A Brief Survey of Hyperspaces and a Construction of a Whitney Map Except where reference is made to the work of others, the work described in this thesis is my own or was done in collaboration with my advisory committee. This thesis does not include proprietary or classi ed information. Sean O?Neill Certi cate of Approval: Stewart L. Baldwin Professor Mathematics and Statistics Michel Smith, Chair Professor Mathematics and Statistics Geraldo de Souza Professor Mathematics and Statistics Gary Gruenhage Professor Mathematics and Statistics George T. Flowers Dean Graduate School A Brief Survey of Hyperspaces and a Construction of a Whitney Map Sean O?Neill A Thesis Submitted to the Graduate Faculty of Auburn University in Partial Ful llment of the Requirements for the Degree of Master of Science Auburn, Alabama August 10, 2009 A Brief Survey of Hyperspaces and a Construction of a Whitney Map Sean O?Neill Permission is granted to Auburn University to make copies of this thesis at its discretion, upon the request of individuals or institutions and at their expense. The author reserves all publication rights. Signature of Author Date of Graduation iii Vita Sean Andrew O?Neill was born. After being born, he grew and grew and grew, until one day, he stopped growing. Along the way, he befriended a green iguana. He has a mother, a father, a brother and a girlfriend. iv Thesis Abstract A Brief Survey of Hyperspaces and a Construction of a Whitney Map Sean O?Neill Master of Science, August 10, 2009 (B.S., Ohio University, 2006) 31 Typed Pages Directed by Michel Smith This paper is a brief survey of hyperspaces of topological spaces. In particular, the hyperspace of all nonempty compact subsets of a space and the hyperspace of all nonempty subcontinua of a space with the Vietoris topology. An example of a Whitney map on the hyperspace of any metric space is then constructed. v Acknowledgments Many are those whom I would like to acknowledge in this section. In fact, it would be more prudent to try to list those which I would not like to acknowledge. However, I am a traditionalist and as such will adhere to the custom of naming those to be recognized. To start, my family must be thanked. My mother and father, Mary DeStefano and Timothy O?Neill, and my brother, Padraic O?Neill, for all of their love, support, and time wasted putting up with and entertaining me. I would also like to thank all of my extended family, Aunts, Uncles, Cousins. I would especially like to thank my Uncle Larkin for convincing me that the empty set is indeed in the power set of any set. Many of the professors at Ohio University are deserving of a hearty thanks. Some of those are Dr. Todd Eisworth, Dr. Kaufman, and Dr. Arhangel?skii. I would also like the thank the faculty and sta in the Department of Mathematics and Statistics here at Auburn. Especially my comittee members and my advisor, Dr. Smith, for any ideas they have contributed to this thesis and all their other e orts in helping me achieve my degree. Further, any students of mine here at Auburn should also consider themselves thanked. The employees of Little Italy Pizzeria. 301D My friends, past, present, and future, must all must be thanked for participating in and/or putting up with my antics. Finally, Christi Morrow is very deserving of acknowledgment. Thank you for putting up with the distance and being loving and understanding these recent years. Please note that this is only a partial list. Add yourself if you feel that you are deserving. I have left you space: vi Style manual or journal used Journal of Approximation Theory (together with the style known as \aums"). Bibliograpy follows van Leunen?s A Handbook for Scholars. Computer software used The document preparation package TEX (speci cally LATEX) together with the departmental style- le aums.sty. vii Table of Contents 1 Preliminaries 