We live in a 3-dimensional world. We know 2-dimensional objects as those we can see on paper. But what would an object with a nonnatural number (a number that isn\u27t a whole positive number) dimension look like? These dimensions exist and some unique mathematical sets have such dimensions, like log2/log3 . We call such dimensions and its tool Hausdorff dimension. In my talk, I examine Hausdorff dimension and its famous examples including the Cantor Set and multiple manipulations of it

Ellis, Amy

English

Sacred Heart University: DigitalCommons@SHU

Sacred Heart UniversityDigitalCommons@SHUAcademic FestivalApr 21st, 12:30 PM - 1:45 PMExploring Hausdorff DimensionAmy EllisFollow this and additional works at: https://digitalcommons.sacredheart.edu/acadfestThis is brought to you for free and open access by DigitalCommons@SHU. It has been accepted for inclusion in Academic Festival by an authorizedadministrator of DigitalCommons@SHU. For more information, please contact ferribyp@sacredheart.edu, lysobeyb@sacredheart.edu.Ellis, Amy, "Exploring Hausdorff Dimension" (2017). Academic Festival. 18.https://digitalcommons.sacredheart.edu/acadfest/2017/all/18Exploring Hausdorff DimensionAMY ELLISSacred Heart UniversityApril 21, 2017AbstractWe live in a 3-dimensional world. We know 2-dimensional objects asthose we can see on paper. But what would an object with a nonnaturalnumber dimension look like? These dimensions exist and some unique setshave such dimensions, like log2log3 . We will examine Hausdorff dimension, theCantor Set, and multiple manipulations of it.1 IntroductionIn this paper, we will primarily look at the middle third Cantor set. This sethas dimension log2log3 , using calculations of Hausdorff measure and Hausdorffdimension. We fully prove the middle third Cantor set has such dimension in a11Ellis: Exploring Hausdorff DimensionPublished by DigitalCommons@SHU, 2017thorough proof. My research consists of changes within the Cantor set and howthose changes affect dimension. We will change: its first interval’s length, itsremoval size at each interval, and its shape. All of these factors differently affectthe set’s dimension. Lastly, we try to visualize how Hausdorff dimension wouldhold in hyperbolic space.2 BackgroundHausdorff dimension is a concept from topology. Theorized by Felix Hausdorff in1918, "the Hausdorff dimension is a counting number agreeing with a dimensioncorresponding to its topology." In order to fully understand this, we must firstdiscuss our basic knowledge of dimension. We learned in grade school thedimension of a line is 1, a flat square 2, and a cube 3. Hausdorff argued thedimension of a single point is zero. He wanted to further examine complexsets in order to refer to its size. Such "complex sets" are usually irregular andextended from simple shapes or intervals in Rn.3 BasicsIn order to discuss Hausdorff Dimension, we first need to look at measures.Definition 3.1. A measure m is a nonnegative real function from a set F → R22Academic Festival, Event 18 [2017]https://digitalcommons.sacredheart.edu/acadfest/2017/all/18such that m(∅) = 0 where ∅ is the empty set, andm(A) = ∑nm(An)for any infinite or countable collection of pairwise disjoint sets (An) in F suchthat A =⋃(An) is also in F.In other words, a measure gives a number to a set in order to reference its size.Hausdorff measure is more complicated and therefore requires more machinery.Definition 3.2. Let U be a non-empty subset of n-dimensional Euclidean space.Then the diameter, |U| is equal to|U| = sup{|x− y| : x, y ∈ U}.Definition 3.3. A cover is any family of subsets of a given set B whose union is B.A cover is different from a measure, though the two are similar. A cover doesnot assign a number – rather, it assigns a set. This is how you can physically"cover" a set where the union is the