Assignment #10 Morgan Schreffler 1 of 7 Lee, Chapter 4 Exercise 10 Let S be the square I I with the order topology generated by the dictionary topology. (a) Show that S has the least uppper bound property. (b) Show that S is connected. (c) Show that S is locally connected but not locally path-connected. Proof. (a) Let A S. For each a A, (0, 0) a (1, 1), so A is bounded (above). Now, for each t I, if (I {t}) A, then (I {t}) A has a least upper bound α t. Similarly, for each t I, if ({t} I) A then ({t} I) A has a least upper bound β t. Now, {α t } t I and {β t } t I are bounded subsets of R, so they have least upper bounds α and β, respectively. It follows that for each (a 1, a 2 ) A, a 1 α and a 2 β, so (a 1, a 2 ) (α, β), making (α, beta) an upper bound of A. Now, suppose (γ, δ) < (α, beta) is an upper bound of A. Then for all (a 1, a 2 ) A, a 1 γ < α or γ = α and a 2 δ < β. But, since α is the least upper bound of the x-coordinates of A and β is the least upper bound of the y-coordinates of A, both of these statements are contradictions. As a result, (γ, δ) (α, β) for all upper bounds (γ, δ) of A, making (α, β) the least upper bound of A by definition. (b) Suppose S is not connected. Then there exist disjoint, nonempty open subsets H and K such that H K = S. Without loss of generality, assume (1, 1) K. Now, K contains some neighborhood of (1, 1), so (α, β) = sup(h) 1. Since H K = S and H and K are both open, (α, β) is an element of some neighborhood which is a subset of either H or K. But, any neighborhood of (α, β) contains a set of the form {s S r < s < t} for some r, t S, which intersects both H and K. This is a contradiction, so S is connected. (c) For any (s 1, s 2 ) S, the closed neighborhoods {(t 1, t 2 ) S (s 1 1, s n 2 1 ) (t n 1, t 2 ) (s 1 + 1, s n 2 + 1 )} n n N of s are homeomorphic to either I or S, each of which is connected. Consequently, each neighborhood of s contains a connected neighborhood of s, making S locally connected. 1
Assignment #10 Morgan Schreffler 2 of 7 Exercise 11 Let X be a topological space, and let C(X) be the cone on X. (a) Show that C(X) is path-connected. (b) Show that C(X) is locally (path)-connected iff X is. Proof. (a) Let {C α } α A be the path-components of X. Notice that each C(C α ) is a pathconnected subspace of C(X). Now, let ξ be the tip of the cone, and let x, y C(X). If x, y C α, then the proof is trivial, so assume x C(C α ) and y C(C β ), where α β. Then there exists a path γ : I C(C α ) such that γ(0) = x and γ(1) = ξ, and a path δ : I C(C β ) such that δ(0) = ξ and δ(1) = y. If we concatenate these paths, we get a path ɛ : I C(X), where ɛ(t) = γ(2t) when 0 t 1 2, and ɛ(t) = δ(2t 1) when 1 2 t 1. Hence, C(X) is path-connected. (b) First we will prove that C(X) is locally connected iff X is. ( ) 2
Assignment #10 Morgan Schreffler 3 of 7 Exercise 13 Let T be the topologist s sine curve. (a) Show that T is connected but not path-connected or locally connected. (b) Determine the components and the path-components of T. Proof. (a) Suppose T is disconnected. Then there exist disjoint, nonempty open subsets H and K such that H K = T. Notice that T 0 and T + are both connected and path-connected, so there is no disjoint union of open sets equal to either piece of T. Thus, without loss of generality, suppose T 0 H. But any open set which contains T 0 also contains an open subset H T+ of T + such that H T+ K = T +. This is a contradiction, so T is connected. However, suppose a path γ : I T exists such that γ(0) = ( 1, 0) and γ(1) = (0, 0). Now, consider 2π the point τ = inf{t I γ(t) T 0 }. It follows that the image γ([0, τ]) has at most one element of T 0, but notice that T 0 γ([0, τ]), so γ([0, τ]) γ([0, τ]) and thus γ([0, τ]) is not closed. But γ([0, τ]) is the continuous image of a compact set of R, which is closed in R. This is a contradiction, so no such path can exist and T is not path-connected. Finally, T is not locally connected, because the intersection of T and the ball of radius 1 2 connected neighborhood. contains no (b) Since T is connected, T has only one component - namely, T itself. Now, notice in the discussion above that the non-existence of the path γ did not depend on the exact location of the point in T + ; it only depended on the fact that ( 1 2 π) was in T + and not in T 0. Now, T 0 and T + are path-connected and, in fact, they are the path-components of T. 3
