Topological dimensions, Hausdor dimensions & fractals Yuval Kohavi, Hadar Davdovich May 2006 1 Abstract We present some basic facts about topological dimension, the motiva- tion, necessary denitions and their interrelations. Finally we discuss the Hausdor dimension and fractals. 1 Topological dimension 1.1 Motivation Topological dimensions denes the basic dierence between related topological sets such as In and Im when n 6= m. The lack of that denition is especially highlighted because of the easy explanation of the geometric dimension. When trying to dene a dimension for a topological space, you might run into several diculties - Unlike a vector space, you can not state that the dimension is the maximum of linear independent vector. Therefore in order to get an intuitive denition for a topological dimension we should look for dierent properties that has the eects of the geometrical dimension. 1.2 General A topological dimension has values in ¡1; 0; 1; 2; 3;::: , and is topological, i.e: if X and Y are homeomorphic then they have the same "dimension". It will also be nice, if Rn gets value n, for each n. There are 3 commonly used denitions: 1. Small inductive dimension (ind) 2. Large inductive dimension (Ind) 3. Lebesgue covering dimension (dim) All these denitions have the required properties, and we will see that they are the same for separable metrizable spaces (ind = Ind = dim). Remainder: metrizable space is a topological space that is homeomorphic to a metric space. 2 2 The inductive dimension 2.1 Intuition Lets look at a cube. What is the (geometrical) dimension of the cube? 3, and what is the (geometrical) dimension of its boundary [square]? 2. Now let's look at the square. What is the dimension of its boundary? well its boundaries are lines, so their dimension is 1. And what about the boundary of the lines? Well their boundary is composed of dots, there dimension is 0. We get a pattern - the dimension of a spaces, is 1 + the dimension of its boundary. As a boundary is well dened in topology, this notion can be easily applied to topological spaces, to create a topological dimension. 3 The small inductive dimension Denition 3.1. The small inductive dimension of X is notated ind(X), and is dened as follows: 1. We say that the dimension of a space X (ind(X)) is -1 i X = ; 2. ind(X) ·n if for every point x 2 X and for every open set U exists an open V, x 2 V such that V ⊆ U, and ind(@V ) · n ¡ 1. Where @V is the boundary of V . 3. ind(X) = n if (2) is true for n, but false for n ¡ 1. 4. ind(X) = 1if for every n, ind(X) · n is false. Remark 3.1. The ind dimension is indeed a topological dimension: X is home- omorphic to Y implies that ind(X) = ind(Y ). This can be shown easily (and inductively) as the denition relays only of open\closed sets. Remark 3.2. An equivalent condition to condition (2) is: ² The space has a base B, and every U 2 B has ind(@U) · n ¡ 1 [You can construct this base using the sets U from condition (2)]. Theorem 3.1. As we would expect, if Y ⊆ X then ind(Y ) · ind(X). Proof. By induction, it is true for ind(X) = ¡1. if ind(X) = n, for every point y 2 X there is an nbd V , with an open set U ⊆ V , such that ind(@V ) · n ¡ 1. Note that VY = Y \ V is a nbd in Y , and UY is open in Y . By induction, it is enough to show that @UY ⊆ @U, Because then ind(@UY ) · n ¡ 1, and therefore ind(Y ) · n = ind(X) (by denition). And indeed Uy ⊆ U; @UY = UY n UY ⊆ U n U = @U Example 3.1. Let's show that ind(R) is 1. for each x 2 R, lets select a nbd, V and a set U = (a; b) ⊆ V , ind(@U) = ind(fa; bg) = 0. This implies that ind(R) · 1. 