whole space. For example, the set [0,1] hasmultiple covers, including:a) [0, 1n ] where n ∈Nb) [0, 12 ]∪(12 , 1]and so on.33Ellis: Exploring Hausdorff DimensionPublished by DigitalCommons@SHU, 2017Definition 3.4. If {Ui} is a countable or finite collection of sets of diameter atmost δ that cover F, i.e. F ⊂ ⋃∞i=1 Ui with 0 ≤ |Ui| ≤ δ for each i, we say that{Ui} is a δ-cover of F.In other words, imagine covering a set of small dots with circles of diameterat most δ (see Figure 1). We enhance the image of F to see two possible δ-coversof F. Let’s look at the cover on the lower left of Figure 1. Clearly, the union ofall these smaller circles is equal to F. Because 0 ≤ |Ui| ≤ δ, then this set withdiameters {Ui} is a possible δ-cover of F. We use the word "possible" becausethere are many other ways to cover F with other sizes of δ.Figure 1: A set F with two possible δ-covers. This figure is taken from [1].Now that we have the necessary tools, we can begin discussing Hausdorff44Academic Festival, Event 18 [2017]https://digitalcommons.sacredheart.edu/acadfest/2017/all/18measure.Definition 3.5. Hausdorff Measure. If F exists in Rn and s > 0, then for any δ > 0,Hsδ(F) = in f{∞∑i=1|Ui|s : {Ui} is a δ -cover of F}.Essentially, the Hausdorff measure is the smallest number of the sums of thesmallest diameters less than δ capable of covering F.Therefore, considering each δ-cover of F, we try to minimize the summationof the diameters by minimizing the size of s. Then as s gets smaller, so does theHausdorff measure of F. But when δ becomes smaller, the amount of coversof F also decreases. This is intuitive. However, the infimum of Hs(F) is non-decreasing. This isn’t as intuitive. Here is a clarification. Consider the set N. Theinfimum of this set is 1. Now, take away one number in N but don’t say which.What can we say about the infimum? It is either the same or higher, hence theinfimum is non-decreasing. The same thing is happening with the amount ofvalid covers of F as δ decreases.Going back to the measure, as δ→ 0 and the infimum ofHsδ(F) increases (isnon-decreasing), thenHs(F) = limδ→0Hsδ(F). (1)This equality is called the s-dimensional Hausdorff measure of F. We find the limitin order to extract δ to get a general Hausdorff measure to help us find s. Any55Ellis: Exploring Hausdorff DimensionPublished by DigitalCommons@SHU, 2017set F (where F ⊂ Rn) has a Hausdorff measure and its limit exists, though it isusually zero or ∞.Hausdorff measure is difficult to calculate. However, we can get close to auniversal equation for whole number, low-dimensional subsets of Rn.1. The number of points in a set F is given byH0(F), where dimHF = s = 0 andF is countable.2. The length of a smooth curve F is given byH1(F), where dimHF = s = 1.3. A smooth surface F is given by H2(F) = (4/π)× area(F), where dimHF =s = 2.4. A shape F is given byH3(F) = (6/π)× vol(F), where dimHF = s = 3.Hausdorff DimensionThe s-dimensional Hausdorff measure exists and we are able to find the value ofs for any set F ⊂ Rn. We can take this a step further to show if t > s and {Ui} isa δ-cover of F, then∑i|Ui|t ≤∑i|Ui|t−s|Ui|s.By the triangle inequality, the above equation works. We can go one step furtherto say∑i|Ui|t−s|Ui|s ≤ δt−s ∑i|Ui|s.Since {Ui} is a δ-cover of F, we know |Ui| ≤ δ by definition 3.4.66Academic Festival, Event 18 [2017]https://digitalcommons.sacredheart.edu/acadfest/2017/all/18Using definition 3.5, the infima showHtδ(F) ≤ δt−sHsδ(F).As δ → 0 and if Hs(F) < ∞, then Ht(F) = 0 when t > s. This follows fromDefinition (3.5), equation (1), and substitution.Figure 2: A graph ofHs(F) against s.We can create a graph ofHs(F) against s (Figure 2) to see there exists a criticalvalue of s where Hs(F) is neither 0 nor ∞. In Figure 2, this is the value dimHF.There is only one value of s for each and any unique set F. We