The Sierpinski Triangle: An Aesthetically Pleasing Limit Point

Jim Bell

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Table of Contents

Page 3: Section 0: Many Methods With One Result

Page 5: Section 1: The Hausdorff Metric Space

Page 11: Section 2: Fractal Dimensions

Page 14: Section 3: The Contraction Mapping Theorem

Page 22: Section 4: The Hutchinson Operator

Page 24: Section 5: An Aesthetically Pleasing Limit Point

Page 35: Bibliography

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Section 0: Many Methods With One Result

Over the years people have developed many methods to create the Sier-pinski triangle (or gasket), some of which are more visual and intuitive andsome of which are more technical and demanding. We will start with someof the more intuitive strategies and build into a rigorous definition accordingto equations and their limits.

To begin, we will play a game called Sir Pinski’s Game, cleverly namedin Manfred Schroeder’s Fractals, Chaos, Power Laws. This game requirestwo people. Consider an equilateral triangle. Each person picks a startingpoint in the triangle and the goal is to keep your point in the triangle longerthan your opponent can. The rules are simple: on each turn the player mustmove his or her point to twice the distance from the nearest corner alongthe line that the point and the corner make. It must be moved along theline connecting the corner and original point so that the new point is on thesame side of the corner as the original point. The player who keeps the dotin the triangle for the most moves is the victor. Some interesting phenomenaoccur in this game, the result of which is that there are actually infinitelymany starting points such that if the player chooses them the player’s dotwill remain in the triangle for all of eternity. Ironically and unfortunately, ifthe player chooses the dot at random, the player has a probability of zero ofchoosing one of these points.

As can be seen from a few examples in Figure 1, any point chosen withinthe large, centered, upside down triangle will move out of the triangle in onemove. This is therefore a poor choice for a player and it is recommendedthat a player start from one of the three surrounding right-side up triangles.It also turns out that if a centered equilateral upside-down triangle is placedinside of each of the three smaller triangles, any dot inside the smaller upsidedown triangles will in one move be transported to the large upside downtriangle in the center. From there the triangle will be moved outside of themain triangle and the player’s game is over.

This pattern can be continued indefinitely so a player can find the numberof moves he or she has left simply by finding the number of iterations neededto place the original dot in an upside down triangle. On the other hand, if aplayer strategically places his or her original point precisely in the shape thatremains after an infinite number of these iterations, the player can make anunlimited number of moves and the resulting point will always be within the

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Figure 1: Jumping out

bounds of the original triangle. The shape where the original point shouldbe strategically placed is the Sierpinski triangle and can be seen on the coverof this paper (only a finite number of its lines can be seen of course). It isobvious that with every step of removing triangles there is still a positivearea for the remaining shape. Interestingly enough, the resulting shape froman infinite amount of iterations has an area of 0.

Yet another method for creating the Sierpinski triangle is via the ChaosGame, described by Michael F. Barnsley in Fractals Everywhere. Createan equilateral triangle with three vertices, V0, V1, and V2. Now put a pointanywhere in the plane in which the triangle exists. A method which selectsone of the three vertices at random is needed. A die works well, with one andtwo corresponding to V0, three and four corresponding to V1, and five and sixcorresponding to V2. Essentially, we have created a 3-sided die. Each timethe die is rolled, the point is moved to one half of its original distance to thevertex corresponding to the number on the die. The new point must remainon the line connecting the vertex and the original point. Suprisingly enough,if this process is repeated enough times and the first few (few meaning upto the hundreds if necessary) points created from this iterated process are

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removed the Sierpinski Triangle is the result! A few steps of the process canbe seen in Figure 2. The triangle at the left of the figure has the least amountof iterated points and each triangle to its right has more points created fromthe Chaos Game.

Figure 2: Chaos Game Iterations

Interestingly enough, not only does it not matter where the original pointis, it does not even matter what order the vertices are in as long as they arechosen randomly and with equal probability. This method of creating theSierpinski triangle is more similar to the method using equations and limitswhich is the focus of this paper. The sequence of points created in thismethod is called the orbit of x0 with x0 being the first point in the sequence.Note that no matter what the starting point, x0, is, if the “function” we areusing is iterated enough times we find ourselves with the same result.

Section 1: The Hausdorff Metric Space

So what is the Sierpinski triangle? How can we define it? In how manydimensions does it exist? These are all questions which should be answeredbut to do so we must first delve into the world of fractals and iterated functionsystems. A very general and loose definition of a fractal is an image thatcontains copies of itself at infinitely many different scales. This means it is“self-similar” which is a term we will visit later. The Sierpinski triangle isa fractal; if any of the smaller right-side-up triangles within the Sierpinskitriangle were to be blown up to the size of the Sierpinski triangle itself, thesame image would be viewed. It should be noted now that in many of theexamples and much of the discussion to follow, unless otherwise noted wewill be working in (R2, d).

To more rigorously define fractals it is necessary that we define somethingcalled the Hausdorff metric denoted H(X) for some metric space X.

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Definition 1 If (X, d) is a complete metric space then, excluding the nullset, H(X) is the space whose points are the compact subsets of X.

This means, for example, that if we are working in H(R), then [0, 1] and [-1,5] are each different points in H(R). We must now define distance from apoint to a set so that we can later properly define distance in the Hausdorffspace.

Definition 2 If (X, d) is a complete metric space, x ∈ X, and B ∈ H(X)then the distance d from x to B is d(x, B) = min{d(x, b) : b ∈ B}.

It is not obvious that such a point b ∈ B would necessarily exist. The proofthat follows is similar to the one found in Barnsley’s Fractals Everywhere.

Proof:Let (X, d) be a complete metric space, x ∈ X, and B ∈ H(X). Rememberthat ∀ y ∈ B, d(x, y) ≥ 0. This means that d(x, y) is bounded below.Let D = inf{d(x, y) : y ∈ B}. From the definition of infimum, we knowthat for all n ∈ N, ∃ p ∈ B such that d(x, p) − D ≤ 1

n. We can now

create an infinite sequence of points, (bn)n≥1 in B such that d(x, bn) −D ≤1n. Because B is compact we know (bn) has a subsequence (bni

)i≥1 thatconverges to a point in B. We will call it b. Because ( 1

ni) converges to

0 and 0 ≤ d(x, bni) − D ≤ 1

ni, we know (d(x, bni

) − D) converges to 0 solimi→∞(d(x, bni

)) = D and d(x, limi→∞(bni)) = D. Because limi→∞(bni

) = bit must be true that d(x, b) = D. Therefore, there exists b ∈ B such thatd(x, b) = min{d(x, y) : y ∈ B}.

Now we must define distance in the Hausdorff metric.

Definition 3 If (X, d) is a complete metric space and A, B ∈ H(X) then thedistance d from A to B is defined as d(A, B) = max{d(a, B) : a ∈ A}.

The proof that such an a exists is similar to the proof above.

Proof:Let (X, d) be a complete metric space and let A, B ∈ H(X). Because A iscompact we know A is bounded. Then there exists D = sup{d(a, B) : a ∈ A}.By the definition of supremum we can create an infinite sequence (an)n≥1 suchthat an ∈ A and D− d(an, B) ≤ 1

n∀ n ∈ N. Because A is compact, (an) has

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a subsequence (ani)i≥1 such that the limit point of (ani

) is in A. We will callthis limit point a. Because ( 1

ni) converges to 0 and 0 ≤ D − d(ani

, B) ≤ 1ni

,we know (D − d(ani

, B)) converges to 0. Then (d(ani, B)) converges to D

so for the same reasons as in the previous proof, d(a, B) = D. Then thereexists a ∈ A such that d(a, B) = max{d(x, B) : x ∈ A}.