1 2 Basic Properties of Hyperspaces 6 3 More Properties of Hyperspaces 12 4 Construction of a Whitney Map 17 Bibliography 23 viii Chapter 1 Preliminaries De nition 1.1 A topological space, (X;T), is a set X together with collection,T, of subsets of X such that: i. ;;X2T, ii. If U;V 2T then U\V 2T, iii. If U T then [U2T. The elements of T are called the open sets. De nition 1.2 A collection of subsets, B, of X is a base for a topology, T, on X if: i. For every x2X there is some B2B such that x2B, ii. If x2B1\B2 where B1;B2 2B, then there is some B3 2B such that x2B3 and B3 B1\B2. The base B is said to generate the topology on X. De nition 1.3 C X is closed if XnC is open. De nition 1.4 x2X is a limit point of Y X if every open U with x2U has U\Y 6=;. De nition 1.5 The closure of M X, denoted M, is the set M together with its limit points. De nition 1.6 A topological space is Hausdor if for every pair of distinct points there exists a pair of disjoint open sets, each containing one point respectively. 1 De nition 1.7 A topological space is regular if for every point and closed subset not containing that point, there exist disjoint open sets such that one contains the point and the other contains the closed subset. De nition 1.8 A topological space is normal if for every pair of disjoint closed subsets there exist disjoint open sets such that each contain one of the sets. De nition 1.9 A function d : X X!R is called a metric if: i. d(x;y) 0 for every x;y2X and d(x;y) = 0 if and only if x = y, ii. d(x;y) = d(y;x) for every x;y2X, iii. (Triangle Inequality) For every x;y;z2X, d(x;z) +d(z;y) d(x;y). De nition 1.10 The -ball about x2X, denoted b(x; ) is the set of all points y2Y such that d(x;y) < . De nition 1.11 A topological space X is said to be metrizable with metric d if the set of -balls generated by d form a base for the topology. X is also said to be a metric space. De nition 1.12 An open cover of a topological space X is a collection, mathcalU, of open subsets of X such that X [U. A subset of U that contains X in its union is called a subcover of U. De nition 1.13 A topological space X is compact if every open cover has a nite subcover. De nition 1.14 Let X be a topological space. Two subsets H and K of X are called mutually separated if neither set contains a point or a limit point of the other. 2 De nition 1.15 A topological space X is connected if it is not the union of two non-empty mutually separated subsets. De nition 1.16 A topological space X is a continuum if X is Hausdor and both connected and compact. De nition 1.17 A topological space X is a metric continuum if it is a continuum and metrizable. De nition 1.18 If X and Y are topological spaces, a function f : X!Y is continuous if for every open U Y and x2X with f(x)2U, there exists an open V X with x2V such that f(V) U. Equivalently, for every open U Y, f 1(U) is open in X. Also, for metric spaces in particular, for every x2X and >o there exists > 0 such that for every y2X with d(x;y) < , f(x) and f(y) are within of each other in Y. De nition 1.19 A function f : X!Y is open if for every open V X, f(V) is open in the image of X. De nition 1.20 A function f : X!Y is a homeomorphic embedding of X into Y if f is one-to-one, continuous and open. X is said to be homeomorphic to its image in Y. If f is onto then X and Y are said to be homeomorphic. Most of the following basic theorems may be found in one or more of [1], [2], and [3]. The proofs of these theorems are omitted, but may be found in [1]. Theorem 1.21 A closed subset of a compact space is compact. Theorem 1.22 A compact subset of a Hausdor space is closed. 