Bucket handles and Solenoids Notes by Carl Eberhart, March 2004

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1 Bucket handles and Solenoids Notes by Carl Eberhart, March Introduction A continuum is a nonempty, compact, connected metric space. A nonempty compact connected subspace of a continuum X is called a subcontinuum of X. A useful theorem about subcontinua, hich every first year topology student should prove is: 1.1. Theorem. The intersection of a toer of subcontinua of X is a subcontinuum of X. A continuum is indecomposable if it cannot be ritten as the union of to proper subcontinua. The composant of a point x in a continuum X is the union of all proper subcontinua of X hich contain x. Here is a nice theorem. 1.. Theorem. The composants of a continuum are dense. The composants of an indecomposable continuum are pairise disjoint. A classical example by contstructed by B. Knaster in the early 190 s is still of interest The Bucket handle continuum. Let B 0 be the set of all closed semicircles in the upper half plane centered on c 0 =.5 hose diameters have enpoints in the Cantor middle third set. Let (B 0 ) denote the reflection of B 0 about the x-axis. For 1 n, let B n = c n c 0 + B 0 3 n, here c n =.5 3 n. Then K = B 0 B 1 B is an indecomposable continuum. The union of the semicircles hose endpoints are endpoints of the Cantor set is called the visible composant. In the figure belo, a portion of the visible composant is shon. It is a union of a toer of arcs all ith (0,0) as one endpoint. (An arc is a continuum hich is homeomorphic ith the unit interval I = [0,1].)

2 It is possible to see the bucket handle as an intersection of a toer of -cells, each one obtained from the preceeding one by digging a canal out of it. Why is K indecomposable? Well, if you can convince yourself that K (1) has only arcs for proper subcontinua, and () has no arcs ith interior (in K ), then it is pretty easy to argue that K is indecomposable, for otherise it ould be the union of to arcs A and B and A - B ould be a nonempty subset of the interior of K. On the other hand there is another ay to construct indecomposable continua hich makes it possible to prove indecomposability rather easily.. The inverse limit construction The inverse limit X = lim (X i, f i ) of a sequence (X i ) 1 of continua and surjective maps f i : X i+1 X i is defined as the intersection of the subsets Q n = {(x i ) 1 such that x i = f i (x i+1 ) for all i = 1,, n} of the product Π i=1 X i. The spaces X i are called the factor spaces of X ; the maps f i are called the bonding maps of the inverse limit. For each positive integer i, the map π i : X X i given by π i ((x i ) 1 ) = x i is called the i th projection map. It is continuous since it is the restriction of the projection ρ i : Π n=1 X n X i to X..1. Theorem. The inverse limit X of continua is a continuum. Further if A is a subcontinuum of X then A = lim (A i, g i ), here A i = π i (A) and g i = f i Ai+1. Proof. X = n=1 Q n is a continuum by 1.1, since for each n, Q n is homeomorphic ith Π i=n. Further, (a i) 1 A if and only for each i, a i A i and g i+1 (a i+1 ) = f i+1 (a i+1 ) = a i. One easy ay to decide if to inverse limits are homeomorphic is described in the next theorem... Theorem. To inverse limits X = lim (X i, f i ) and Y = lim (Y i, g i ) are homeomorphic if there is a sequence of homeomorphisms h i : X i Y i such that h i f i (x) = g i h i+1 (x) for each i and all x X i+1. Proof. Define a function h : X Y by h ((x i ) i=1 ) = (h i(x i )) i=1 for each positive integer i. h is into Y because g i h i+1 (x i+1 ) = h i f i (x i+1 ) = h i (x i ), for each i. h is continuous since π i h = h i π i is continuous for each i. Call h the map induced by the commutative diagram. f 1 f X 1 X Xi X h 1 h h i h g 1 g Y 1 Y Yi f i g i Y In the same ay, using the inverse homeomorphisms h 1 i : Y i X i, e can define an induced map from Y to X and sho that it is the inverse of h. Thus a map induced by homeomorphisms is a homeomorphis.

