Ohrimenko+ / Barnsley. Superfractals

R → to the picture on the left. Transformations of pictures are defined in Chapter 2. The transformation here is not a metric transformation and allows extreme stretching on very small scales. It is actually an example of a fractal transformation, as discussed in Chapter 4.

Figure 1.19 This represents an elliptical-shaped subspace X R2 before and after a homeomorphism is applied. Some open sets belonging to the natural topology are represented by the white and coloured regions. Of course there are vastly many more open sets, of endless diversity. The homeomorphism is of the form A ◦ M ◦ A −1, where A is projective and M is a Mobius¨ transformation (see Section 2.6).

We say that a property of a set S X is invariant under or is preserved by a transformation f : X → Y iff {the property is true of S iff it is true for f (S)}. For example, we have already mentioned that fractal dimension is invariant under any metric transformation. Homeomorphisms preserve topological properties, i.e. properties that can be defined in terms of being open and closed. Figures 1.18 and 1.19 show the result of applying certain homeomorphisms f : R2 → , where the points in a space have been assigned colours.

T h e o r e m 1.8.8 The metric spaces ( A, d ) and ( A, d|A|) have the same natural topology; that is, Td ( A) = Td|A| ( A).

P r o o f Let e : ( A, d ) → ( A, d|A|) denote the identity map, so that e(x) = x for all x A. Then we show that e is a homeomorphism. The continuity

where m is a positive integer and x, y A.

To prove that e−1 is continuous we use the ‘reverse’ inference,

Similar arguments show that ( A, d ) and ( A, d|A|) have the same natural topology, as do ( A A, d ) and ( A A, d|A|). We will refer to this topology as the natural topology T on any subset of = A A. Now that you know this, you should not fuss much about which metric we use, d or d|A|.

A map may be one-to-one, onto and continuous, but not open. For example, the map f : ([0, 1), deuclidean) → (S1, dshortest), where S1 is the circle of radius 1 centred at the origin in R2, defined by f (x) = (cos 2π x, sin 2π x) for all x [0, 1) is one- to-one, onto and continuous but not open. To see this, note that [0, 0.5) is an open subset of [0, 1) in the relative topology (see below) in say R1 or R2 but f ([0, 0.5)) is not an open subset of the circle.

Notice that if two metric spaces are equivalent then they are homeomorphic, but the converse is not true. For example f : ((0, 1], deuclidean) → ([1, ∞), deuclidean) defined by f (x) = 1/x is a homeomorphism but not a metric transformation.

1.9Important basic topologies

We have already introduced the natural topology associated with a metric space. But there are five other key topologies that are easy to build and that we will need for our discussion of fractals.

Discrete topology

Let X be a space. Then the discrete topology Tdiscrete on X is obtained by defining all the subsets of X to be open. It follows that all subsets of X are also closed.

The discrete topology is the natural topology associated with (X, d) for which d(x, y) = 1 whenever x, y X, with x =y, and of course d(x, x) = 0. This is the ultimate Hausdorff space! Every point x X lives in its own private open-and- closed set {x}, nicely separated from every other point. Such a space may seem very artificial, but we will soon use the discrete topology on an alphabet A to build a natural topology on A.

Relative topology

Let (X, T) be a topological space and let S X. Then we can convert S into a topological space, which we may denote by (S, T|S ) and which is a subspace of the original space, by defining any set O S to be open in the relative topology T|S iff it is the same as the intersection of some open set O T with S, i.e. O T|S O = O ∩ S for some O T. The relative topology never ceases to surprise me, because S X may be neither open nor closed in (X, T) yet it is both open and closed in (S, T|S ).

E x e r c i s e 1.9.1 Let (X, d) be a metric space, let Td (X) be the associated natural topology and let S X. Show that the natural topology on the metric space (S, d|S ) is the same as the relative topology T|S , i.e. that Td|S (S) = Td (X)|S .