3 Theorem 1.23 If H and K are disjoint compact subsets of a Hausdor space X then there is a pair of disjoint open sets, each containing one. Theorem 1.24 The continuous image of a compact set is compact. Theorem 1.25 If f is a continuous one-to-one map from a compact space to a Hausdor space then f is a homeomorphic embedding. Theorem 1.26 (Tychono ) Any product of compact sets is compact. Theorem 1.27 Let B be a basis for a topological space X. Then every open set of X is a union of members of B. Theorem 1.28 The following are equivalent: i. f : X!Y is a continuous function from topological space X to topological space Y. ii. If O is a (basic) open set in Y, then f 1(O) is open in X. Theorem 1.29 If X is a compact Hausdor space, then X is regular. Theorem 1.30 If X is a compact Hausdor space, then X is normal. Theorem 1.31 If X is metrizable, then X is Hausdor , regular, and normal. Theorem 1.32 If X is regular, then X is Hausdor . Theorem 1.33 If X is normal, then X is regular. 4 Theorem 1.34 (Zorn?s Lemma) Let A be a nonempty partially ordered set. Then if every chain has an upper bound, A has a maximal element. Theorem 1.35 If X is a metric space and every sequence has a convergent subsequence, then X is compact. 5 Chapter 2 Basic Properties of Hyperspaces De nition 2.1 Suppose that X is a topological space. Then the hyperspace of X, denoted by 2X is the space of nonempty compact subsets of X together with the following types of sets forming a base for its topology. Suppose that fU1;U2;::;Ung=U is a nite collection of open subsets of X, then R(U) = R(U1;U2;:::Un) = fK 2 2X : K [ni=1Ui and for all 1 i n;K\Ui 6= ;g. The topology which this base generates is called the Vietoris topology. Before moving on, let us indeed see that these sets form a base for a topology on 2X. Theorem 2.2 The sets of the form R(U) = fK 2 2X : K [ni=1Ui and for all 1 i n;K\Ui6=;gwhereU =fU1;U2;:::Ungis a nite collection of open sets of X form a base on 2X. Proof: First, observe that every K 2 2X is in R(X). Now let U = fU1;:::;Ung and V = fV1;:::;Vmg be nite collections of open subsets of X and suppose K 2 2X with K R(U)\R(V). From this we can see that K [[ni=1Ui]\[[mj=1Vj]. For every i n and j m let Oi;j = Ui\Vj. De ne O to be those Oi;j that have K\Oi;j6=;. It is clear that R(O) is a member of the collection of subsets of 2X in the hypothesis. To see that K 2R(O), rst note that K\O6= ; for all O2O by de nition. Next, K [O for if x2K then for some i n and j m, x2Ui and x2Vj, hence x2K\Oi;j which implies that Oi;j2O, and so x2Oi;j [O. Finally, we must see that R(O) R(U)\R(V). First show R(O) R(U). Let H 2R(O) then H [O [U since each O2O is a subset of some V 2V. Now, if i n then since K\Ui6=;there exists j m so that K\Ui\Vj6=; 6 and so Oi;j2O. Since H2R(O), ;6= H\Oi;j H\Ui, and so H2R(U). By a similar argument, H2R(V) and so R(O) R(U)\R(V). Theorem 2.3 If X is Hausdor then 2X is Hausdor . Proof: Suppose H;K 2 2X such that H 6= K. Without loss of generality, assume there exists x2HnK. Then there exist disjoint open U;V X with x2U and K V. Then H2R(U;X) and K2R(V). If L2R(U;X) then L\U6=; and L*V, hence L =2R(V). If L2R(V) then L V implying that L\U =;, hence L =2R(U;X). From this it can be concluded that R(U;X)\R(V) =;, thus 2X is Hausdor . Example 2.4 2I contains a Hilbert Cube. It can be shown that X = Q1i=1[ 12i; 12i 1] is embeddable in 2I. One might ask, \How?" Well this is how: de ne f : X!2I by f(x) =fxig1i=1[f0g where x = (xi)1i=1 2X. It is clear that f(x) I for each x2X and is closed since 02f(x) and 0 being its only limit point. Hence f(x) is compact for every x2X and is in 2I. It is also clear that f is one-to-one. To see that f is continuous, let U = fU1;U2;:::;Ung and suppose that f(x) 2R(U). Assume 0 2U1, hence there exists an N such that if i N then [ 12i; 12i 1] U1. For each i < N de ne Ui = fU 2U : xi 2Ug and Xi = [ 12i; 12i 1]\[\U]. Now, if for some 1 < j n, Uj =2Ui for any i < N, then since f(x) 2R(U) there is some i such that xi 2Uj since the x0is converge to 0, so in this case de ne Xi = [ 12i; 12i 1]\Uj. For all other i?s de ne Xi = [ 