not the values. I think the article should be reverted to version of August 26 [1].-- CSTAR 14:44, 12 October 2006 (UTC)
"For instance R has dimension 1 and its 1-dimensional Hausdorff measure is infinite." The first reason I would imagine that R is able to have infinite Hausdorff measure is that it is not compact. Is there a compact set whose Hausdorff measure is infinite, or a proof that such a set cannot exist? It would be a nice little comment to add after the quoted sentence either way. 69.215.17.209 16:21, 31 March 2007 (UTC)
Hausdorff dimension eh, has anyone got a fractal with a hausdorff dimension = golden ratio? 83.70.46.38 06:21, 23 April 2007 (UTC)
Yes: Beardon (1965, On the Hausdorff dimension of general Cantor sets) showed that generalised Cantor sets with arbitrary dimension can be generated. Colin Rowat ( talk) 07:07, 24 October 2011 (UTC)
The two concepts are related but are not the same: Hausdorff measure of smooth sets coincides with Lebesgue measure, while the Hausdorff dimension coincides with the "algebraic" dimension (very rougly, it is the cardinality of the set of linearly independent unit vectors needed to span the smallest linear subspace enclosing the given set). The confusion derives from the fact that in classical references, Hausdorff dimension is also called Hausdorff dimension measure: I have reported two links from the online Springer Enciclopedya of Mathematics clarifyng the two conceps.
I also remember that the fundamental reference in the field, i.e. Herb Federer's masterpiece, clearly states the two concepts and their difference (I'm quite sure and I will check later). In my opinion it is necessary to give rise to two separate voices. Daniele.tampieri ( talk) 14:43, 7 April 2008 (UTC)
As far as I can see the earliest reference was published in 1919. The fact that Hausdorff submitted the paper in 1918 doesn't mean the theory was "introduced" then, as the introduction currently states. The current text could be retained if Hausdorff was presenting his work at conferences in 1918 before publication of the article. —DIV ( 128.250.80.15 ( talk) 02:44, 28 April 2008 (UTC))
Whoa. The "Informal Discussion" is anything but. Someone needs to bring this down to a level where someone who isn't a math major can understand. Or perhaps consider changing the name. Timothyjwood ( talk) 04:37, 7 July 2008 (UTC)
[In the Section entitled "Informal discussion," i]s the construction given in the paragraph beginning, "Consider the number N(r) of balls of radius at most r ..." not that used in deriving Minkowski dimension, rather than Hausdorff? The key detail is the use of regular elements, the r-balls. If so, this gives an upper bound for Hausdorff dimension, which is typically much more easily derived than the Hausdorff dimension itself. 147.188.129.99 ( talk) 20:30, 27 October 2011 (UTC)
Anyone have a proof (or sketch) that the dimension of a product is at least the sum of the dimensions?
I'm not an expert in this area, but I'm pretty sure the given definition of Hausdorff content is incorrect. It disagrees with the books I've looked at, and moreover does not make sense. The given definition places no restriction on the balls used to cover the set. The typical definition of Hausdorff content is to say something like: H^d_delta is the infimum of sums of (r_i)^d, where the set can be covered by balls B_i of radius r_i LESS THAN DELTA. Then define H^d of the set to be the limit of H^d_delta as delta goes to 0. Otherwise, for instance you get that the 1-dimensional Hausdorff content of the unit ball in R^3 is finite. Kier07 ( talk) 16:36, 6 February 2009 (UTC)
Never mind, sorry. Though the words "content" and "measure" are very similar, so I think the article should be clearer about the fact that these are different ideas. Kier07 ( talk) 16:44, 6 February 2009 (UTC)
Hausdorff content (defined here without delta) and Hausdorff measure (defined in many sources with delta) are, indeed, different. For example, the unit ball in Euclidean space R^n is covered by some ball of some radius, so the d-dimensional content is finite for all d. But of course the d-dimensional Hausdorff measure is finite only if d >= n. G A Edgar ( talk) 11:21, 26 July 2009 (UTC)
Furthermore, defining Hausdorff dimension using the Hausdorff content is wrong. It must be defined from the Hausdorff measure. G A Edgar ( talk) 11:21, 26 July 2009 (UTC)
OK, the definition of Hausdorff dimension is not wrong. Although the Hausdorff content and the Hausdorff measure are different, when one is zero so is the other. G A Edgar ( talk) 13:34, 30 July 2009 (UTC)
From the definitions at the beginning at the paragraph, it seems clear that we are interested in the Hausforff dimension of S. However, the definition at the end is given with X, which is the enclosing metric space and not what we were interested in in the first place. Wouldn't it be clearer and more correct to give it with S instead? Joelthelion ( talk) 10:22, 17 August 2010 (UTC)
This assertion appears in the article. I can't tell if it's wrong or just confusing. None of the space-filling curves listed on the Space-filling curves page intersect themselves and the ones pictured have a Hausdorff dimension of 2.