Intuitively, the Hausdorff distance h is the same as putting an “epsilonnet” around one set so you can capture the other. This means if there aretwo sets, A and B, and you are finding the distance from A to B then youmust find the point b in B closest to A. Then, as shown in Figure 3, find thesmallest radius possible that, when one end is fixed at b, will inscribe A. This

Figure 3: Epsilon Net from A to B where d(A, B) = Epsilon and Epsilon2Net from B to A where d(B, A) = Epsilon2

radius is the distance from A to B and is sometimes called an epsilon net.

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Note that this distance is not commutative; d(A, B) is not necessarily thesame as d(B, A). We thus come to the definition of the Hausdorff distance:

Definition 4 If (X, d) is a complete metric space and A, B ∈ H(X) then theHausdorff distance, h, is defined as follows; h(A, B) = max{d(A, B), d(B, A)}.

This just means you make an epsilon net from A to B and from B to A andthe largest of the two is h(A, B). We must now show that h is a metric forH(X).

Theorem 5 If X is a complete metric space then h is a metric for H(X).

Proof:Assume X is a complete metric space. We must show that h is a metric forH(X). To do this we have to prove four statements are true.

This first is that for all A, B ∈ H(X), h(A, B) ≥ 0. We know the followingthree statements are true:

h(A, B) = max{d(A, B), d(B, A)}

d(A, B) = max{d(a, B) : a ∈ A}d(B, A) = max{d(b, A) : b ∈ B}.

Delving once more into a more basic definition, d(a, B) = min{d(a, b) : b ∈B} and d(b, A) = min{d(b, a) : a ∈ A}. We know because d is a metric on Xthat min{d(a, b) : b ∈ B} ≥ 0 and that min{d(b, a) : a ∈ A} ≥ 0. This meansthat d(a, B) ≥ 0 and that d(b, A) ≥ 0 which implies that max{d(a, B) : a ∈A} ≥ 0 and that max{d(b, A) : b ∈ B} ≥ 0. Finally, we know that d(A, B) ≥0 and that d(B, A) ≥ 0 so we find that h(A, B) = max{d(B, A), d(A, B)} ≥0.

The second item on our list to prove is that if A, B ∈ H(X), then h(A, B) = 0iff A = B. To begin, assume A = B. Then

d(A, B) = max{d(a, B) : a ∈ A}.

Because A = B, if a ∈ A then a ∈ B as well. Then because

d(a, B) = min{d(a, b) : b ∈ B}

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and for all a ∈ A, a is in B as well, d(a, B) = 0. This implies that d(A, B) = 0.The same argument can be made for the fact that d(B, A) = 0. Then itmust be true that max{d(A, B), d(B, A)} = 0 so h(A, B) = 0. Now assumeh(A, B) = 0. Then it must be true that max{d(A, B), d(B, A)} = 0. Con-sider the fact that d(A, B) = 0. Then max{d(a, B) : a ∈ A} = 0. This meansif we choose some point a ∈ A that min{d(a, B) : b ∈ B} = 0. Thus, for anya ∈ A, there is some b ∈ B such that a = b. Therefore, for all a ∈ A, a ∈ Bas well. The same argument can be made for the fact that for all b ∈ B,b ∈ A as well. Therefore, A = B.

Thirdly, we must show that if A, B ∈ H(X) then h(A, B) = h(B, A). This isrelatively simple. Because both the statements

h(A, B) = max{d(A, B), d(B, A)}

h(B, A) = max{d(B, A), d(A, B)}

are true, it must be true that either h(A, B) = d(A, B) and h(B, A) =d(A, B) or h(A, B) = d(B, A) and h(B, A) = d(B, A). Then it must be truethat h(A, B) = h(B, A).

Finally we must show that if A, B, C ∈ H(X) then

h(A, C) ≤ h(A, B) + h(B, C).

Assume (X, d) is a metric space. Let A, B, C ∈ H(X). We know

h(B, C) = max{d(B, C), d(C, B)}.

Assume

h(B, C) = d(B, C) = max{d(b, C) : b ∈ B} = d(b1, C)

for some b1 ∈ B. We know

d(b1, C) = min{d(b1, c) : c ∈ C} = d(b1, c1)

for some c1 ∈ C. Also, h(A, B) = max{d(A, B), d(B, A)}. Assume

h(A, B) = d(A, B) = max{d(a, B) : a ∈ A} = d(a1, B)

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for some a1 ∈ A. We know

d(a1, B) = min{d(a1, b) : b ∈ B} = d(a1, b2)

for some b2 ∈ B. Finally, we know h(A, C) = max{d(A, C), d(C, A)}. As-sume

h(A, C) = d(A, C) = max{d(a, C) : a ∈ A} = d(a2, C)

for some a2 ∈ A. We know

d(a2, C) = min{d(a2, c) : c ∈ C} = d(a2, c2)

for some c2 ∈ C.From the above we know that there exists b3 ∈ B such that d(a2, b3) ≤

d(a1, b2) because d(a1, b2) = max{d(a, B) : a ∈ A}. Also, there exists c3 ∈ Csuch that d(b3, c3) ≤ d(b1, c1) because d(b1, c1) = max{d(b, C) : b ∈ B}.Finally, we know that d(a2, c2) ≤ d(a2, c3) because d(a2, c2) = min{d(a2, c) :c ∈ C}.

From all this we can deduce that

h(A, C) = d(a2, c2)

≤ d(a2, c3)

≤ d(a2, b3) + d(b3, c3)

≤ d(a1, b2) + d(b3, c3)

≤ d(a1, b2) + d(b1, c1)

= h(A, B) + h(B, C).

Then it is true that h(A, C) ≤ h(A, B)+h(B, C). The cases when h(A, B) =d(B, A), h(B, C) = d(C, B), and when h(A, C) = d(C, A) are all similar.From this and the previous three results we know that h is a metric onH(X).

Now that we know that h is a metric on H(X) we know that (H(X), h)is a metric space. Fortunately for us, this is a metric space where it will bepossible for us to define distances between fractals and convergence of shapesin R2. Hence, the Hausdorff metric space, (H(X), h), is a metric space wherewe can truly define and understand fractals. Many fractals, such as theSierpinski triangle, are the limit points of iterated functions. It does notmake sense, however, to speak of an entire shape being a limit point in a

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metric space such as (R2, d). How would we decide whether an iterated pointwas within a certain distance of the limit point? The Hausdorff metric spaceis a space where we can treat entire compact sets as points which simplifiesdefinitions and equations immensely.

Section 2: Fractal Dimensions

According to Robert L. Devany in his book A First Course in ChaoticDynamical Systems, “A fractal is a subset of Rn which is self-similar andwhose fractal dimension exceeds its topological dimension” (178). What inthe world does that mean? To understand this definition we must learn aboutself similarity and magnification. It must be noted at this point that thissection is a bit of a tangent from the ultimate goal of this paper. It seemednecessary, however, to at least give fractals a more rigorous definition andtouch on their complex dimensions. This is a very large field so what followsis only a scratch on the surface.