3 Here is the theorem describing the indecomposability of a continuum constructed as an inverse limit. A map f : X Y is called indecomposable provided henever X = A B, then f(a) = Y or f(b) = Y. So a continuum is indecomposable if its identity map is indecomposable..3. Theorem. X is indecomposable provided each of its bonding maps is indecomposable. Proof. Suppose X is the union of to proper subcontinua A and B. Then there is an n 1 so that π n1 (A) X n1, otherise by.1, X = A. Note also that for all n n 1, π n (A) X n. In the same ay, there is an n so that π n (B) X n. So if e let n be the maximum of n 1 and n, then π n (A) and π n (B) are proper subcontinua of X n. Hoever, π n+1 (X ) = X n+1 = π n+1 (A) π n+1 (B), f n (π n+1 (A)) = π n (A), and f n (π n+1 (B)) = π n (B), a contradiction to the assumed property of the bonding maps f i. Another classical indecomposable continuum, studied somehat later [] than the bucket handle is the dyadic solenoid S = lim (X i, f i ), here each factor space X i is the unit circle S 1 and each bonding map f i is the squaring map z..4. Corollary. S is indecomposable. Proof. The squaring map is indecomposable, so the theorem follos from.3. Note that any inverse limit of circles here the bonding maps are poer maps z i, i is indecomposable by the same argument. Such continua are referred to as solenoids. Solenoids are the only indecomposable continua hich admit a group mulitiplication[]. Another class of indecomposable continua are the Knaster continua, hich are defined as the continua obtained as inverse limits by using the unit interval I as the factor space and standard maps n, n > 1, defined by { nx i if i is even and 0 i n (x) = n x i + n 1 1, i + 1 nx if i is odd and 0 < n i x i + n 1 1 Clearly, the map n : I I is indecomposable hen n > 1, and so one has the corollary..5. Corollary. Any Knaster continuum is indecomposable..6. Exercise. Sho that the dyadic solenoid has the property that each proper subcontinuum is an arc, and has a basis of neighborhoods consisting of sets homeomorphic ith the Cantor set crossed ith an arc An embedding theorem A homeomorphism of a space X onto a subspace of a space Y is called an embedding of X into Y Theorem. (A version of the Anderson-Choquet embedding theorem) Suppose the factor spaces X i of X are all embedded in a common continuum X (ith metric d, say) and the bonding maps f i satisfy the conditions: (1) There is a positive number K so that

4 4 for each i and each x X i+1, d(x, f i (x)) < K i, and () for each factor space X i and each positive δ, there is a positive δ so that if k > i and p, q X k ith d(f ik (p), f ik (q)) > δ then d(p, q) > δ. Then for each x = (x i ) i X, the sequnce of coordinates x i converges to a unique point h(x) X and the function h : X X is an embedding. Proof. Condition (1) suffices to guarantee that the sequence of coordinates of a point in X is Cauchy, and so converges. So the function h is ell defined. To see that h is continuous, let ɛ > 0 be given. Choose N so large that K i=n i < ɛ/4 and and hence by (1) d(x N, h(x)) d(x N, x N+1 ) + d(x N+1, h(x)) K/ N + d(x N+1, h(x)) < ɛ/4 for all x X. Hence if x X and U is an open set in X N about x N of diameter less than ɛ/4. Then π 1 N (U) is a basic open set in X. Further, if y = (y i ) i π 1 N (U), then y N U, and so d(h(x), h(y) d(h(x), x N ) + d(x N, y N ) + d(y N, h(y)) < ɛ. This shos h is continuous. To see that h is 1-1, suppose x y here x and y are in X. Since x y, there is an i 0 so that x i0 y i0. Let δ = d(x i0, y i0 )/, and use () to get a δ \prime > 0 so that for k > i 0 and p, q X k, if d(f i0 k(p), f i0 k(q)) > δ, then d(p, q) > δ. Hence for k > i 0, d(x k, y k ) > δ, and so d(h(x), h(y)) δ. This shos h is an embedding. 3.. Theorem. K is homeomorphic ith an inverse limit of arcs ith indecomposable bonding maps. Proof. To apply 3.1 to K, for each positive integer i, let a i = ( i 6, i 6 ), the bottom of the (i + 1) st set of semicircles B i counting from the right, and let A i be the subarc of the visible composant ith endpoints a 0 and a i. These arcs are the factor spaces. The bonding map r i : A i+1 A i is the retraction hich projects a point on the first quarter circle starting at a i+1 horizontally to the right onto the matching point on the end quartercircle of A i, and otherise projects a point radially onto the nearest point of A i, except near a i+1. There the last quarter circle of A i+1 is mapped homemorphically onto the circular arc from a 0 to (.5 + cos((1 (1/) i )π),.5 + sin((1 (1/) i )π)), and the circular arc from (1 (1/) i )π to π on the last semicircle of A i+1 is mapped homeomorphically onto the circular arc from (.5 + cos((1 (1/) i )π),.5 + sin((1 (1/) i )π)) to (.5 + cos((1 (1/) i )π),.5 + \sin((1 (1/) i )π)). It is easy to see that r i is indecomposable. Note that d(x, r 1 (x)) < 1/3, d(x, r (x)) < 1/3, and in general d(x, r i (x)) < 1/3 i, so condition (1) of.3 is satisfied ith K = 1. So the function h : X K is ell-defined and continuous. Since K = i=1 A i, the continuity of h guarantees that h is onto. We verify that h is 1-1 directly. Suppose h(x) = h(y). Then for some n, x n and y n lie in the same circular arc of A n. Then from then on the coordinates x n+i and y n+i lie along the outer third of the radii dran from the center of that circular arc. But since h(x) = h(y), for some m > n, x i and y i lie in the same radius for all i > m. But then x i = r i (x i+1 ) = r i (y i+1 ) = y i for all i > m. So x = y, and h is 1-1. So e have realized K as the inverse limit lim (A i, r i ) of arcs ith indecomposable bonding maps. No e ant to prove the 3.3. Theorem. K is homeomorphic ith the Knaster continuum lim (I, ).