12i; 12i 1]. So, Q1i=1Xi X is open and if y 2 Q1i=1Xi, f(y) 2 R(U) based on the construction, hence f is continuous. Since X is compact and 2I is Hausdor , f is a homeomorphic embedding. De nition 2.5 Let C(X) denote the subspace of 2X consisting of all nonempty subcontinua in X. 7 Example 2.6 C(I), where I is the unit interval. Note that every nonempty subcontinuum of I is of the form [x;y] where x y. So, it makes sense that perhaps C(I) might be homeomorphic to X =f(x;y)2I I : x yg I I, and hence be homeomorphic to I I. To see this, de ne f : X!C(I) by f((x;y)) = [x;y] for each (x;y)2I I. It is clear that f is one-to-one and onto. To see that f is continuous, let R(U) C(I) be a basic open set and [x;y] 2R(U). Let Ux \fU 2U : x2Ug and Uy \fU2U : y2Ug, be connected open intervals in I each containing x or y respectively. Further, assume the upper boundary of Ux is less than some element in [x;y]\U and the lower boundary of Uy is greater than the same element in [x;y]\U for every U2U. Now, (x;y)2Ux Uy I I which is open. Let (w;z)2Ux Uy. If t2[w;z] then at least one of the following is true: t2 [w;x], t2 [x;y], t2 [y;z], all of which are covered by U and so [w;z] [U. Next, for every U 2U such that either x or y2U then U\[w;z]6=;. If U2U contains neither x nor y, then by the construction of Ux and Uy, there is some t2U such that w < t < z, i.e. t2 [w;z] hence [w;z]\U 6= ; and [w;z] 2R(U). From this we can conclude that f is continuous and thus a homeomorphism. Example 2.7 C(S1) If each proper subcontinuum is identi ed with its length and midpoint and the entire circle as a point at the tip it is not too di cult to see that C(S1) is a cone, and thus homeomorphic to I I. Theorem 2.8 C(X) is a closed subset of 2X. Proof: We shall see this by showing that 2X rC(X) is open. Let K 2 2X rC(X), then since K is not connected, there exist U;V X open and disjoint such that K U[V and 8 K has nonempty intersection with each. It is clear that K2R(U;V) and that R(U;V) 2X rC(X), hence 2X rC(X) is open, and C(X) is closed. Theorem 2.9 If X is a metric space then 2X is also a metric space. Proof: Suppose X is a metric space with bounded metric d. De ne D on 2X by D(H;K) = maxfsupx2Hfd(x;K)g;supx2Kfd(x;H)ggfor every H;K22X. Hence forth, B will denote a ball in 2X and b will denote a ball in the underlying space X. First let us see that D de nes a metric on 2X. If H = K, d(x;K) d(x;x) = 0 for every x2H and d(x;H) d(x;x) = 0 for every x2K, so D(H;K) = 0. Since each is compact and hence closed, the implication reverses, so D(H;K) = 0 if and only if H = K. It is also clear that symmetry holds from the de nition of D. Now, suppose H;K;G22X, and assume D(H;K) = 1 and D(K;G) = 2. Since D(H;K) = 1, d(h;K) 1 for every h 2 H. So for every h 2 H there exists kh 2 K with d(h;kh) 1. Similarly, there exists gh 2 G with d(kh;gh) 2. So for every h 2 H there exists gh 2 G such that d(h;gh) 1 + 2 and so d(h;G) 1 + 2 for every h2H. Similarly d(g;H) 1 + 2 for every g2G and so D(H;G) 1 + 2. Hence D is a metric. Now to see that D generates the topology on 2X, let U 2X be open in the metric topology, K 2 U and > 0 such that B (K) U. Then fb =3(k) : k 2 Kg is an open cover of K in X. By compactness of K, there exist k1;k2;:::;kn 2 K such that fb =3(k1);b =3(k2);:::;b =3(kn)g=V cover K. Clearly K2R(V) = R which is a basic open set in the topology on 2X. Now, let H2R then if k2K, k2b =3(ki) for some i n. Since b =3(k1)\H6=; let h2b =3(k1)\H then d(k;h) d(k;ki) + d(ki;h) < =3 + =3 = 2 =3 and so supk2Kfd(k;H)g 2 =3 < . Also, since H [ni=0b =3(ki), suph2Hfd(h;K)g< hence D(H;K) < . So, H2B (K) U which means U is open in the Vietoris topology. Next let U 2X be open in the Vietoris topology and let K 2U. Then there is a nite collection,U, of open subsets of X so that K2R(U) U. For every k2K let k > 0 such that b k(k) U for every U 2U containing k. Since K