I suggest removing or clarifying this assertion, but I don't have the mathematical expertise to know which is appropriate.
-
That statement is correct and here's why: A space filling curve maps a line to a filled square or cube or hypercube. The definition gives us that the curve is both continuous and onto. If it were to not intersect itself, then the curve would also be 1-1, which in turn become an isomorphism. Since a line and a square are not isomorphic, the curve should hit some points multiple times.
It seems counter-intuitive; but it's correct nonetheless. 212.175.32.133 ( talk) 09:26, 7 April 2015 (UTC)
Yet another mathematical encyclopedia completely inaccessible to a lay reader? Tag added because not even the lede is approachable. This is a typical maths article problem, that mathematicians are writing for themselves. I challenge contributing editors to write a lede and figure legends (if not the whole of the introductory portions of the article) that any high school-educated reader can read—without chasing every wikilink that appears. Wikilinks and citations are intended to allow interested readers to go deeper in particular subject areas, not to replace the need for good, understandable exposition. 165.20.108.150 ( talk) 23:20, 5 March 2015 (UTC)
These are the sources that appear for the Hausdorff dimension at its entry at the Encyclopedia of Mathematics, see [ http://www.encyclopediaofmath.org/index.php/Hausdorff_dimension], accessed 5 March 2014:
71.239.87.100 ( talk) 01:33, 6 March 2015 (UTC)
Given the WP:PSTS policy strictures on use of primary sources, the following two bits of text and their sources are moved here for discussion. In my view they add nothing to the article, and begin a pattern of dumping of trivial results here that will in long run, move the article to a status even further from GA than it currently is:
Le Prof 71.239.87.100 ( talk) 04:56, 6 March 2015 (UTC)
Why is the examples section flooded with "citation needed"s? These examples aren't some hard problems which took hundreds of years to solve, at maximum they would take an average person ten minutes to verify with a pen and paper. 212.175.32.133 ( talk) 11:28, 6 April 2015 (UTC)
The famous formula is absent!
If the article should be reduced to one formula, it would be that formula, and is not clearly and explicitly exposed. — Preceding unsigned comment added by 206.132.109.103 ( talk) 14:07, 5 January 2017 (UTC)
In the "Formal Definition" section, I'm assuming that the "balls" are balls under the supremum measure? I don't think that would be at all obvious to the average reader... Twin Bird ( talk) 03:26, 5 April 2020 (UTC)
Hi Editors; I suggest adding this Theorem to the page for readers' reference. It shows the existence of deterministic fractals for given Hausdorff dimension and Lebesgue measure. Thanks\\
P.S.\\
Theorem 6 (The Generalized Hausdorff Dimension Theorem). For any real r>0 and l≥0, there are aleph-two (symmetric) fractals with the Hausdorff dimension r.1{0}(l)+n.1(0,∞)(l) and Lebesgue measure l in Rn where (⌈r⌉≤n).
Link:
https://www.mdpi.com/2227-7390/9/13/1546 https://doi.org/10.3390/math9131546 — Preceding unsigned comment added by MohsenSoltanifarNA ( talk • contribs) 02:21, 9 July 2021 (UTC)
The approximation of the British Coast Line looks like an approximation of the Minkowski dimension (that often coincides with the Hausdorff dimension). For Hausdorff dimension balls of different size < delta are allowed. Maybe that should be clarified. — Preceding unsigned comment added by 2A02:908:621:3520:FFDE:36A8:F977:BA99 ( talk) 18:47, 16 April 2022 (UTC)
This article is rated B-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | ||||||||||||||||||||||||
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"If M is a metric space, and d > 0 is a real number, then the d-dimensional Hausdorff measure Hd(M) is defined to be the infimum of all m > 0 such that for all r > 0, M can be covered by countably many closed sets of diameter < r and the sum of the d-th powers of these diameters is less than or equal to m."