E x e r c i s e 1.9.2 Verify that the discrete topology on a space X is same as the natural topology associated with the metric space (X, d) where d(x, y) = 1 whenever x, y X, with x =y, and d(x, x) = 0.

Topology generated by a basis

Let {Oi : i I} be a collection of subsets of a space X. Then the smallest topology T on X such that Oi T for all i I is called the topology generated by the basis {Oi : i I}. It can be proved (by you) that

that is, the open sets of T are exactly those that can be written as unions of members of the basis. Of course the sets in the basis, the individual Oi , are open in T.

I

It turns out to be very useful to have a countable basis for a topological space. In Chapter 5 we will describe certain ergodic properties associated with fractals, which are established by first showing that they hold for each member of a countable basis for Rn . So we note the following:

T h e o r e m 1.9.4 A countable basis for Rn is provided by the set of all open balls with rational radii and centres at rational points.

In the following exercise we introduce a useful collection of decompositions of a closed rectangle in R2. These will be used in Chapter 2 to describe ‘pixel functions’.

E x e r c i s e 1.9.5 Show that a countable basis for ( interior, deuclidean) is provided by the interiors of the set of squares

w,W,hH : w {1, 2, . . . , W }, h {1, 2, . . . , H }, W N, H N

where

Product topology

Let {(Xi , Ti )}∞n=1 be an infinite sequence of topological spaces. Let X = X1 × X2 × · · · denote the space whose points are sequences of the form x = {xn Xn : n = 1, 2, . . . }. Then the product topology for the space X is defined as the topology that is generated by sets of the form

O = O1 × O2 × · · ·

where On Tn for all n = 1, 2, . . . and for only finitely many values of n is it true that On = Xn . Similarly we define the product topology on the finite product space X = X1 × X2 × · · · × XN to be the topology generated by sets of the form O1 × O2 × · · · × ON , where now the only constraint is that On Tn for all n =

1, 2, . . . , N .

The case that interests us is where Xn = A for all n = 1, 2, . . . and Tn = Tdiscrete(A) is the discrete topology on the alphabet A. In this case we note that

A∞ := A × A× · · · = A.

In general, if X is a space then we write X∞ to denote the product space X × X × · · · . Thus we obtain, in a very simple way, the product topology Tproduct( A) on code space. We have the following observations, which we leave as either an exercise or an act of faith for the reader.

T h e o r e m 1.9.6 The product topology on code space is the same as the natural topology associated with the metric d (and with the metric d|A|). That is,

( A, Tproduct( A)) = A, Td|A| = A, Td .

Henceforth we refer to the natural topology on code space A as the topology on code space, and it should be assumed that this topology is the one meant when no other assertion is made. It turns out that there is a wonderful basis for the product topology on code space. To describe it we need the following definition.

D e f i n i t i o n 1.9.7 A cylinder set of the code space A is a subset of A that can be written in the form

C(σ ) := {ω A : ωn = σn for all n = 1, 2, . . . , |σ |},

for some σ A.

We will also refer to C(σ ) as a cylinder subset of A. Notice that the set of cylinder subsets of A is addressed by the code space A.

E x e r c i s e 1.9.8 Show that for each m = 1, 2, . . .

%

A = {C(σ ) : σ A, |σ | = m}.

Cylinder sets are used in the construction of fractals. Indeed, this is one reason why we introduced A. We mention the following because of its relevance to the existence of certain invariant measures on fractals.

T h e o r e m 1.9.9 A countable basis for ( A, Tproduct) is the set of all cylinder sets {C(σ ) A : σ A}.

P r o o f A basis for ( A, Tproduct) is, from the definition of Tproduct = Tproduct( A), the set of sets that can be written in the form

O1 × O2 × · · · × ON × A × A × · · ·

for some finite integer N , where Ok can be written as {ak,1, ak,2, . . . , ak,nk } A with nk {1, 2, . . . , |A|} for k = 1, 2, . . . , N . ( Note that some of the Ok may be equal to A, and recall that |A| is finite.) It follows that every set in Tproduct( A) can be written as a union of sets of the form

%

{a1,k1 } × {a2,k2 } × · · · × {aN ,kN } × A × A × · · · .

k1=1,...,n1

k2=. 1,...,n2

.