is compact let k1;k2;:::;kn 2K 9 such that fb k i=2 (k) : i ng cover K and without loss of generality, assume that for every V 2U that for some i n, b k i=2 (ki) V. Let = minf ki4 : i ng. If D(H;K) < then for every i n there is an h2H such that d(h;ki) < < ki2 and so V \H 6= ; for every V 2U. Also, for every h2H there exists a k2K such that d(h;k) < . Since k2b k i=2 (ki) for some i n, d(h;ki) d(h;k) +d(k;ki) < + ki2 < i, which implies that H [ni=0b ki(ki) [U, hence H2R(U) U. So U is open with respect to the metric D, and so D generates the topology on 2X. Note that this also makes C(X) a metric space if X is. Example 2.10 The following metrics are equivalent to that de ned in Theorem 2.9: Sup- pose that X is a metric space with a bounded metric d . If H;K22X de ne D1(H;K) = inff jH [x2Kb (x) and K [x2Hb (x)g; and D2(H;K) = supf j there is a point p 2H so that B (p)\K = ; or there is a point p2 K so that B (p)\H =;g: Not only do these metrics generate the same topology on 2X as D, but in fact for each pair H;K22X, these metrics produce the exact same value as D. Suppose D(H;K) = . Without loss of generality, assume there is some h2H such that d(h;K) = . Then for every k2K, h =2b (k) hence, D1(H;K) = D(H;K). Next, suppose D1(H;K) = . Without loss of generality, assume for every < , H * [x2Kb (x). Then for each < there is some h 2 H such that d(h;K) and so b (h)\K =;. Hence D2(H;K) for each < , so D2(H;K) = D1(H;K). 10 Now, suppose D2(H;K) = . Without loss, assume that for every < , there is some h 2 H such that b (h)\K = ;. Then for each < , there is some h 2 H such that d(h;K) , and so D(H;K) for every < , hence D(H;K) = D2(H;K). Theorem 2.11 If F1 is the subset of 2X consisting of all the singleton sets, then F1 is homeomorphic to X. Proof: De ne f : X ! 2X by f(x) = fxg for every x2X. f is clearly one to one and f(X) = F1. Now, if U X is open, f(U) = R(U)\F1 which is open in f(X), hence f is open. Also, if R(U) is a basic open set in 2X then R(U)\F1 = ffxg : x 2 U for every U 2Ug and so f 1(R(U)) = \U which is open X. So f is continuous and thus an embedding of X. Corollary 2.12 If 2X is metrizable, then so is X. Proof: Follows from Theorem 2.11. Note that this also holds if 2X is replaced by C(X) since F1 C(X). 11 Chapter 3 More Properties of Hyperspaces De nition 3.1 Suppose thatfMigi2N is a sequence of nonempty sets. Let M be the set to which the point p belongs if and only if every open set containing p intersects in nitely many sets of the sequence fMigi2N, then M is called the limiting set of the sequence fMigi2N. De nition 3.2 The set M is called the sequential limiting set of the sequence of fMigi2N if and only if it is the limiting set of every in nite subsequence of fMigi2N. Lemma 3.3 If M is the limiting set of a sequence fMigi2N in 2X then M is closed Proof: Let x2M and U X be open with x2U. Then, there is some m2M such that m2U, hence U intersects in nitely many members of fMigi2N and x2M. Thus, M is closed. Theorem 3.4 Let X be a compact space and suppose that M is an element of 2X. Then the sequence of elements fMigi2N of 2X converges to M in the Vietoris topology on 2X if and only if M is the sequential limiting set of the sequencefMigi2N of sets in the topology of X. Proof: Suppose that fMigi2N converges to M in the topology of 2X, and let fMikgk2N be a subsequence of fMigi2N with limiting set M0. If p2M and U X is open with p2U then R(U;X) 2X and M 2R(U;X). By hypothesis, fMikgk2N converges to M in 2X, hence there exists a J 2N such that if j J Mij 2R(U;X), in particular Mij \U 6= ; for every j J. Hence p2M0 and so M M0. Now suppose that p2M0nM. Since 12 M is compact there are disjoint open U1;U2 X such that p2U1 and M U2. Clearly M 2R(U2) 2X which is open, hence there exists a J 2N such that for every j J, Mij 2R(U2), in particular Mij U2 for every j J. From this, U1\Mij = ; for every j