I read infimium, so I'm clear on that, but this definition is still opaque to me. Please clarify. --BlackGriffen
The box-counting dimension is not the same thing as the Housdorff dimension! -- Miguel
Hausdorff dimension isn't the only fractal dimension...it should have its own entry, and fractal dimension to talk about fractal dimensions collectively, it seems.
The banner is ridiculous and I am going to remove it. CSTAR 17:43, 25 Nov 2004 (UTC)
I put in a condition at the beginning which is basically the doubling property (which is of course much weaker). I'm not even sure I stated it correctly. Will the experts object? I perhaps should do my homework and consult my copy of Gromov-Semmes et-al. But hey, I'll let Tosha do that. CSTAR 17:49, 26 Nov 2004 (UTC)
Haha, ironic for the page-doubling bug to strike here. X) --[[User:Eequor| η υωρ]] 21:24, 26 Nov 2004 (UTC)
It's called Hausdorff dimension. Policy is that usual names should be used. This page should be moved back. Charles Matthews 19:31, 26 Nov 2004 (UTC)
It's User:Eequor on a stampede. See Besicovitch. Charles Matthews 20:08, 26 Nov 2004 (UTC)
If you really want to be accurate, you should say something like
ALso Hausdorff dimension is not a measurement and it also would be misleading to call it a measure. It's just a number assigned to a set. CSTAR 22:07, 26 Nov 2004 (UTC)
Why the name change? Virtually every authoritative work on the subject refers to the concept as Hausdorff dimension. For example
CSTAR 19:56, 26 Nov 2004 (UTC)
Firstly, Mandelbrot is much more of a publicist than a serious mathematician. I'm not prepared to take that book as an authority on mathematics. We use fractal the way he does (for better or for worse), because it's his term.
Second point: Wikipedia naming policy is (generally) to use the most usual name. The article can discuss credit where due; the title should be the normal way of naming anything. Charles Matthews 20:19, 26 Nov 2004 (UTC)
I think to split it. Tosha 06:09, 28 Dec 2004 (UTC)
They are not tied, one is defined using the other Tosha 21:39, 28 Dec 2004 (UTC)
I think it is a good idea to split them. The simplest definition of Hausdorff dimension does not involve the Hausdorff measure, but involves the Hausdorff content. Anyway, the Hausdorff content should be discussed somewhere. I think there is a lot to say about Hausdorff measure and it would not be good to put all this stuff here. Oded ( talk) 19:52, 9 May 2008 (UTC)
I think this does not work correctly! Take a look at the image at right, there you can see three examples of Hausdorff dimensions. The first is a 2D Cantor set, four main attractors and scale factor , no rotation. This gives a dimension of 2.0 and the result is also a two dimensional, 45º tilted square. The second image are the same but here a rotation of 45º is also used. From the Hausdorff method point of view this is still 2.0 dimensions but it's far from a two dimensional surface we see. Now if one choose to use a scale factor of then we vill get 4.0 dimensions and a perfec two dimensional surface. Is the Hausdorff method of counting dimensions wrong? Or did I miss something? ;-) // Solkoll 12:35, 1 Mar 2005 (UTC)
Well, you are computing the similarity dimension (for an iterated function system), not the Hausdorff dimension. There is a theorem that says you get the Hausdorff dimension when you do this, provided the Open Set Condition (OSC) holds. Basically, this means there is not too much overlap of the images. In the first case, the OSC is true. You get a solid square, and it does have dimension 2. In the other two cases, the OSC is not true. In the second case, because of the overlap, the Hausdorff dimension could be less than 2. From the picture, it looks like it is. In the third case, the image is a solid square, but there is overlap again (can you see it in the picture?)... so the Hausdorff dimension (2) could be less than the similarity dimension (4).
-- G A Edgar 15:19, 17 Mar 2005 (UTC)
I don't want to get into an discussion of the merits of pictures, but I don't particularly like the new SIrepinski image. CSTAR 12:49, 1 Mar 2005 (UTC)
"The Hausdorff outer measure Hs is defined for all subsets of X. However, we can in general assert additivity properties, that is
for disjoint A, B, only when A and B are both Borel sets. From the perspective of assigning measure and dimension to sets with unusual metric properties such as fractals, however, this is not a restriction."