.

kN =1,...,nN

i.e. it is a union of cylinder sets. So every set in Tproduct( A) can be written as a union of cylinder sets, which is obviously countable.

Identification topologies

Let (X, T) be a topological space, say a Hausdorff space. Let x1, x2 X, with x1 = x2. Define a new topology T on X as follows: remove from T all those sets that contain either x1 or x2 but not both x1 and x2; then T consists of the sets that remain. It is readily verified that T is a topology. But it is no longer a Hausdorff topology, for there is no open set that contains x1 but not x2. According to the topology T the set {x1, x2} behaves like a single point in the sense that whenever O T we have: x1 O {x1, x2} O.

E x a m p l e 1.9.10 Let X = {x1, x2, x3, x4} and Let T = { , X, {x1, x2, x3},

{x1, x2, x4}, {x1, x3, x4}, {x2, x3, x4}, {x1, x2}, {x1, x3}, {x1, x4}, {x2, x3}, {x2, x4},

{x3, x4}, {x1}, {x2}, {x3}, {x4}}. Then T = { , X, {x1, x2, x3}, {x1, x2}, {x3, x4},

{x3}, {x4}}. In this case we have started with the discrete topology on X and have ended up with a new topology T. It looks quite like the discrete topology on X = {{x1, x2}, x3, x4}. Notice how the topology T is coarser than T, that is,

T T.

D e f i n i t i o n 1.9.11 Let f : X → Y be a mapping from a topological space (X, T) to a space Y. Let T f (X) = T f :X→Y (X) be the topology on Y specified by: O X is open iff O T and f −1 f (O) = O. Then T f (X) is called the identification topology on X induced by f : X → Y.

Here we use the notation f −1 f to denote the mapping obtained by first applying f and then applying f −1. We might also have written f −1( f (O)) or f −1 ◦ f (O). We prove now that Definition 1.9.11 is a good one.

P r o o f Let T f (X) denote the set of subsets of X specified in the definition. We need to demonstrate that it is a topology for X.

(i) Since X T and f −1 f (X) = X it follows that X T f (X). Since T and f −1 f ( ) = it follows that T f (X).

(ii) Suppose that {Oα T f (X) : α I} is a collection of sets in T f (X). Then

(iii)Suppose that O1, O2 T f (X). Then O1, O2 T and so O1 ∩ O2 T. It remains to prove that f −1 f (O1 ∩ O2) = O1 ∩ O2. This follows at once

from Exercise 1.3.2(iv) provided that f (O1 ∩ O2) = f (O1) ∩ f (O2). But from Exercise 1.3.2(ii) we know that f (O1 ∩ O2) f (O1) ∩ f (O2). So the proof is complete if we can show that f (O1 ∩ O2) f (O1) ∩ f (O2). Suppose that y f (O1) ∩ f (O2); then there exists x1 O1 such that f (x1) =

E x e r c i s e 1.9.12 In Example 1.9.10 choose Y = {{x1, x2}, x3, x4} and define f : X → Y by f (x1) = {x1, x2}, f (x2) = {x1, x2}, f (x3) = x3, f (x4) = x4. Verify that the identification topology on X induced by f : X → Y is exactly T.

D e f i n i t i o n 1.9.13 Let f : X → Y be a mapping from a topological space (X, T) onto a space Y. Let T f (Y) (= T f :X→Y (Y)) be the topology on X specified by: O Y is open iff f −1(O) T. Then T f (Y) is called the identification topology on Y induced by f : X → Y.

The proof that T f (Y) is indeed a topology is similar but easier than the proof (above) that T f (X) is a topology, and we leave it to the reader.