Assignment #10 Morgan Schreffler 4 of 7 Exercise 15 Suppose G is a topological group. (a) Show that every open subgroup of G is also closed. (b) For any neighborhood U of 1, show that the group U generated by U is open and closed in G. (c) For any connected subset U G containing 1, show that U is connected. (d) Show that if G is connected, then every connected neighborhood of 1 generates G. Proof. (a) Suppose H G is an open subgroup, and consider G H. This is equal to g G H (gh). Further, since H is open and left translation is a homeomorphism, each gh is open, so G H is open. Therefore, H is closed. (b) The group U is equal to g G (gu), which is an open subgroup, so by (a), U is open and closed in G. 4
Assignment #10 Morgan Schreffler 5 of 7 Willard, Section 23 Problem A Exercise 1 Prove that the looped line is metrizable. Proof. By exercise 14A(4), the looped line is T 3.5, so it is T 1 and (completely) regular. Further, B = {(x q, x + q) x Q {0}, q Q} {(, n) ( q, q) (n, ) q Q, n N} is a countable basis for the looped line, so it is second countable. Finally, by Urysohn s Metrization Theorem, the slotted line is metrizable. Exercise 2 Prove that the scattered line is not metrizable. Proof. Clearly the scattered line is T 1, but it is not second countable, since the irrational numbers with the discrete metric have no countable basis. Hence, by the Urysohn Metrization Theorem, the scattered line is not metrizable. Exercise 3 Prove that the disjoint union of metrizable spaces is metrizable. Proof. Let {X α } α A be a collection of metrizable spaces. Construct metrics ρ α bounded above by 1 for each space X α. Now, define a metric ρ on α A X α as follows: ρ(x, y) = ρ α (x, y) if x, y X α and ρ(x, y) = 1 if x X α, y X β, α β. The metric ρ generates the topology of α A X α, so it is metrizable. Exercise 4 Let A be an infinite set, and let X be the hedgehog space of spininess A. Does the metric ρ(x, y) = x a + a y, x I α, y I β, α β ρ(x, y) = x y, x, y I α, where a is the common point of each I α, generate the topology of X? Proof. An open subset of X is a union of open subsets {U β } β B, where each U β is open in some Iα. Thus, each U β is of the form [a, x), (a, b), or (b, 1]. Each of these sets can be expressed as an open set in the set X with the metric ρ, so ρ generates the topology on X. 5
Assignment #10 Morgan Schreffler 6 of 7 Problem C Prove that, for a locally compact space X, the following are equivalent: (a) X is separable. (b) X = n=1 K n, where K n is compact and K n (K n+1 ). (c) The one-point compactification X of X is metrizable. Proof. 6
Assignment #10 Morgan Schreffler 7 of 7 Willard, Section 27 Problem B Exercise 1 Prove that the continuous image of a path-connected space is path-connected. Proof. Let f : X Y be continuous, and let X be path-connected. Without loss of generality, we can assume that f is surjective. Then for distinct y 1, y 2 Y, there are points x 1, x 2 X such that f(x 1 ) = y 1 and f(x 2 ) = y 2. Since X is path-connected, there is a path γ : I X such that γ(0) = x 1 and γ(1) = x 2. Now, because γ is continuous, the composition f γ : I Y is also continuous. Further, f(γ(0)) = f(x 1 ) = y 1 and f(γ(1)) = f(x 2 ) = y 2. Hence, f γ is a path from y 1 to y 2, and therefore Y is path-connected. Exercise 2 Prove that the nonempty product of finitely many spaces is path-connected iff each factor space is connected. Proof. ( ) Suppose X 1, X 2,..., X n are topological spaces, and X = n i=1 X i is pathconnected. Let i, 1 i n be arbitrary, and consider distinct points x i, y i X i. Then x = (x 1, x 2,..., x i,..., x n ) and y = (x 1, x 2,..., y i,..., x n ) are distinct points in X for arbitrary x j X j, j i, so there is a path γ : I X such that γ(0) = x and γ(1) = y. Since π i is continuous, so too is π i γ : I X i. Further, π i (γ(0)) = x i and π i (γ(1)) = y i, making π i γ a path from x i to y i. Because i, x i, and y i were arbitrary, it follows that each X i is path-connected. ( ) Suppose each X 1, X 2,..., X n is path-connected. Let x = (x 1, x 2,..., x n ) and y = (y 1, y 2,..., y n ) be distinct points in X. Now, {a 1 } {a 2 }... X i... {a n } is homeomorphic to X i for each i and for each a = (a 1, a 2,..., a n ) X, and path-connectedness is preserved by homeomorphism. In short, there exist paths γ i : I ({y 1 } {y 2 }... {y i 1 } X i {x i+1 }... {x n }) such that γ(0) = (y 1, y 2,..., y i 1, x i, x i+1,..., x n ) and γ(1) = (y 1, y 2,..., y i 1, y i, x i+1,..., x n ). Because X is a finite product, we can concatenate the γ i s into a well-defined, continuous path γ : I X from x to y, proving that X is path-connected. 7