J and p2M0, hence M0 M thus M0 = M and so M is the sequential limiting set of fMigi2N. Conversely, assume M is the sequential limiting set of fMigi2N in the topology of X. It is also clear the M contains all of it?s limit points, hence M is closed, and since X is compact, M is compact and so M 22X. Now let R(U) 2X be open with M 2R(U). Note that if p2M and U X is open with p2U then there is some N 2N such that if n N then Un\Mn6=; since M is the sequential limiting set of the sequence. So, since for every U2U, U\M6=; there is some NU 2N such that if n NU then Mn\U6=;. Further, there exists an N0 2N such that if n N0 then Mn [U. If not, then there exists an in nite subsequence fMikgk2N of fMigi2N with Mik 6 [U for every k. For every k pick xk2Mik n[U. Then fxikgk2N is a sequence in X, and by compactness has a limit point or point that repeats in nitely many times, x2X, not in[U. It is clear that x would then be in the limiting set of fMikgk2N contradicting M being the sequential limiting set of our sequence. Let N = maxffNU : U2Ug[fN0gg. Then, if n N, Mn2R(U) and so fMigi2N converges to M in 2X. Lemma 3.5 If X is a compact space and fMigi2N is a sequence in 2X, then the limiting set M of fMigi2N is nonempty. Proof: Choose xi2Mi for every i2N. If one xi is repeated in nitely many times, then it is in M and we are done. Suppose this is not the case. Then fxigi2N is in nite and has a limit point x2X. Then x2M, in particular M6=;. Lemma 3.6 If X is a compact space and fMigi2N is a sequence in 2X with limiting set M. Then if no proper subset of M is the limiting set for some subsequence of fMigi2N, fMigi2N converges to M in 2X. 13 Proof: Let fMikgk2N be a subsequence of fMigi2N with limiting set M0. Then M06=; and it is clear that M0 M. By hypothesis the containment cannot be proper so M0 = M, hence M is the sequential limiting set of fMigi2N. By Theorem, fMigi2N converges to M. Lemma 3.7 Let X be a compact metric space and fMigi2N be a sequence in X. Then M = fM : M is the limiting set for some subsequence of fMigi2Ng ordered by reverse inclusion is a partially ordered set and contains a maximal element, i.e. there exists a subsequence of fMigi2N with limiting set satisfying the hypothesis of the preceding lemma and hence is convergent. Proof: Let fMigi2N be a sequence in X and M = fM : M is the limiting set for some subsequence offMigi2Ng. Then reverse inclusion is re exive, anti-symmetric and transitive, hence (M; ) is a partially ordered set. Let C M be a chain. Then C = \C 6= ; and for every n 2 N there is some Cn 2C such that Cn [x2Cb(x; 1n) = Bn. Then C \n2NCn \n2NBn = C, i.e. \n2NCn = C. For every n2N let Kn N such that Mn =fMkgk2Kn is a subsequence of fMigi2N with limiting set Cn. Now, for every n2N since Cn Bn and Bn open, there exists Nn 2N such that if k2Kn with k Nn and Mk2Mn then Mk Bn. Choose Mk1 2M1 such that k1 N1. For n> 1 pick Mkn 2Mn such that kn Nn. Let C0 be the limiting set offMkngn2N. Note that Bn 1 Bn for each n2N so if x62Bn then x2XnBn+1 an open set missing Mkj for every j n+ 1 and so C0 \n2NBn = C. So C02M with C0 N for every N 2C, i.e. C0 is an upper bound for C. By Zorn?s lemma, there exists a maximal M2M with associated subsequence MM satisfying the hypothesis of the previous lemma. Theorem 3.8 If X is a compact metric space then 2X is compact. 