call this theHausdorff dimension of F, written dimHF. Recall Definition (3.5) and see the values = dimHF. ThereforeHs(F) will always satisfy 0 ≤ Hs(F) ≤ ∞.Let’s look at a calculation of Hausdorff dimension.Example. Let F be a flat disk with unit radius in R3. We know the dimension of77Ellis: Exploring Hausdorff DimensionPublished by DigitalCommons@SHU, 2017this disk is greater than zero since it’s more than one single point. We also knowthe length of a disk is infinite because it doesn’t have endpoints. Therefore:H1(F) = length(F) = ∞0 < H2(F) = (4/π)× area(F) = 4 < ∞H3(F) = (6/π)× vol(F) = 0Therefore dimHF = 2 andHs(F) = ∞ when s < 2 andHs(F) = 0 when s > 2.Later, we will see how the length, area, and volume of a set may not alwaysbe consistent or easily determined. This yields a dimension between two wholenumbers.As with all new mathematical terms, Hausdorff dimension holds similarproperties to that of most dimension definitions. They are:Scaling Property: Let S be a similarity transformation of scale factor λ > 0. IfF ⊂ Rn, thenHs(S(F)) = λsHs(F).Monotonicity: If E ⊂ F then dimHE ≤ dimHF, which directly comes from equa-tions above.Countable Stability: If F1, F2, ... is a countable sequence of sets then dimH⋃∞i=1 Fi =sup1≤i<∞{dimHFi}. Essentially, dimension of the infinite union of Fi is equal tothe dimension supremum Fi88Academic Festival, Event 18 [2017]https://digitalcommons.sacredheart.edu/acadfest/2017/all/184 Hausdorff Dimensions of SetsOne of the most famous examples using Hausdorff dimension is the middlethird Cantor set. Picture the segment [0, 1]. Call this line E0. For E1, right belowE0, remove the middle third of E0, such that E1 is the segments [0, 13 ], [23 , 1]. Thisrepeats to infinity.Figure 3: The middle third Cantor set.The middle third Cantor set F is composed of the numbers in En for all n ∈N.Formally, F is the intersection⋂∞n=0 En. F is so infinitely small that it would behard to physically draw the set. Therefore, we can consider the middle thirdCantor set F to be limn→∞ En and pictures of F to be pictures of some En.What can we say about the dimension of the Cantor set? F is clearly morethan one point by construction. However, the length of F is infinite because F as99Ellis: Exploring Hausdorff DimensionPublished by DigitalCommons@SHU, 2017a set is infinite (though restricted between [0, 1]). Furthermore, every subintervalEn is a subset of a one-dimensional line [0, 1]. Hence, we know 0 < s < 1. This isa clear sign we need to use Hausdorff dimension!Theorem 4.1. The Hausdorff dimension of the Cantor Set is log2log3 = 0.6309...An intuitive proof would go as follows.Proof. The Cantor Set, call it F, becomes two parts: a left side FL = F ∩ [0, 13 ] anda right side FR = F ∩ [23 , 1]. We know both FL and FR are the same length and area ratio to F by(13). They are geometrically similar so F = FL ∪ FR. Therefore, forany value s we can see:Hs(F) = Hs(FL) +Hs(FR) =(13)sHs(F) +(13)sHs(F). (2)Because we are looking for a critical value 0 < s < ∞ we are able to divide byHs(F) to see 1 = 2(13)s. Applying log properties we get s = log2log3 .This explanation yields the Hausdorff dimension, but we need a rigorousproof in order to prove the Hausdorff measure supports the dimension we found.Proof. Let En be the intervals that make up the Cantor set F. We call En level-nintervals. Then for any n ≥ 0 we have En consisting of 2n level-n intervals, eachwith length 3−n. Take a 3−n-cover over the intervals of En. We see:Hs3−n(F) ≤ 2n3−ns. (3)1010Academic Festival, Event 18 [2017]https://digitalcommons.sacredheart.edu/acadfest/2017/all/18Letting s = log2log3 in equation (3), we findHs3−n(F) ≤ 1. As n→ ∞ thenHs(F) ≤ 1.Now we proveHs(F) ≥ 12 . We want to ultimately end up with the equation:∑ |Ui|s ≥12= 3−s (4)for any cover {Ui} of F. This equation will reduce toHs(F) ≥ 