The Sierpinski triangle is self-similar. One can look at the figure on thetitle page of this paper and see that any “right side up” equilateral triangle, ifblown up to the right size, will yield the original triangle. To be more explicit,one of the ways to create the Sierpinski triangle is to take the midpoint ofeach segment of an equilateral triangle, connect them to make a triangle, andremove that triangle from the set. This leaves three identical triangles, eachwith sides that are half the length of the sides of the original triangle. Forthe rest of this paper I will use the fact that the sides become half as longto mean that the triangle becomes half as big, in spite of the fact that thenewly created triangle actually has only a quarter of the original triangle’sarea. This process can be repeated with the three new triangles and onewould have nine new triangles, each one fourth the size of the original andone half the size of the second generation of triangles. If this process isrepeated an infinite number of times, the Sierpinski triangle is the result.

This has certain interesting implications. One can therefore take theSierpinski triangle and find 3n identically-sized triangles within it and zoomany one of them up to 2n times its original size to yield the original Sierpinskitriangle! From this we can define the term affine self-similar.

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Definition 6 A set is considered affine self-similar if it can be dividedinto some integer m matching subsets. The matching subsets must be able tobe magnified by some positive constant n to yield the original set.

To understand Devany’s definition of a fractal, we must still define topo-logical dimension.

Definition 7 A set has a topological dimension of 0 if all its pointshave arbitrarily small boundary ovals such that the boundary ovals do notintersect the set.

In this situation, “boundary ovals” just means small circles or ellipses cir-cumscribing points. For instance, in Figure 4, the set depicted by the pointshas a topological dimension of zero because around each point can be placedan ellipse that does not intersect any of the other points.

Figure 4: 0 Topological Dimensions

Now we can define topological dimension in general.

Definition 8 A set has topological dimension m if at each point an ar-bitrarily small boundary oval can be created which intersects the original setin a topological dimension of m− 1 and m must be the smallest nonnegativeinteger for which this is true.

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This basically means that if the boundary ovals around the points in someset K intersect K so that the intersection has a topological dimension of 0,the topological dimension of K is 1 because the intersection had a topologicaldimension of 0 = 1 - 1. This is illustrated in the top figure in Figure 5. Foranother object, say a filled in circle, any boundary oval around any point inthat circle will yield an intersection with a topological dimension greater thanone. If we take a boundary oval around a point in the intersection we get anew set with a topological dimension of 0. This means that the intersectionof the circle with the original boundary oval had a topological dimension of1 which implies the filled in circle had a topological dimension of 2. This isillustrated in the bottom figure in Figure 5.

Figure 5: Topological Dimensions

This basically indicates that finding the topological dimension of a set canbe an iterative process. The Sierpinski triangle therefore has a topologi-cal dimension of 1 because no matter where a boundary oval is created theboundary oval will intersect the Sierpinski triangle in non-connected points.This intuitively makes sense because the Sierpinski triangle is composed en-tirely of straight lines while all boundary ovals are some form of an ellipse.Therefore, no two consecutive points of the ellipse will intersect any lines of

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the Sierpinski triangle because of the curvature of the ellipse so the intersec-tion of the Sierpinski triangle with a boundary oval will yield non-connectedpoints. Keep in mind that this is not a rigorous proof that the Sierpin-ski triangle has a topological dimension of 1; it hopefully is just a helpfulillustration of why it has a topological dimension of 1.

Finally, we must define fractal dimension.

Definition 9 If a set K can be broken into some integer m identical partsand those parts can each be magnified by some factor R to yield K then Khas a fractal dimension D of D = log(m)

log(R).

Let us apply this new knowledge to the Sierpinski triangle. We know that forany n ∈ N, the Sierpinski triangle can be broken into 3n identical pieces, eachof which may be magnified by a factor of 2n to yield the Sierpinski triangle.This means it has a fractal dimension of D = log(3n)

log(2n)= log(3)

log(2)≈ 1.584962501.

We may come to the conclusion with our result that the Sierpinski triangleis, according to Devany, a fractal because it is a subset of R2, it is affine self-similar, and it has a fractal dimension which is greater than its topologicaldimension.

So what does a fractal dimension of 1.584962501 mean when normally wedeal with dimensions that are integers? We know a line is one dimensional.Intuitively if we were placed on the line, we could either move forward orbackward but that is it. We would only have choices in one dimension. Afilled in circle is two dimensional with the same idea. We could move up,down, left, right, or some combination which involves two dimensions. Withthe Sierpinski Triangle, however, at each point there are more choices thanjust back and forth, however, not as many choices as if we were in an infiniteplane. This is why its dimension is stuck between 1 and 2.

Section 3: The Contraction Mapping Theorem

The Sierpinski triangle is actually a limit point of an iterated map ofa function. Odd as it is to think of it this way, it is true. How can anentire shape be considered a point? Remember that we are working in theHausdorff metric space so compact sets are actually considered points. Notonly is it a limit point of the iterated map of this function, but it is the limit

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point no matter what shape you start off with or where in R2 this shapeis located. How can we show the Sierpinski triangle is the limit point ofany initial point’s orbit under this special function? We can do this by thecontraction mapping theorem.

Definition 10 Let X be a metric space and let 0 < k < 1. Then if f : X →X is a function such that d(f(a), f(b)) ≤ k(d(a, b)) for all a, b ∈ X then fis a k-contraction of X.

Theorem 11 (The Contraction Mapping Theorem) If X is a completemetric space and f : X → X is a k-contraction of X then for all x ∈ X, theorbit of x under f converges to some unique fixed point x0.

This is a very powerful theorem to have. To use it, however, we must prove it.The following proof was outlined in Carol Schumacher’s Closer and Closer.Proof:Let X be a complete metric space and f : X → X be a k-contraction of X.The first step in proving this theorem is to show that f cannot have morethan one fixed point. This is relatively simple and we shall prove this bycontradiction. Assume f has more than one fixed point. Let two of them bex and y. Then f(x) = x and f(y) = y. Then we know d(f(x), f(y)) = d(x, y)so f is not a contraction. This is a contradiction so f can have at most onefixed point.

The second step is to show that for all n ∈ N and for some x ∈ X,

d(fn(x), fn+1(x)) ≤ knd(x, f(x)).

This will be a proof by induction. We know because f is a k-contraction that

d(f(x), f2(x)) ≤ kd(x, f(x)).

Now assume that

d(fn(x), fn+1(x)) ≤ knd(x, f(x)).

Then because f is a k-contraction, it must be true that

d(fn+1(x), fn+2(x)) = d(f(fn(x)), f(fn+1(x)))

≤ k · knd(x, f(x))

= kn+1d(x, f(x)).

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Induction holds and this completes this part of the proof.Finally, we must prove that if X is complete then for any x ∈ X, the

orbit of x under f converges to some fixed point x0. Because X is complete,every Cauchy sequence in X must converge to some point in X. We knowthat because 0 < k < 1 that limn→∞(kn) = 0. Set ε ≥ 0 and let x ∈ X. Weknow from the properties of geometric series that for all N ∈ N that

∞∑i=N

kid(x, f(x)) = d(x, f(x))∞∑

i=N

ki

= d(x, f(x))kN

∞∑i=0

ki

= d(x, f(x))kN(1

1− k).