T f (Y) = { , Y, {{x1, x2}, x3}, {{x1, x2}}, {x3, x4}, {x3}, {x4}}.

Identification topologies are rather simple in the case of finite sets of points, but they become decidedly interesting in the case of non-denumerable spaces. For example, let X = [0, 1] R and Y = S1, the circle in R2 of radius 1 centred at the origin, let T be the natural topology on R2 and let f : (X, T) → Y be defined by

Then the two points x1 = 0 and x2 = 1 in [0, 1] are mapped onto the single point P := (1, 0) R2. In the identification topology on S1 induced by the mapping f : [0, 1] → S1, the point P is an element of each of the many open sets that consist of arcs of the circle that contain P but do not terminate at P and do not contain the points that define their extent. Indeed, the identification topology on S1 induced by f is just the natural topology as a subset of R2. But the corresponding open sets in [0, 1], which contain both the points x1 and x2, are of the form [0, a) (b, 1] where a, b (0, 1). See Figure 1.20(i).

Figure 1.20 This figure illustrates some open sets (purple and green) and some sets that are not open (red) in various identification topologies. In (i) the identification topology is induced by a mapping f from the real closed interval [0, 1] to a circle; see Equation (1.9.1). Neither purple subinterval of [0, 1] is open on its own (in the induced topology) but their union is open. The single green subinterval is also open.

In (ii) the transformation is from the interval [0, 1] onto a sideways figure-eight in R2; see Equation (1.9.2). None of the three purple subintervals is open, nor any pairwise union of them, but the union of all three is open. The image of this union is the purple X-shaped segment of the sideways figure eight.

In (iii) is shown a model of the projective plane; it consists of a disk centred at the origin, with opposite points on its circular boundary identified (via an appropriate mapping from the disk onto itself minus half its circular boundary). The two purple regions comprise a single open set (a bucket) in the identification topology, but neither on its own is open. The red region represents a set that includes part of the circle but since none of its points expands across the opposite side it cannot represent an open set.

In (iv) each point in the side A D of the filled square A B C D is identified with the point ‘vertically below it’ in B C . Each point on A B is then identified with the opposite point (through the centre of the square) on D C . The purple car, which becomes inverted as it ‘drives through the barrier D C to emerge through A B ’, represents an open set, as does the purple girl and the green dog. The red region, however, does not represent an open set because although it touches B C the bottom of the flower-pot does not extend below A D .

A related example is obtained by changing the definition of f : [0, 1] → S1 from Equation (1.9.1) to

See Figure 1.20(ii).

E x e r c i s e 1.9.15 Let f : X → Y be a mapping from a topological space (X, T) to a space Y. Let C Y. Show that C is closed in the identification topology on Y induced by f : X → Y iff f −1(C) is closed in the topology T.

The following theorem tells us that f is almost but not quite a homeomorphism with respect to the identification topologies it induces because in general f is not one-to-one as a point map.

T h e o r e m 1.9.16 Let f : X → Y be a mapping from a topological space (X, T) to a space Y. Then, as a mapping from subsets of X to subsets of Y, restricted to T f (X), f is one-to-one from T f (X) onto T f (Y).

= f −1( f (B)). But A = f −1( f (A)) since A T f (X), and B = f −1( f (B)) since B T f (X). So A = B.

We will be particularly interested in identification topologies on code spaceA that are associated with mappings from A onto subspaces of Rn such as fractals. For example, a general theorem in Section 4.14 implies in particular that the natural topology on [0, 1] R is the identification topology induced by the continuous mapping f : ( A, T ) → [0, 1] defined as follows. Take A to be {0, 1, 2, . . . , N − 1} and set

This tells us that the real interval can indeed be thought of, from a topological point of view, as being code space ‘joined to itself’ at those pairs of points, namely addresses, , ω A, of the form

= σ1σ2 · · · σM−1σM 0 and ω = σ1σ2 · · · σM−1(σM − 1)(N − 1)