14 Proof: Follows from the preceding lemmas and since sequential compactness is equivalent to compactness in metric spaces. Note that since C(X) is closed in 2X, it is also compact. De nition 3.9 If n is an integer then Fn denotes the set of all elements of 2X that have at most n points. Theorem 3.10 Fn 2X is closed for every n. Proof: We shall show that 2XnFn is open in 2X. Let K22XnFn, then K has at least n+1 distinct points, x1;:::;xn+1. Since X is Hausdorf, there exist disjoint open U1;::;Un+1 2X each containing the respective points in K. Then K2R(U1;::;Un+1;X) and it is clear that ifH2R(U1;::;Un+1;X), thenH has at leastn+1 points and soR(U1;::;Un+1;X) 2XnFn, i.e. 2XnFn is open. Corollary 3.11 If 2X is compact, so is X. Proof: X is homeomorphic to F1. Note that this also holds if 2X is replaced by C(X). De nition 3.12 If n is an integer then Kn denotes the set of all elements of 2X that have at most n components. Theorem 3.13 Kn 2X is closed for every n. 15 Proof: We shall again show that 2X nKn is open in 2X. Let H 2 2X nKn, then K has at least n + 1 di erent components, H1;:::;Hn+1, each closed as a subset of H and hence compact. Since X is Hausdor and Hi is compact for every i n + 1, there exist disjoint open U1;::;Un+1 2X each containing the respective components of H. Then H 2 R(U1;::;Un+1;X) and it is clear that if J 2 R(U1;::;Un+1;X), then J has at least n+ 1 components and so R(U1;::;Un+1;X) 2XnKn, i.e. 2XnKn is open. Theorem 3.14 The set F =[1n=1Fn is dense (F ) in 2X. Proof: Let R 2X be open. Then there exists a nite collectionU of open subsets of X such that R(U) R. Now, for each U 2U let xU 2U. Then Y = fxU : U 2Ug2R(U) R, and Y 2FjUj F, thus F is dense in 2X. Theorem 3.15 If W C(X) is a continuum then [fHjH2Wg is a continuum. proof: Let U be an open cover of W0 =[fHjH2Wg. Then for every H2W there exists a nite UH U covering H and so that each member of UH has nonempty intersection with H. So, U2X = fR(UH) : H 2W is an open cover of W in 2X. Since W is compact, there exist H1;H2;::;Hn so that W [ni=1R(UHi). Clearly, [ni=1UHi is a subset of U that is nite and covers W0, hence W0 is compact. To see that W0 is connected, assume by way of contradiction that it is not. So, let W0 = A[B where A and B are non-empty and mutually separated. Notice that each A and B are closed, and hence, compact and so there exist disjoint open U;V X such that A U and B V. Since each H 2W is connected, either H U or H V and so W R(U)[R(V) each open and having non-empty intersection with W, a contradiction. Hence, W0 is connected. 16 Chapter 4 Construction of a Whitney Map Lemma 4.1 Let : Y !X be continuous. Then : 2Y !2X, de ned by (K) = (K) =f (k) : k2Kg for each K22Y , is continuous. Proof: For each K 2 2Y , (K) = (K) is compact and hence in 2X. Let V 2X be open, (K) 2 V and (K) 2 B(U1;:::;U2) V. Since (K) = (K) [ni=1Ui, K [ni=1 1(Ui), each of which is open in Y. Also, since for each i n, Ui\ (K) = Ui\ (K) 6= ;, we have 1(Ui)\K6= ;, i.e. K2B( 1(U1);::; 1(Un)). Further, it is clear that if H2B( 1(U1);::; 1(Un)) then (H)2B(U1;:::;U2). Lemma 4.2 If K X is compact, then 2K is homeomorphic to fH22X : H Kg. Proof: Let be the identity map. Then is clearly one to one, onto and continuous by Lemma 4.1. Since K is compact, 2K is compact and so is an embedding. In the following, let (X;d) be a metric space and assume d 1 De nition 4.3 For every n 2 de ne wn : Fn ! [0;1) by wn(K) = 0 if jKj < n or wn(K) = minfd(x;y) : x;y2K and x6= yg if jKj= n. Lemma 4.4 If H;K2Fn and D(H;K) < 2 then jwn(H) wn(K)j< 17 Proof: Case 1: jHj;jKj 0. Suppose H 2 2X with D(H;K) < 2. Let K0 K such that K0 2 Fn and wn(K0) = wn(K). Since D(H;K) < 2, for each k 2 K0 there exists h2H such that d(h;k) < 2. Let H0 be the set of those elements in H. Then H02Fn and D(H0;K0) < 2, so by preceding lemma, wn(K) = wn(K0) 0 and let K 2 2X be in nite. Then since K is compact in X there is a nite cover, U, of K by balls of diameter less than . Then if H K is nite andjHj= n>jUj= N, there exist distinct h;h02H such that h;h0 are in the same member of U and so wn(H) 0 so that b(k; )\H = ;. By preceding lemma, wn(H) < for all n su ciently large. Note that wn(H) > 0 for every n since H is in nite, this coupled 21 with the preceding lemma give us that for some m;m 1 2N, wm(H) < wm 1(H) < . Let H0 H be in Fm 1 such that wm 1(H0) = wm 1(H). Then K0 = H0[fkg K is in Fm. Also, since d(k;h) for each h2H0, wm(H)