Because d(x, f(x)) and 11−k

are constant and (kn) converges to 0 we know

we can fix N ∈ N such that d(x, f(x))kN( 11−k

) < ε. Fix N such that this is

true. Then∑∞

i=N kid(x, f(x)) < ε. Now choose m, n > N such that m < nand m, n ∈ N. Then from the triangle inequality we know

d(fm(x), fn(x)) ≤ d(fm(x), fm+1(x)), +d(fm+1(x), fm+2(x)) + ... + d(fn−1(x), fn(x))

≤n−1∑i=m

kid(x, f(x))

<∞∑

i=N

kid(x, f(x))

< ε.

This means (fn(x)) is Cauchy so it converges to some point x0 in X. x0 mustbe a fixed point because (fn(x)) converges to it. This completes our proofof the contraction mapping theorem.

Now we should use this powerful new theorem to show that we may iteratea certain function an infinite number of times on any starting point in H(X)and the result is the Sierpinski triangle. To apply this theorem properlythough, we must show that H(X) is complete.

This proof is not simple. It is similar to the one proven in Barnsley’sFractals Everywhere. To prove it, however, we need to prove a few lemmasfirst. The first is called the Dilation Lemma so we should define dilation.

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Definition 12 Let X be a metric space and let A ∈ X and let δ ∈ R. Then

A + δ = {x ∈ X : d(x, a) ≤ δ ∀ a ∈ A}.

This is the dilation of A by a ball of radius δ.

Now we are ready to work with the Dilation Lemma.

Lemma 13 (The Dilation Lemma) If (X, d) is a complete metric space,and A, B ∈ H(X), and ε > 0 then h(A, B) ≤ ε iff A ⊆ B + ε and B ⊆ A+ ε.

Proof:Let (X, d) be a complete metric space, and A, B ∈ H(X), and ε > 0. Toprove this lemma we will begin with the fact that we know that h(A, B) =max{d(A, B), d(B, A)}. Assume h(A, B) = d(A, B). Assume d(A, B) ≤ ε.Then max{d(a, B) : a ∈ A} ≤ ε so d(a, B) ≤ ε ∀ a ∈ A and therefore∀ a ∈ A, a ∈ B + ε and finally A ⊆ B + ε. We know d(B, A) ≤ d(A, B)so d(B, A) ≤ ε so max{d(b, A) : b ∈ B} ≤ ε and d(b, A) ≤ ε ∀ b ∈ B.Hence, ∀ b ∈ B, b ∈ A + ε so B ⊆ A + ε. A similar argument can be madeif h(A, B) = d(B, A). Now suppose A ⊆ B + ε and B ⊆ A + ε. BecauseA ⊆ B+ε we know ∀ a ∈ A, ∃ b ∈ B such that d(a, b) ≤ ε. Let a ∈ A. Thend(a, B) ≤ ε. Because we know this is true ∀ a ∈ A, it must be true thatd(A, B) ≤ ε. A similar argument can be made for the fact that d(B, A) ≤ ε.Then because h(A, B) = max{d(A, B), d(B, A)} we know h(A, B) ≤ ε. Thush(A, B) ≤ ε iff A ⊆ B + ε and B ⊆ A + ε.

Another lemma we must prove before trying to tackle the fact that(H(X), h) is complete is the Extension Lemma which is as follows.

Lemma 14 (The Extension Lemma) Let (X, d) be a metric space. Let(An) be a Cauchy sequence of points in (H(X), h). Let (ni) be a sequence ofintegers such that 0 < n1 < n2.... Assume there exists a Cauchy sequenceof points {ani

∈ Ani: i = 1, 2, 3...} in (X, d). Then there exists a Cauchy

sequence of points {a∗n ∈ An : n = 1, 2, 3, ...} such that a∗ni= ani

∀ i =1, 2, 3....

Proof:Let (X, d) be a metric space. Let (An) be a Cauchy sequence of points in(H(X), h). Let (ni) be a sequence of integers such that 0 < n1 < n2....

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Assume there exists a Cauchy sequence of points {ani∈ Ani

: i = 1, 2, 3...}in (X, d). To prove this lemma, we must first show such a sequence (a∗n)exists. To do this we will create it. For all n ∈ {1, 2, ..., n1}, pick a∗n ∈ {a ∈An : d(a, an1) = d(an1 , An)}. We know a∗n exists because An is compact.Parallel proofs to this are the proofs corresponding to Definitions 2 and3. Now for all i ∈ {2, 3, 4...} and all n ∈ {ni−1 + 1, ni−1 + 2, ..., ni} leta∗n ∈ {a ∈ An : d(a, ani

) = d(ani, An)}. Intuitively, each a∗n is just the closest

point in An to its respective ani. This means that for a∗ni

, that the closestpoint in Ani

to aniis of course ani

itself so a∗ni= ani

.Now we must show that (a∗n) is Cauchy. Let ε > 0. Because (ani

) isCauchy, there exists N1 ∈ N such that if nr, ns > N1 then d(anr , ans) < ε

3.

Also, because (An) is Cauchy, there exists N2 ∈ N such that if e, f > N2

then h(Ae, Af ) < ε3. Now let N = max{N1, N2}. Let m, n > N such that

m ∈ {nr−1 + 1, nr−1 + 2, ..., nr} and n ∈ {ns−1 + 1, ns−1 + 2, ..., ns}. Becauseh(Am, Anr) < ε

3we know there exists a∗m ∈ Am such that d(a∗m, anr) < ε

3.

A similar argument can be made for d(ans , a∗n) < ε

3. Now we know that

d(a∗m, a∗n) ≤ d(a∗m, anr)+ d(anr , ans)+ d(ans , a∗n) < ε

3+ ε

3+ ε

3= ε. Hence, (a∗n)

is Cauchy.

Lemma 15 Let X be a metric space. Let A be a compact subset of X. Thenif δ > 0 it follows that A + δ is closed.

Proof:Let X be a metric space and let A ⊆ X be compact. Set δ > 0. Let (xn) bea sequence in A + δ that converges to some x. Then, from the definition ofA+ δ, there exists a sequence (an) in A such that d(an, xn) ≤ δ for all n ∈ N.Because A is compact, we know there exists a subsequence (ani

) of (an) suchthat limn→∞(ani

) = a and a ∈ A. We can now create (xni) such that (xni

)is a subsequence of (xn) and for all i ∈ N, d(ani

, xni) ≤ δ. Set ε > 0. We can

find I1 ∈ N such that for all i > I1, d(ani, a) < ε

2. We can find I2 ∈ N such

that for all i > I2, d(xni, x) < ε

2. Set I = max{I1, I2}. Now we know for all

i > I that

d(a, x) ≤ d(a, ani) + d(ani

, xni) + d(xni

, x) <ε

2+ δ +

ε

2= ε + δ.

Since ε was arbitrarily chosen we know d(a, x) ≤ ε. Because a ∈ A, we knowx ∈ A + δ. Therefore A + δ contains all its limit points and is closed.

We now need a definition and one more lemma.

18

Definition 16 If X is a metric space and A is a subset of X then A istotally bounded if for any ε > 0, A can be covered by a finite number ofclosed balls with a radius of ε.

Lemma 17 Let X be a complete metric space and let A be complete and atotally bounded subset of X. Then A is compact.