This is wrong, it is for measurable A, B, not only Borel. The set of all measurable sets is strictly larger than the set of all Borel sets. This is true for at least Hausdorff and Lebesgue measure (proof can be by construction using inverse images of the Cantor ternary function, or by cardinality). 220.245.178.131 05:37, 28 February 2006 (UTC)
Why are these recent edits an improvement? Any subset of a metric space is a metric space. Moreover, it's thge behavior of the measure which is unintuitive, not the values. I think the article should be reverted to version of August 26 [1].-- CSTAR 14:44, 12 October 2006 (UTC)
"For instance R has dimension 1 and its 1-dimensional Hausdorff measure is infinite." The first reason I would imagine that R is able to have infinite Hausdorff measure is that it is not compact. Is there a compact set whose Hausdorff measure is infinite, or a proof that such a set cannot exist? It would be a nice little comment to add after the quoted sentence either way. 69.215.17.209 16:21, 31 March 2007 (UTC)
Hausdorff dimension eh, has anyone got a fractal with a hausdorff dimension = golden ratio? 83.70.46.38 06:21, 23 April 2007 (UTC)
Yes: Beardon (1965, On the Hausdorff dimension of general Cantor sets) showed that generalised Cantor sets with arbitrary dimension can be generated. Colin Rowat ( talk) 07:07, 24 October 2011 (UTC)
The two concepts are related but are not the same: Hausdorff measure of smooth sets coincides with Lebesgue measure, while the Hausdorff dimension coincides with the "algebraic" dimension (very rougly, it is the cardinality of the set of linearly independent unit vectors needed to span the smallest linear subspace enclosing the given set). The confusion derives from the fact that in classical references, Hausdorff dimension is also called Hausdorff dimension measure: I have reported two links from the online Springer Enciclopedya of Mathematics clarifyng the two conceps.
I also remember that the fundamental reference in the field, i.e. Herb Federer's masterpiece, clearly states the two concepts and their difference (I'm quite sure and I will check later). In my opinion it is necessary to give rise to two separate voices. Daniele.tampieri ( talk) 14:43, 7 April 2008 (UTC)
As far as I can see the earliest reference was published in 1919. The fact that Hausdorff submitted the paper in 1918 doesn't mean the theory was "introduced" then, as the introduction currently states. The current text could be retained if Hausdorff was presenting his work at conferences in 1918 before publication of the article. —DIV ( 128.250.80.15 ( talk) 02:44, 28 April 2008 (UTC))
Whoa. The "Informal Discussion" is anything but. Someone needs to bring this down to a level where someone who isn't a math major can understand. Or perhaps consider changing the name. Timothyjwood ( talk) 04:37, 7 July 2008 (UTC)
[In the Section entitled "Informal discussion," i]s the construction given in the paragraph beginning, "Consider the number N(r) of balls of radius at most r ..." not that used in deriving Minkowski dimension, rather than Hausdorff? The key detail is the use of regular elements, the r-balls. If so, this gives an upper bound for Hausdorff dimension, which is typically much more easily derived than the Hausdorff dimension itself. 147.188.129.99 ( talk) 20:30, 27 October 2011 (UTC)
Anyone have a proof (or sketch) that the dimension of a product is at least the sum of the dimensions?
I'm not an expert in this area, but I'm pretty sure the given definition of Hausdorff content is incorrect. It disagrees with the books I've looked at, and moreover does not make sense. The given definition places no restriction on the balls used to cover the set. The typical definition of Hausdorff content is to say something like: H^d_delta is the infimum of sums of (r_i)^d, where the set can be covered by balls B_i of radius r_i LESS THAN DELTA. Then define H^d of the set to be the limit of H^d_delta as delta goes to 0. Otherwise, for instance you get that the 1-dimensional Hausdorff content of the unit ball in R^3 is finite. Kier07 ( talk) 16:36, 6 February 2009 (UTC)
Never mind, sorry. Though the words "content" and "measure" are very similar, so I think the article should be clearer about the fact that these are different ideas. Kier07 ( talk) 16:44, 6 February 2009 (UTC)
Hausdorff content (defined here without delta) and Hausdorff measure (defined in many sources with delta) are, indeed, different. For example, the unit ball in Euclidean space R^n is covered by some ball of some radius, so the d-dimensional content is finite for all d. But of course the d-dimensional Hausdorff measure is finite only if d >= n. G A Edgar ( talk) 11:21, 26 July 2009 (UTC)
Furthermore, defining Hausdorff dimension using the Hausdorff content is wrong. It must be defined from the Hausdorff measure. G A Edgar ( talk) 11:21, 26 July 2009 (UTC)
OK, the definition of Hausdorff dimension is not wrong. Although the Hausdorff content and the Hausdorff measure are different, when one is zero so is the other. G A Edgar ( talk) 13:34, 30 July 2009 (UTC)
From the definitions at the beginning at the paragraph, it seems clear that we are interested in the Hausforff dimension of S. However, the definition at the end is given with X, which is the enclosing metric space and not what we were interested in in the first place. Wouldn't it be clearer and more correct to give it with S instead? Joelthelion ( talk) 10:22, 17 August 2010 (UTC)
This assertion appears in the article. I can't tell if it's wrong or just confusing. None of the space-filling curves listed on the Space-filling curves page intersect themselves and the ones pictured have a Hausdorff dimension of 2.