12 when s =log2log3 .That will allow us to always have an infimum greater than 12 and thereforealways yields a Hausdorff measure greater than 12 .For each {Ui}, let n be an integer so that:3(−n+1) ≤ |Ui| < 3−n. (5)Equation (5) just uses definition 3.4 to ensure each cover is less than the next.From equation (5), we see Ui can intersect at only one level-n integer becauseeach level-n integer is separated by a distance of at least 3−n.In equation (3), we saw 2n3−ns = 1 when log2log3 . Then 2−n = 3−ns.Let j ≥ n. Then by construction {Ui} intersects at most2j−n = 2j2−n = 2j3−sn ≤ 2j3s|Ui|s (6)level-j intervals of Ej. Equation (6) comes from what we know about inequalitiesand that 2n3−ns = 1 when s = log2log3 . The right side of the equation, which wehave not yet considered, comes from equation (5). We know |Ui| < 3−n, so whenwe raise these to the sth power, we have the opposite.1111Ellis: Exploring Hausdorff DimensionPublished by DigitalCommons@SHU, 2017Continuing on with the proof, we can find a j big enough to satisfy 3(−j+1) ≤|Ui| for all {Ui}, just as we did in equation (4). Then the selected {Ui} intersecteach 2j basic intervals with length 3−j. When we count these intervals, we get:2j ≤∑i2j3s|Ui|s. (7)Divide by 2j and divide again by 3−s to see equation (5) is equal to equation (2).It may seem like we only proved something about the Hausdorff measure,that it is somewhere between or equal to 12 and 1. Remember that this worksbecause we found an s to satisfy 12 ≤ Hs(F) ≤ 1, which is also true about dimHF.From the definition of Hausdorff dimension, we saw that we need to satisfy0 ≤ Hs(F) ≤ ∞. The critical value of s whereHs(F) is not 0 or ∞ is how we findthe Hausdorff dimension of F, just as we have proven above.5 Manipulations of Hausdorff Dimension SetsLengthWhat happens if we examine the Cantor set and change its size? Let F be themiddle third Cantor set where E0 = [0, 3]. We will still remove the middle third,such that E1 is the set [0, 1] ∪ [2, 3]. Therefore, we still have a scaling factor of(13)from Ek to Ek−1. We set up an intuitive calculation:1212Academic Festival, Event 18 [2017]https://digitalcommons.sacredheart.edu/acadfest/2017/all/18Hs(F) = Hs(FL) +Hs(FR) =(13)sHs(F) +(13)sHs(F).Notice this is exactly equation (2). Hence, we still have s = log2log3 and therefore theHausdorff dimension stays the same despite our change of endpoint from 1 to 3.Now let X = [0, π]. Remove the middle third once again, and continueremoving the middle third of each line from Ek to Ek+1. Then E1 = [0, π3 ]∪ [2π3 , π].We have the exact same idea with this set as the Cantor set, just with length πinstead of length 3. Again, we have a scaling factor of(13)from Ek to Ek+1 andnote we again have equation (2), based on the math in the previous paragraph.Therefore, Hausdorff dimension of the Cantor set does not change based onlength of the first interval (in these cases, the length of E0). This may seem hardto believe, so let me explain it in a different, less-mathematical way. Chooseany natural number, say 42. Let’s take a look at E42 in the normal middle thirdCantor set. We can find the length of E42 by using 3−n, so the length is(1342).Furthermore, there are 242 intervals at this level of the Cantor set. Imagine thatwe relabel E42 as the beginning of an entirely new middle third Cantor set A.Then E42 = A0.Though we have a separate set A, the dimension of A does not change basedon the length of A0 or E42. We still take away the same scaled amount(13)fromAn and have the same amount of intervals (2n). Hence, the dimension of A is1313Ellis: Exploring Hausdorff DimensionPublished by DigitalCommons@SHU, 2017still log2log3 . Furthermore, nothing limits the middle third Cantor set from beingpart of some bigger set B, just looking at the [0, 1] portion.Hyperbolic LengthsHyperbolic geometry is