This lemma is essentially proven in Carol Schumacher’s standard proofof the Heini-Borel Theorem in Closer and Closer. She helpfully points outthat any totally bounded subset of a compact metric space that is also closedcan be shown to be compact as well. We will not go into the details of thisproof in this paper, but again, the intricacies of the argument can be foundin Closer and Closer.

Now we should have the information we need to prove that if (X, d) is acomplete metric space then (H(X), h) is a complete metric space.

Theorem 18 Let (X, d) be a complete metric space. Then (H(X), h) is acomplete metric space. Thus, if (An)n≥1 is a Cauchy sequence in H(X) thenA = limn→∞(An) ∈ H(X) can be rewritten as A = {x ∈ X : there is asequence, (xn) with xn ∈ An, that converges to x}.

Proof:Let (X, d) be a complete metric space and let (An) be a Cauchy sequence inH(X). Let A = {x ∈ X : there is a sequence (xn) such that xn ∈ An thatconverges to x}. We must prove the following:(1) A 6= ∅.(2) A is closed which implies it is complete since X is complete.(3) For all ε > 0, ∃ N ∈ N such that for all n ≥ N, A ⊆ An + ε.(4) A is totally bounded which, combined with the results of (2) and Lemma17 implies A is compact.(5) limn→∞(An) = A.

Proof of (1): A 6= ∅.

We must show there exists a Cauchy sequence (an) such that an ∈ An.We know there exists a sequence {n1, n2, n3, ...} such that if we pick someni from this sequence then h(Am, An) < 1

2i ∀ m, n ≥ ni. Pick an1 ∈ An1 .Because h(An1 , An2) < 1

2we know from the Dilation Lemma that there exists

an2 ∈ An2 such that d(an1 , an2) < 12. Because of the way we created the

19

sequence {n1, n2, n3, ...} we know that we can pick any nj and similarly,h(Anj

, Anj+1) < 1

2j . Then, again from the Dilation Lemma, we know thatthere exists anj

∈ Anjand anj+1

∈ Anj+1such that d(anj

, anj+1) < 1

2j . In thisfashion we can create an entire sequence which we will call (ani

). Now wehave a sequence (ani

) such that ani∈ Ani

and d(ani, ani+1

) < 12i . We must

now show (ani) is Cauchy in X. Let ε > 0. Pick Nε such that

∑∞i=Nε

12i < ε.

Then for all t > s ≥ Nε we know

d(ans , ant) ≤ d(ans , ans+1) + d(ans+1 , ans+2) + ... + d(ant−1 , ant)

<

t∑i=s

1

2i

<∞∑

i=Nε

1

2i

< ε.

From the Extension Lemma we know there exists a Cauchy sequence (a∗n)such that a∗n ∈ An and a∗ni

= ani. Because (a∗n) is in X and X is complete,

(a∗n) converges to some a in X. By the definition of A, a ∈ A and thereforeA 6= ∅.

Proof of (2): A is closed which implies it is complete since X is complete.

Let (xi) be a sequence in A that converges to some point x. Because ofthe way we defined A, we know for each xi, there exists a sequence (aij) suchthat aij ∈ Aj and (aij) converges to xi. For each n ∈ N, we can pick xin suchthat d(xin , x) < 1

2n. We can then pick aijn such that d(aijn , xin) < 1

2n. Then

d(aijn , x) ≤ d(aijn , xn) + d(xn, x) < 12n

+ 12n

= 1n. Thus, for any ε > 0, we can

find N ∈ N such that for all n > N, d(aijn , x) < ε. Then (aijn) converges tox. From the Extension Lemma, we know there exists a sequence (a∗in) suchthat (a∗in) converges to x and a∗in ∈ An and a∗in = aijn . Then from the waywe created A, x ∈ A and A is closed. Because A is a closed subset of X andX is complete, A is complete as well.

Proof of (3): For all ε > 0, ∃ N ∈ N such that for all n ≥ N, A ⊆ An + ε.

Let ε > 0. Let a ∈ A and let (ai) be a sequence such that ai ∈ Ai and (ai)converges to a. Pick N ∈ N such that for all m,n, and j > N, h(Am, An) ≤ ε

20

and d(aj, a) < ε. Set n > N . If m > n then Am ⊆ An + ε and am ∈ An + ε.Because An is compact, An is closed. We can then deduce from Lemma 15that An + ε is closed as well. This fact combined with the fact that for allj > m, aj ∈ An +ε shows that (ai) must converge to a point in An +ε. Thena ∈ An+ε. Because a was arbitrarily chosen, it must be true that A ⊆ An+ε.Hence, for all ε > 0, ∃ N ∈ N such that for all n > N, A ⊆ An + ε.

Proof of (4): A is totally bounded which, combined with the results of(2) and Lemma 17 implies A is compact.

This is a proof by contradiction. Let A not be totally bounded and setε > 0. Then we can find a sequence (ai) in A such that for all i and j,d(ai, aj) ≥ ε for i 6= j. From part (3) we know that there exists n such thatA ⊆ An + ε

3. Then for all ai, there exists an xi ∈ An such that d(ai, xi) < ε

3.

We know An is compact so a subsequence (xin) of (xi) converges in An.Then we can find two points xin and xim such that d(xin , xim) < ε

3. For these

points xin and xim , there exists ain , aim ∈ A such that d(ain , xin) < ε3

andd(aim , xim) < ε

3. Thus

d(ain , aim) ≤ d(ain , xin) + d(xin , xim) + d(xim , aim) <ε

3+

ε

3+

ε

3= ε.

This is a contradiction so A is totally bounded. Therefore, because A is alsoclosed, A is compact so A ∈ H(X).

Proof of (5): limn→∞(An) = A.

From the Dilation Lemma, we know we must show that for all ε >0, ∃ N ∈ N such that for all n > N, An ⊆ A+ε. Let ε > 0 and let N be suchthat for all m, n > N, h(Am, An) < ε

2. Then for all m,n > N, Am ⊆ An + ε

2.

Set n > N . Let y ∈ An. Then because (An) is Cauchy and because ofthe Dilation Lemma we know there exists an increasing sequence (Ni) in Nsuch that n < N1 < N2 < ... < Nk < ... and for all r, s > Nj for somej, Ar ⊆ As + ε

2j+1 . Then it must be true that An ⊆ AN1 + ε2. Because

y ∈ An, ∃ aN1 ∈ AN1 such that d(y, aN1) < ε2. We also know from the con-

struction of (ANi) that AN1 ⊆ AN2 + ε

22 . Because aN1 ∈ AN1 , ∃ aN2 ∈ AN2

such that d(aN1 , aN2) < ε22 . Through induction we can make a sequence

aN1 , aN2 , ... such that aNj∈ ANj

and d(aNj, aNj+1

) < ε2j+1 . Then

d(y, aNj) ≤ d(y, aN1) + d(aN1 , aN2) + ... + d(aNj−1

, aNj) <

j∑i=1

ε

2j< ε.

21

It is true that (aNi) is Cauchy so because X is complete, (aNi

) convergesto some a. In addition, because of the way we chose n and the way (Ni)was constructed, ANi

⊆ An + ε. Then (aNi) is in An + ε. We know from

Lemma 15 An + ε is closed so it must be true that a ∈ An + ε. We knowd(y, aNi

) < ε ∀ Ni and (aNi) converges to a so it must be true that d(y, a) ≤ ε.