I suggest removing or clarifying this assertion, but I don't have the mathematical expertise to know which is appropriate.
-
That statement is correct and here's why: A space filling curve maps a line to a filled square or cube or hypercube. The definition gives us that the curve is both continuous and onto. If it were to not intersect itself, then the curve would also be 1-1, which in turn become an isomorphism. Since a line and a square are not isomorphic, the curve should hit some points multiple times.
It seems counter-intuitive; but it's correct nonetheless. 212.175.32.133 ( talk) 09:26, 7 April 2015 (UTC)
Yet another mathematical encyclopedia completely inaccessible to a lay reader? Tag added because not even the lede is approachable. This is a typical maths article problem, that mathematicians are writing for themselves. I challenge contributing editors to write a lede and figure legends (if not the whole of the introductory portions of the article) that any high school-educated reader can read—without chasing every wikilink that appears. Wikilinks and citations are intended to allow interested readers to go deeper in particular subject areas, not to replace the need for good, understandable exposition. 165.20.108.150 ( talk) 23:20, 5 March 2015 (UTC)
These are the sources that appear for the Hausdorff dimension at its entry at the Encyclopedia of Mathematics, see [ http://www.encyclopediaofmath.org/index.php/Hausdorff_dimension], accessed 5 March 2014:
71.239.87.100 ( talk) 01:33, 6 March 2015 (UTC)
Given the WP:PSTS policy strictures on use of primary sources, the following two bits of text and their sources are moved here for discussion. In my view they add nothing to the article, and begin a pattern of dumping of trivial results here that will in long run, move the article to a status even further from GA than it currently is:
Le Prof 71.239.87.100 ( talk) 04:56, 6 March 2015 (UTC)
Why is the examples section flooded with "citation needed"s? These examples aren't some hard problems which took hundreds of years to solve, at maximum they would take an average person ten minutes to verify with a pen and paper. 212.175.32.133 ( talk) 11:28, 6 April 2015 (UTC)
The famous formula is absent!
If the article should be reduced to one formula, it would be that formula, and is not clearly and explicitly exposed. — Preceding unsigned comment added by 206.132.109.103 ( talk) 14:07, 5 January 2017 (UTC)
In the "Formal Definition" section, I'm assuming that the "balls" are balls under the supremum measure? I don't think that would be at all obvious to the average reader... Twin Bird ( talk) 03:26, 5 April 2020 (UTC)
Hi Editors; I suggest adding this Theorem to the page for readers' reference. It shows the existence of deterministic fractals for given Hausdorff dimension and Lebesgue measure. Thanks\\
P.S.\\
Theorem 6 (The Generalized Hausdorff Dimension Theorem). For any real r>0 and l≥0, there are aleph-two (symmetric) fractals with the Hausdorff dimension r.1{0}(l)+n.1(0,∞)(l) and Lebesgue measure l in Rn where (⌈r⌉≤n).
Link:
https://www.mdpi.com/2227-7390/9/13/1546 https://doi.org/10.3390/math9131546 — Preceding unsigned comment added by MohsenSoltanifarNA ( talk • contribs) 02:21, 9 July 2021 (UTC)
The approximation of the British Coast Line looks like an approximation of the Minkowski dimension (that often coincides with the Hausdorff dimension). For Hausdorff dimension balls of different size < delta are allowed. Maybe that should be clarified. — Preceding unsigned comment added by 2A02:908:621:3520:FFDE:36A8:F977:BA99 ( talk) 18:47, 16 April 2022 (UTC)