3 So, it is enough to prove that ind(R) is not 0. But that is easy! if ind(R) = 0, that means that a set U exists, such that: ind(@U) = ¡1 , @U = ;, U is clopen! but if U is clopen, R is disconnected! Because R is connected, we get that ind(R) > 0. So nally, ind(R) = 1 ¦. The same proof can be done for the Sphere S1 and the interval I1. Corollary 3.2. A zero-dimensional space is disconnected. Corollary 3.3. if for every nbd V of x 2 X exists a clopen set U ⊆ V then ind(X) = 0 Example 3.2. What is the dimension of C the cantor set? Lets look at C in R, C has no interior points (fact). That means, that for each nbd U = (a; b) of x 2 C, we can nd two points c; d such that a · c · x · d · b, and c; d2 = C. Lets look at C as a subspace, V = (c; d) is open, and closed ) ind(C) = 0 Note that these results are not exactly what we expect. The cantor set has the same 'size' as R: 2@0 , so we should expect it to have the same dimension. But then again, the cantor set has no interval in it. Its dimension should be then, somewhere between 0 and 1. We shall later learn of a dimension denition that suits our expectations. It is worth to mention, that the cantor set is universal 0-dimensional space of separable metrizable spaces. meaning that every other space separable metric space , is homeomorphic to a sub-space of the cantor set. X; ind(X) = 0 P Remember that a number is in the Cantor set It has the form xi , i2N 3i ; xi 2 f0; 2g. for example: 2 0 2 0 . 3 + 9 + 27 + 81 + ¢ ¢ ¢ We can display this number simply as a fraction of radix 3 0:2020:::. This set of fractions has one-to-one and onto mapping to the set of binary fractions 0:1010:::. We simply replaced all the 2's with 1's. The latter set of fractions is more familiar to topology, it is the space:D@0 ;D = f0; 1g, D with the discrete topology. Proposition 3.4. The D@0 is homeomorphic to the Cantor set. Now it is enough to show that every zero dimensional separable metrizable space is homeomorphic to a sub-spaces of D@0 . Remark 3.3. Every zero dimensional spaces has a clopen base (by the alternative denition). If the space is also metrizable and separable, it has a countable clopen base (because then every base has a countable base contained in it). Theorem 3.5. The Cantor set is a universal space for all the zero-dimensional metrizable separable space. Proof. Let X be a separable metrizable space. It is enough to show that X is homeomorphic to a sub-set of D@0 . Let be its clopen countable base. We dene to be: ½ B = fBig f fi(x) = 1 x 2 B i . Dene f(x) = (f (x); f (x);:::). f is the homeomorphism 0 otherwise 1 2 that we wanted. ¦ 4 An interesting question we can ask about this dimension, is wether ind(Rn) = n. The answer is yes,we will start by showing that ind(Rn) · n. Proposition 3.6. The dimension of Rn;Sn;In · n. We can show this by induction. we have already shown that the dimension of R1;S1;I1is 1. For every point x 2 Rn;Sn;In, there is a nbd U with an open set V which is homeomorphic to Sn¡1, and by induction the proof is complete ¦ In order to show that ind(Rn) ¸ n, We shall have some denitions and theorems. Let's start by dening a partition between sets: Denition 3.2. A Partition L on between A and B exists, if there are open sets U; W , A ⊆ U; B ⊆ W , such that W \ U = ; and L = U c \ W c. Theorem 3.7. ind(X) · n ,For every point x and every closed set B there is a partition L, such that:ind(L) · n ¡ 1. Proof. ()) If x 2 X and B is closed, Then there is a set V , such that V \ B = ;. By denition of ind, there is a U ⊆ V , x 2 U, ind(@U) · n ¡ 1 now, let W = U and L = @U. (() Let x 2 X and V an nbd of x. The set B = V c is closed, and therefore there are U; W such that x 2 U, B ⊆ W . Note that by denition of B, U ⊆ V . Now W c is closed, therefore U ⊆ W c, and obviously @U ⊆ U ⊆ W c, and by denition @U = U n U_ = U n U ⊆ U c. Therefore @U ⊆ U c \ W c = L ind(U) · n ¡ 1 Theorem 3.8. If X is a metric space and Z is a zero dimensional separable subspace of X, then for all closed set A; B of XA \ B = ;, there is a partition between them L, such that L \ Z = ; Theorem 3.9. X is a separable metrizable space, then ind(X) · n , X is the union of two sub spaces Y; Z such that ind(Y ) · n ¡ 1; ind(Z) · 0. From this we can easily see, that for a separable metrizable space X ind(X) · n , X is the union of Z1;:::;Zn+1, ind(Zi) · 0.