We can prove this by contradiction. Assume d(y, a) > ε. If d(y, a) > ε thenwe can rewrite d(y, a) as ε + b where b > 0. Then it is true that for all Ni,d(y, a) ≤ d(y, aNi

) + d(aNi, a) so d(y, a)− d(y, aNi

) ≤ d(aNi, a). There exists

q > 0 such that q < ε + b − d(y, aNi) because d(y, aNi

) < ε. Hence, for allNi, q ≤ d(aNi

, a) and this is a contradiction because (aNi) converges to a.

We know that, according to the Extension Lemma, there is some sequence(a∗n) that converges to a where a ∈ A and a∗ni

= aniand a∗n ∈ An. Hence,

A ⊆ An + ε ∀ n > N . This shows limn→∞(An) = A and because An was anarbitrary Cauchy sequence in (H(X), h), it is true that (H(X), h) is complete.

Section 4: The Hutchinson Operator

We have now successfully shown that the Hausdorff Metric Space is com-plete so we can apply the contraction mapping theorem to k-contractionswithin H(X). This means it is time for us to become more familiar with theSierpinski triangle itself, and not just general formulas. As has been men-tioned before, there are many ways to create the Sierpinski triangle. Thereare Sir Pinksi’s Game, the Chaos Game, and others. There is also the methodwhich iterates functions on points in the Hausdorff Metric Space. A specificfunction exists which, when iterated an infinite number of times, yields theSierpinski triangle. It can be described as a Hutchinson operator.

Definition 19 Let X be a metric space and let A be a subset of X. Letf0, f1, ... , fn be a finite number of affine transformations. The Hutchin-son Operator of f0, f1, ... , fn on A is f0(A)

⋃f1(A)

⋃...

⋃fn(A).

This definition basically means that if there is a group of functions where eachfunction changes an image in R2, the Hutchinson operator of these functionscreates their resulting images simultaneously. There are many choices that

22

are possible for the individual affine transformations and lead to differentinteresting fractals. In the case of the Sierpinski triangle, however, we will beworking with the three affine transformations A0 : R2 → R2, A1 : R2 → R2,and A2 : R2 → R2 which are defined as follows:

A0(~z) = 12

(xy

)A1(~z) = 1

2

(x− 1

y

)+

(10

)A2(~z) = 1

2

(x− 1

2

y −√

32

)+

( 12√3

2

)

where ~z =

(xy

).

Now we can extend each of these functions so they act on compact sets.This is relatively simple. They will each perform their respective opera-tions on every point in each set, the only difference is that once they areextended they can act on more than one point at once. Because we are onlyconcerned with compact sets we can define these functions in the HausdorffMetric Space. We, for clarification purposes, will denote each of these affinetransformations when acting on compact sets differently than A0, A1, and A2

which only act on points in R2. We will refer to them as A′0 : H(R2) → H(R2),A′1 : H(R2) → H(R2), and A′2 : H(R2) → H(R2).

From A′0, A′1, and A′2 we can deduce that the Hutchinson operator forthe Sierpinski triangle when acting on Z ∈ H(R2) is

H(Z) = A′0(Z)⋃

A′1(Z)⋃

A′2(Z)

where H : H(R2) → H(R2).Let us take a minute and try to understand intuitively what each of these

functions actually does to sets in R2 or to points in H(R2). We will startwith A′0. A′0 basically just takes sets in R2, contracts them to half theiroriginal size, and shifts them linearly toward (0, 0). A′1 takes sets in R2,contracts them to half their original size, and shifts them linearly toward (1,0). A′2 takes sets in R2, contracts them to half their size, and shifts them

linearly toward (12,√

32

). What does this mean when the three functions arecombined and how does this create the Sierpinski triangle? Each function’soutput is an image half the size of the original. That means the Hutchinson

23

operator takes an element in H(R2) and returns the union of three sets, eachthe same shape as the original but only half its size. Figure 6 illustrates thisnicely. The figure is a simple case because, in reality, the original set canbe anywhere in R2 but for the sake of making it easier to understand thepicture, the original set is in the center of the future Sierpinski triangle.

Figure 6: Hutchinson Operator Iterations

Section 5: An Aesthetically Pleasing Limit Point

Figure 6 helps us understand how if the Sierpinski triangle can be createdfrom an angry face with fangs then the Sierpinski triangle can be created fromany compact subset of R2. It is an iterated process which occurs in H(R2).With each iteration of the specific Hutchinson operator, which will from nowon be referred to as H, the set in R2, or the point in H(R2) becomes more likethe Sierpinski triangle. A more mathematical way of phrasing this is thatlimn→∞(Hn(S)) = S where S ∈ H(R2) and S denotes the Sierpinski triangle.Once again, even though the idea of an entire geometrical shape being a limitpoint is fairly abstract, it must be remembered that we are working in H(R2)and shapes are in fact points.

We of course cannot be satisfied with the idea that figures converge to Swith just some equations and a figure illustrating what the equations do toa poorly drawn face. We must prove that no matter what the initial point

24

S0 ∈ H(R2) is, the limit of (Hn(S0)) is S. This will be the focus of therest of this paper, but how can we accomplish such a task? We can use thecontraction mapping theorem. We have already shown that because R2 iscomplete, H(R2) is complete as well. Now we must show H is a k-contractionand we must show that S is a fixed point of H.

We will begin showing that H is a k-contraction on H(R2) by showingthat each of the “point versions” of its component functions is a k-contractionon R2. To do this we will need to prove the following lemma.

Lemma 20 Let

(x1

y1

),

(x2

y2

),

(x3

y3

),

(x4

y4

)∈ R2. If

d

((x1

y1

),

(x2

y2

))≤ d

((x3

y3

),

(x4

y4

))then it must be true that

d

(A0

(x1

y1

), A0

(x2

y2

))≤ d

(A0

(x3

y3

), A0

(x4

y4

)).

Proof:

Let

(x1

y1

),

(x2

y2

),

(x3

y3

),

(x4

y4

)∈ R2. We know that for any(

xy

)∈ R2, A0

(xy

)=

(x2y2

)so we know that

d

(A0

(x1

y1

), A0

(x2

y2

))=

√(x1

2− x2

2

)2

+(y1

2− y2

2

)2

=

√(1

2(x1 − x2)

)2

+

(1

2(y1 − y2)

)2

=

√1

4((x1 − x2)2 + (y1 − y2)2)

=1

2

√(x1 − x2)2 + (y1 − y2)2

=1

2d

((x1

y1

),

(x2

y2

)).

A parallel argument can be made to show that

d

(A0

(x3

y3

), A0

(x4

y4

))=

1

2d

((x3

y3

),

(x4

y4

)).

25

This means that if

d

((x1

y1

),

(x2

y2

))≤ d

((x3

y3

),

(x4

y4

))then it must be true that

d

(A0

(x1

y1

), A0

(x2

y2

))≤ d

(A0

(x3

y3

), A0

(x4

y4

)).

Overall this implies that if in a set there is a pair of points such that thedistance between them is the maximum distance between any pair of pointsin the set, the distance between the output of A0 of those two points will bethe maximum of the distance between the output of A0 of any pair of pointsin the set. The same conclusion can be drawn from similar arguments forA1, and A2.

With this result we can show that A′0 is a k-contraction in (H(R2), h).

Lemma 21 A′0 is a k-contraction in (H(R2), h).