a type of geometry where Euclid’s parallel postulatedoes not hold. It is changed to the following: For any given line R and point P noton R, in the plane containing both line R and point P there are at least two distinct linesthrough P that do not intersect R. Lines and shapes look different in hyperbolicspace. Lines are more circular typically.Remark. The Hausdorff dimension of any set G in Euclidean geometry has thesame Hausdorff dimension in hyperbolic geometry.Consider a circle with r = 1. Then the circumference C = 2πr = 2π. Lookat the top half of this circle, and create the middle third Cantor set from it soG0 = [0, π]. Hence, G1 = [0, π3 ] ∪ [2π3 , π]. To find the Hausdorff dimension of G,note the scaling factor of(13)from Gk to Gk+1. ThenHs(G) = Hs(GL) +Hs(GR) =(13)sHs(G) +(13)sHs(G).This is similar to the last section. After dividing byHs(G), we again end up with1 = 2(13)sto see s = log2log3 .Though we need two dimensions to see a half-circle, the removal size andlengths are the same between G and X. But we can fit G in the first dimension1414Academic Festival, Event 18 [2017]https://digitalcommons.sacredheart.edu/acadfest/2017/all/18if we flatten it so that G becomes X. Therefore, the dimensions are the same,where s = log2log3 . If we looked at G as a full circle instead of a half-circle, thenthe dimension would be different because we would always need at least twodimensions to see G. Therefore, s would probably be greater than 1.Removal SizeHow does removing the middle third differ from removing the middle ninth inthe Cantor set? Let Y be the line [0, 1]. Remove the middle ninth from Y andplace the remaining two intervals directly below Y such that Y1 = [0, 49 ]⋃[59 , 1].Repeat this process for Yn to infinity.For a quick intuitive proof, let’s again break apart YL and YR such that YL =Y ∩ [0, 49 ] and YR = Y ∩ [59 , 1]. Each new interval is scaled to Y by a ratio of49 .Therefore, we can use the same set-up for the equation:Hs(Y) = Hs(YL) +Hs(YR) =(49)sHs(Y) +(49)sHs(Y).Divide byHs(Y) and we have 1 = 2(49)s. Then s = −log22(log2−log3) .We can assume that whenever we change the removal size of the middlepiece in the Cantor set, then the dimension of the Cantor set changes. This worksout because the scaling ratio is different, hence creating a large difference in theequation to find s.1515Ellis: Exploring Hausdorff DimensionPublished by DigitalCommons@SHU, 2017Further ResearchThe middle third Cantor set can be manipulated in many different ways. Forexample, what happens if we let k ∈N and remove k segments from the interval[0, 1], leaving a space between the segments of size 3−k? Would the set haveno dimension, for each interval with removal size 3−k would have a differentdimension? I also looked at the plane analogue Cantor Dust set, which happenedto have dimension 1. With this Cantor Dust, the image remains the same sizebut smaller pieces get added with every new subset, whereas the middle thirdCantor set removes pieces with every new subset. The possibilities are endlesswith Hausdorff dimension, for we have no solid foundation for how dimensionis changed based on small changes of a set.References[1] Falconer, K. J. Fractal Geometry: Mathematical Foundations and Applications. 2nd ed. Chichester:Wiley, 1990. Print.[2] Gamelin, Theodore W., and Robert Everist Greene. Introduction to Topology. 2nd ed. Philadel-phia: Saunders College Pub., 1983. Print.[3] MacGregor, Campbell. "5.6 Scaling and the Hausdorff Dimension," at AnnenbergLearner:MATHematics illuminated. 2013. Web. Accessed 30 November 2016.1616Academic Festival, Event 18 [2017]https://digitalcommons.sacredheart.edu/acadfest/2017/all/18

Exploring Hausdorff Dimension

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Exploring Hausdorff Dimension

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