5 5 Proof. Let g 1 be any homeomorphism from A 1 to I such that g 1 (a 0 ) = 0 and g 1 (a 1 ) = 1. Then define g : A I by g 1 (x) g (x) = if x A 1 1 g 1(r 1 (x)) if x A A 1 The function g is a homeomorphism such that g 1 r 1 = g. Using the same method, e successively define homeomorphisms g i : A i I so that g i 1 r i 1 = g i. This sequence induces a homeomorphism g : lim (A i, r i ) lim (I, ). r A 1 r 1 A Ai r i lim (A i, r i ) g 1 g g i g I I I lim (I, ) 4. Where is (x i ) i=1? It is an interesting exercise to ask here a given point in lim (I, ) goes under the homeomorphism h(h ) 1 : lim (I, ) K constructed previously. For example, it is easy to see hat point goes to a 0 = (0, 0): the point (0, 0, ). Where does the point x = (1, 1/, 1/4, )go? First find its image y = (h ) 1 (x) in lim (A i, r i ), then push that on to h(y) in K. By the ay h as constructed, e can compute that y = (a 1, a 1, a 1, ). Then h(y) = a 1. Every inverse limit ith one factor space and one bonding map has a shift homeomorphism, s, defined by s((x 1, x, x 3, )) = (x, x 3, ). So the bucket handle K has a shift homeomorphism, s = h(h ) 1 sh h 1 : K K. Where does the shift homeomorphism take the point a 0? It is pretty easy to see that this point is fixed under the shift. What about s(a 1 )? First push a 1 to (1, 1/, 1/4, ), then shift it to (1/, 1/4, ), then pull it back to (h 1 1 (1/), h 1 1 (1/), h 1 1 (1/), ) in lim (A i, r i ) then push to h 1 1 (1/) under h. Note that the shift homeomorphism on K depends on the homemorphism h 1 of I to A 1. Hoever, the fixed point a 0 of s does not depend on h Exercise: There is another fixed point of the shift s defined above. Where does it come from in lim (I, ) and here is it at in K? Does its location in K depend on h 1? 5. Mapping the dyadic solenoid lightly onto the bucket handle David Bellamy [1] as the first to see this, I think. First you have to think of the group S 1 as the quotient space I/{0 = 1} of I ith the operation of addition mod { 1. In this setting, the squaring map on S 1 becomes the doubling t if 0 t 1/ mod 1 map on I. dbl(t) = t 1 if 1/ t 1

6 6 Note that this is continuous on I/{0 = 1}, and commutes ith. Hence, there is an induced map of the solenoid onto the bucket handle. dbl dbl I/{0 = 1} I/{0 = 1} S I I lim (I, ) 5.1. Exercise. Use the diagram above to sho that S /{x = x} = K. 6. The hyperspaces of subcontinua of K and S If X is any continuum ith metric d say, and A and B are subcontinua of X, the Hausdorff distance beteen A and B, H d (A, B) is defined by H d (A, B) = inf{\epsilon each point of either set is ithin ɛ of some point of the other.} The set of all subcontinua of X is denoted by C(X) Theorem. H d is a metric on C(X), and C(X) topologized by this metric is a continuum. C(X), topologized ith the Hausdorff metric, is called the hyperspace of subcontinua of X 6.. Theorem. C(S ) is homeomorphic ith the cone over S. C(K ) is homeomorphic ith the cone over K. References [1] David Bellamy, A tree-like continuum ithout the fixed point property, Houston Math. J. 6 (1979), [] Anthonie van Heemert, Topologische Gruppen und unzerlegbar Kontinua, Compositio Math. 5 (1937), [3] Sam B. Nadler, Jr. Continuum Theory. Marcel Dekker, 199. [4] K. Kuratoski, Topology, vol. 1, Amsterdam, 1968.