Proof:Let A, B ∈ (H(R2), h). For this proof assume that h(A, B) = d(A, B). Aparallel argument exists for when h(A, B) = d(B, A). We know from our

proof stemming from Definition 3 that there exist some

(ax

ay

)∈ A and

some

(bx

by

)∈ B such that h(A, B) = d

((ax

ay

),

(bx

by

)). Now we can

apply A0 to these points and we know that

d

(A0

(ax

ay

), A0

(bx

by

))=

√(ax

2− bx

2

)2

+

(ay

2− by

2

)2

=

√(1

2(ax − bx)

)2

+

(1

2(ay − by)

)2

=

√1

4((ax − bx)2 + (ay − by)2)

=1

2

√(ax − bx)2 + (ay − by)2

=1

2d

((ax

ay

),

(bx

by

)).

26

We have already proven Lemma 20 which says that since

d

((ax

ay

),

(bx

by

))= max{d

((xy

), B

):

(xy

)∈ A}

it must be true that

d

(A0

(ax

ay

), A0

(bx

by

))= max{d

(A0

(xy

), A′0(B)

):

(xy

)∈ A}

by showing that if A0 is applied to every point in two sets, the distancerelationships between all pairs of points between the two sets will not change.We can apply what we just showed to show that

h(A′0(A), A′0(B)) = d

(A0

(ax

ay

), A0

(bx

by

))=

1

2d

((ax

ay

),

(bx

by

))=

1

2h(A, B).

We therefore know that A′0 is a k-contraction in H(R2) with a contractionfactor of 1

2.

As a side note not relevant to these proofs, we can use the contraction

mapping theorem on A′0 alone. We know that A0

(00

)=

(00

)so

(00

)is

a fixed point of A0 and because A′0 acts in the same way on individual points

as A0,

(00

)is a fixed point of A′0 as well. With that fact and Lemma 21,

we know that if S ∈ (H(R2), h) then (A′n0 (S)) converges to

(00

).

Now back to the focus of the paper. The method used to show A′0 is ak-contraction in H(R2) can be applied to A′1 and A′2 to show that they arealso k-contractions in H(R2) with contraction factors of 1

2. We will now use

these new results to show that H is a k-contraction as well.

Theorem 22 H is a k-contraction in H(R2, h).

27

Proof:Let A, B ∈ (H(R2), h). For this proof assume that h(A, B) = d(A, B).A parallel argument exists for when h(A, B) = d(B, A). We know that

there exist some

(ax

ay

)∈ A and some

(bx

by

)∈ B such that h(A, B) =

d

((ax

ay

),

(bx

by

)). Now we can apply A′0 to these sets and we know that

h(A′0(A), A′0(B)) = d

(A0

(ax

ay

), A0

(bx

by

))=

1

2h(A, B).

This means that

d

(A0

(ax

ay

), A0

(bx

by

))= min{d

(A0

(ax

ay

), A0

(xy

)):

(xy

)∈ B}.

Then either

min{d(

A0

(ax

ay

), H

(xy

)):

(xy

)∈ B} = d

(A0

(ax

ay

), A0

(bx

by

))

or there exists

(xy

)∈ B such that

min{d(

A0

(ax

ay

), H

(xy

)):

(xy

)∈ B} = d

(A0

(ax

ay

), Ai

(xy

))≤ 1

2d

((ax

ay

),

(bx

by

))=

1

2h(A, B)

such that i = 1, 2. Now we know that d

(A0

(ax

ay

), H(B)

)≤ 1

2h(A, B)

and we know that

d

(A0

(ax

ay

), H(B)

)= max{d

(A0

(xy

), H(B)

):

(xy

)∈ A}.

This is true because if any other point in A′0(A) were chosen there would besome point in A′0(B) such that the distance between the two points would

28

be smaller than the distance from A′0(A) to A′0(B). A parallel argument can

be used for A1

(ax

ay

)and A2

(ax

ay

). In any of the three cases,

max{d(

An

(xy

), H(B)

):

(xy

)∈ A} ≤ 1

2h(A, B).

Therefore d(H(A), H(B)) ≤ 12h(A, B). The same argument can be used to

show d(H(B), H(A)) ≤ 12h(B, A). Then

h(H(A), H(B)) = max{d(H(A), H(B)), d(H(B), H(A))} ≤ 1

2h(A, B),

so H is a k-contraction with contraction factor 12

in the metric space (H(R2), h).

We have only one proof left until we can use the contraction mappingtheorem. We must prove S is a fixed point in (H(R2), h). Before we canbegin with the proof, however, we must devise a way of giving each pointin S an “address.” Obviously the Sierpinski triangle can be divided intotriangles within triangles. At each level, there are three new triangles, eachone-half the size of the triangles at the level before. Let us define these “sub-triangles” as R for the bottom right triangle, L for the bottom left triangle,and T for the top triangle. Each point in S can be given an address consistingof an infinite series of these letters such as L, R, R, R, T, L, T, R . . .. TheL at the beginning denotes that the point is located within the bottom lefttriangle. The R that follows denotes the point is within the bottom righttriangle of the bottom left triangle, and so on. Figure 7 helps illustrate thispoint.

To prove S is a fixed point of H we will need to show that if

(xy

)∈ S

then H

(xy

)∈ S as well and we will need to show that if

(xy

)∈ S then

there is some

(x0

y0

)∈ S such that Ai

(x0

y0

)=

(xy

)where i = 0, 1, 2.

Lemma 23 Let ~s ∈ S. Then H(~s) ∈ S.

Proof:

Consider some point

(xy

)∈ S. We know it has some address as described

29

Figure 7: Sierpinski Triangle Addresses

above. For the sake of argument, let us choose the first address letter to be

R and let us apply A0 to

(xy

). There are parallel arguments for if the

first address letter is L or T . We know that because the first address letter

of

(xy

)is R that 1

2≤ x ≤ 1 and 0 ≤ y ≤

√3

4. We can therefore rewrite(

xy

)as

( 12

+ ε1√3

4− ε2

)where 0 ≤ ε1 ≤ 1

2and 0 ≤ ε2 ≤

√3

4. If we apply A0

to our point written with its newly written coordinates we get

( 14

+ ε1

2√3

8− ε2

2

).

The strategy for the rest of this proof is to show that if we apply A0 to ourpoint then we will get a new point located in the same “geometrical” positionof another triangle in S as our original point was located in the triangle withits first address label as R. Because S is self-similar this means the rest ofthe new point’s address labels will be the same as the original point’s addresslabels so it will be an element of S. This concept is illustrated in Figure 8where point a was shifted to point b. Point a has the same geometricallyrelevant position within the bottom right triangle as point b does within thebottom right triangle within the bottom left triangle. Because of the selfsimilarity of S, if each of these triangles were scaled to be the same size theywould be identical. This means that other than the first couple of addresslabels for each point, the two points have identical addresses within S.

Let us examine our new coordinates for A0

(xy

)more closely. We know

30

Figure 8: Proportional Spatial Geometry

our new point will have an address starting with L. This is because the xcoordinate is 1

4+ ε1

2. Because 0 ≤ ε1 ≤ 1

2, it must be true that 0 ≤ ε1

2≤ 1

4.

Therefore, 14≤ 1

4+ ε1

2≤ 1

2, which means either the new point has a first

address letter of either L or T because all of the points in R have an xcoordinate greater than 1

2.

