1701354 Introduction to Topology – Nguyen 2-14-06 Class Handout CURVES IN THE PLANE I. Continuity for Vector Functions from to 2 : Definition: A mapping f : D 2 defined by f (t ) ( f1 (t ), f 2 (t )) in terms of its component functions f1 , f 2 is called a vector function. Example: Now that we know how to measure distance on 2 using the formula defined by (called a metric), it is straightforward to define continuity for vector functions. Write a definition for a function f : D 2 to be continuous at a point a D : Definition A2: Example: Let f : (0, 2) 2 be defined by f (t ) (3t ,1/ t ) : (a) Graph the image of f. (b) Prove that f is continuous at a 1 if and only if each of its component functions, f1 (t ) 3t and f 2 (t ) 1/ t , are continuous at a 1 by using arguments. The example above reveals the following general connection between vector functions and their components in terms of continuity: Theorem: A function f : D 2 is continuous at a D if and only if each of its component functions f1 , f 2 : D where f (t ) ( f1 (t ), f 2 (t )) are continuous at a D . Sequential Continuity Just as before, continuity can also be described in terms of sequences: Theorem: A function f : D 2 is continuous at a D if and only if for every sequence {xn } that converges to a, we have f ( xn ) f (a) . II. Curves Definition: A plane curve is a continuous map :[a, b] 2 . The image of is denoted by *. The initial and end points are given by (a) and (b) , respectively. If (a) (b) , then the curve is said to be closed. Space-Filling Curves 1 1701354 Introduction to Topology – Nguyen 2-14-06 It is naïve to assume that the continuous image of a one-dimensional domain must be onedimensional as the following theorem demonstrates otherwise. Theorem: (Peano) There exist a curve space-filling curve :[0,1] the entire unit square: * [0,1] [0,1] {( x, y) 2 : x, y [0,1]} 2 such that its image is This space-filling curve is called the Hilbert curve. Its construction depends first on defining special subintervals and sub-squares. Part A. Construction of Hilbert Subintervals and Sub-squares Consider the sequence of collections of subsets constructed in stages: Stage 1: Division of [0,1] into 4 closed subintervals (arranged in increasing order): {I 0 , I1 , I 2 , I3} {[0,1/ 4],[1/ 4,1/ 2],[1/ 2,3/ 4],[3/ 4,1]} Division of [0,1] [0,1] into 4 closed sub-squares (arranged counter-clockwise): {S0 , S1 , S2 , S3} {[0,1/ 2] [0,1/ 2],[1/ 2,1] [0,1/ 2], [1/ 2,1] [1/ 2,1],[0,1/ 2] [1/ 2,1]} Stage 2: Division of [0,1] into 16 closed subintervals (based on Stage 1): {I 00 , I 01 , I 02 , I 03 ,..., I 30 , I 31 , I 32 , I 33} Division of [0,1] [0,1] into 16 closed sub-squares (based on Stage 1): {S00 , S01 , S02 , S03 ,..., S30 , S31 , S32 , S33} Stage n: Division of [0,1] into 4 n closed subintervals (based on Stage n-1): {I i1i2 ...in }, i1i2 ...in {0,1, 2,3} Division of [0,1] [0,1] into 4 n closed sub-squares (based on Stage n-1): {Si1i2 ...in }, i1i2 ...in {0,1, 2,3} Nested Property of Hilbert Subintervals and Sub-squares I i1 I i1i2 I i1i2i3 ... Si1 Si1i2 Si1i2i3 ... Base Decimal Representation of Real Numbers in [0,1] Base 10: Let {d n } be a sequence of digits such that dn {0,1, 2,...,9} . We define the base 10 decimal expansion of a real number x having digits d n as 2 1701354 Introduction to Topology – Nguyen 2-14-06 def x 0.d1d 2 d3 ... 0.d1 0.0d 2 0.00d3 ... d1 0.1 d 2 0.01 d3 0.001 ... n d d d1 d 2 2 33 ... lim ii n 10 10 10 i 1 10 Base 4: Let {d n } be such that dn {0,1, 2,3} . Then we define the base 4 decimal expansion of x with digits d n as def x 0.d1d 2 d3 ... n d d1 d 2 d3 2 3 ... lim ii n 4 4 4 i 1 4 Example: (a) Find a base 4 decimal expansion of the following values of x: 0, 1, and 1/2. Is the decimal expansion of a given number unique? (b) Prove that if t [0,1] , then t belongs to infinitely many Hilbert subintervals, i.e. there exists a sequence of digits {an } with an {0,1, 2,3} and such that t I a1a2 ...an for all n. Moreover, prove that t has base 4 decimal expansion whose digits are an , i.e. t 0.a1a2 a3 ... . Part B. Construction of Map (to be explained on the next handout) 3