Now we will turn to the y coordinate. Our new y coordinate is√

38− ε2

2.

We know that 0 ≤ ε2 ≤√

34

so it must be true that 0 ≤ ε2

2≤

√3

8which implies

that 0 ≤√

38− ε2

2≤

√3

8. All of the points with addresses that start with T

have y coordinates greater than√

34

so our point must have an address thatstarts with L.

We can now examine our point’s new address even more microscopically.The triangle with the first two address labels as L, R has x coordinates thatrange from 1

4to 1

2. It has y coordinates that range from 0 to

√3

8. Therefore,

because our new point’s coordinates are

( 14

+ ε1

2√3

8− ε2

2

)with 0 ≤ ε1

2≤ 1

4and

0 ≤ ε2

2≤

√3

8we know that if our new point is an element of S then its first

two address labels will be L and R. The triangle (L, R) is one half the size ofthe triangle (R). In the triangle (R) our original point was shifted a distanceof ε1 to the right of the leftmost point. In the square containing the triangle(L, R) the point resulting from A0 will be shifted by a factor of ε

2to the right

of the leftmost point. In addition, in the triangle (R) our original point wasshifted a distance of ε down from the topmost point. In the square containing

31

the triangle (L, R) our newly created point is shifted a value of ε2

down fromthe topmost point. This means, because the triangle (L, R) is half the size of

the triangle (R), that not only is

( 14

+ ε1

2√3

8− ε2

2

)in the triangle (L, R), but it

is in the same exact geometrically relative position in that triangle as

(xy

)was in the triangle (R). This means that if the first address point of

(xy

)is R, then A0

(xy

)∈ S and its first two address points are L, R followed

by the same sequence of address points

(xy

)had. This same argument can

be used for the functions A1 and A2 as well. They will, however, shift pointsin different directions, which in turn, will add different address labels thanwill A0.

This means that if any one of these functions is applied to any point inS, its output will be an element of S as well. Therefore if H is applied toany point in S the set of all three of its affine transformations’ outputs willbe an element of S.

Now we must show that if ~s ∈ S then there exists ~s1 ∈ S such thatAn(~s1) = ~s.

Lemma 24 Let ~s ∈ S. Then there exists ~s1 ∈ S such that An(~s1) = ~s wheren = 0, 1, or 2.

Proof:This proof will be similar to the previous one. We will use an address thatstarts with the points L and R but any pair of the three possible addresspoints will suffice as a starting point. Also, any of the three functions will

work but we will prove this using A0. Let ~s =

(xy

)and let its first two

address points be L and R. This will allow us to write

(xy

)as

( 14

+ ε1

2√3

8− ε2

2

)with 0 ≤ ε1 ≤ 1

2and 0 ≤ ε2 ≤

√3

4for the reasons explained in the proof of

Lemma 23.The strategy for the rest of this proof is to show that there is some other

point ~s1 such that H(~s1) will yield ~s. To do this we will find a point ~s1 such

32

that A0(~s1) = ~s. We will then need to show ~s1 ∈ S. We can do this throughsimilar methods as the ones used in the proof to Lemma 23. We can showthat ~s1 is in the same geometrically relevant position in one of the trianglesof S as ~s was in another triangle of S and because the triangles are similarand ~s ∈ S, ~s1 ∈ S as well.

Now consider the point

( 12

+ ε1√3

4− ε2

). We know that A0

(( 12

+ ε1√3

4− ε2

))=( 1

4+ ε1

2√3

8− ε2

2

). Now we must show that

( 12

+ ε1√3

4− ε2

)∈ S. We know that

the triangle (R) is twice the size of the triangle (L, R). Let us denote( 12

+ ε1√3

4− ε2

)as s1. The x coordinate of ~s1 is twice the distance to the right

of the leftmost point of the the triangle (R) as the x coordinate of ~s wasin (L, R). The y coordinate of ~s1 is twice the distance down from the top-most point of the triangle (R) as the y coordinate of ~s was in (L, R). Thismeans that ~s1 is in the same geometrically relevant position in (R) as ~s wasin (L, R). Because all the triangles in S are similar and ~s ∈ S it must betrue that ~s1 ∈ S as well. Thus, for every point ~s ∈ S, there exists some otherpoint, ~s1 ∈ S such that H(~s1) will yield ~s (it will also give two other points,but we have already proven that they will be in S also).

Theorem 25 S is a fixed point of H in (H(R2), h).

Proof:From Lemmas 23 and 24 we know that if ~s ∈ S then H(~s) ∈ S and there issome ~s0 ∈ S such that H(~s0) will yield ~s. Therefore, H(S) = S so S is a fixedpoint of H in (H(R2), h).

We can therefore conclude that because (H(R)2, h) is complete, H is ak-contraction with a contraction factor of 1

2, and S is a fixed point of H, if

S ∈ (H(R)2, h) then Hn(S) converges to S. This fact is fascinating in andof itself. More generally, any compact shape in R2 can be taken and moldedinto a very precise and ordered shape with the many interesting propertiesencountered throughout this paper. There are many more interesting fractalssimilar to the Sierpinski triangle such as the Koch Snowflake and the CantorSet. Each can be written as the limit of an iterated map of a Hutchinsonoperator on a point in the Hausdorff metric space. There are also many more

33

complicated fractals which would require a more in-depth study of fractaldimension, complex functions, and other areas to understand. Each of thesefractals has its own special qualities but all of them share qualities such asself-similarity. From what we have learned here, however, we know that theSierpinski triangle is justifiably described as an aesthetically pleasing limitpoint.

34

Bibliography

Howard Anton, Chris Rorres. Elementary Linear Algebra. New York: JohnWiley and Sons Inc., 2000.

Barnsley, Michael F. Fractals Everywhere. San Diego: Morgan Kaufmann,1993.

Devany, Robert L. A First Course In Chaotic Dynamical Systems. Reading,Massachusetts: Addison-Wesley Publishing Company, Inc., 1992.

Devany, Robert L. An Introduction to Chaotic Dynamical Systems.Reading, Massachusetts: Addison-Wesley Publishing Company, Inc., 1989.

Hart, John C. “Iterated Function Systems and Recurrent Iterated FunctionSystems.” May 14, 1996.

Hartmut Jurgens, Heinz-Otto Peitgen, Dietmar Saupe. Chaos and FractalsNew Frontiers of Science. New York: Springer-Verlag, 1992.

Holmgren, Richard A. A First Course in Discrete Dynamical Systems. NewYork: Springer, 1991.

McCauley, J. L. Chaos, Dynamics and Fractals. Cambridge: CambridgeUniversity Press, 1995.

Pickover, Clifford A. Fractal Horizons. New York: St. Martin’s Press, 1996.

Schroeder, Manfred. Fractals, Chaos, Power Laws. New York: W. H.Freeman and Company, 1991.

Schumacher, Carol. Closer and Closer An Introduction to Real Analysis.Jones

and Bartlett Publishers (not yet in print).

The cover art was taken from the website with the addresshttp://www.spsu.edu/math/tile/im/sierp.gif.

35

The triangles in Figure 2 were taken from the website with the addresshttp://mathforum.org/advanced/robertd/chaos game.html

Special thanks in alphabetical order to Professors Judy Holdener, RobertMilnikel, and Carol Schumacher who all helped me with this paper.

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