TOPOLOGY HW 4 CLAY SHONKWILER 24.1 (a) Show that no two of the spaces (0, 1), (0, 1], and [0, 1] are homeomor- phic. Proof. Suppose (0, 1) and (0, 1] are homeomorphic, with the homeomor- −1 phism given by f. If A = (0, 1) − {f (1), then f|A : A → (0, 1) is a homeomorphism, by Theorem 18.2(d). However, the interval (0, 1) is con- nected, whereas A = (0, f −1(1)) ∪ (f −1(1), 1) is not connected, so A and (0, 1) cannot be homeomorphic. From this con- tradiction, then, we conclude that (0, 1) and (0, 1] are not homeomorphic. Similarly, suppose g : (0, 1] → [0, 1] is a homeomorphism. Then, if B = −1 −1 (0, 1] − {g (0), g (1)}, g|B : B → (0, 1) is a homeomorphism. However, g−1(0) 6= g−1(1), so at most one of these can be 1, meaning one must lie in the interval (0, 1). Suppose, without loss of generality, that g−1(0) ∈ (0, 1). Then B = (0, g−1(0)) ∪ (g−1(0), 1] − {g−1(1) is not connected, whereas (0, 1) is, so the two cannot be homeomorphic. From this contradiction, then, we conclude that (0, 1] and [0, 1] are not homeomorphic. A similar argument easily demonstrates that (0, 1) and [0, 1] are not home- omorphic, so we see that no two of the spaces (0, 1), (0, 1], and [0, 1] are homeomorphic. (b) Suppose that there exist imbeddings f : X → Y and g : Y → X. Show by means of an example that X and Y need not be homeomorphic. Example: Let f : (0, 1) → [0, 1] be the canonical imbedding and let g : [0, 1] → (0, 1) such that x 1 g(x) = + . 3 3 1 2 0 Then g([0, 1]) = [ 3 , 3 ]. Obtain g by restricting the range of g to g([0, 1]) = 1 2 0 [ 3 , 3 ]. We claim that g is a homeomorphism. Since multiplication and addi- tion are continuous, as are the inclusion map and compositions of continuous functions, we see that g0 is continuous, as is g0−1, where 1 g0−1(x) = 3 x − . 3 1 2 CLAY SHONKWILER Both of these maps are also bijective, so we see that g0 is indeed a homeo- morphism, meaning g is an imbedding. However, as we saw in part (a) above, (0, 1) and [0, 1] are not homeomor- phic, so two spaces need not be homeomorphic for each to be imbedded in the other. ♣ n (c) Show R and R are not homeomorphic if n > 1. Lemma 0.1. If f : X → Y is a homeomorphism and X is path-connected, then Y is path-connected. Proof. Let x, y ∈ Y . Then there exists a continuous path g :[a, b] → X such that g(a) = f −1(x) and g(b) = f −1(b). Define h := f ◦ g. Then h :[a, b] → Y is continuous and h(a) = (f ◦ g)(a) = f(g(a)) = f(f −1(x)) = x and h(b) = (f ◦ g)(b) = f(g(b)) = f(f −1(y)) = y since f is bijective. Hence, h is a path from x to y, so, since our choice of x and y was arbitrary, Y is path connected. n Proposition 0.2. R and R are not homeomorphic if n > 1. n Proof. Suppose f : R → R is a homeomorphism. Then, restricting the n domain to R −{0} gives a homeomorphism of the punctured euclidean space to R − {f(0)}. However, the punctured euclidean space is path-connected (as shown in Example 4), whereas R − {f(0)} is not even connected, let alone path-connected. To see this, we need only note that R − {f(0)} = (−∞, f(0)) ∪ (f(0), ∞) so the open sets (−∞, f(0)) and (f(0), ∞) give a separation of this space. Hence, by the above lemma, the punctured euclidean space and R − {f(0)} n are not homeomorphic, a contradiction. Therefore, we conclude that R and R are not homeomorphic. 1. 24.8 (a) Is a product of path-connected spaces necessarily path-connected? Answer: Yes. Suppose X and Y are path-connected. Let x1 × y1, x2 × y2 ∈ X×Y . Now, we know that X×y1 is homeomorphic to X and, therefore, is path-connected. Hence, there exists a continuous map f : [0, 1] → X × y1 such that f(0) = x1 × y1 f(1) = x2 × y1. Also, x2 × Y is homeomorphic to Y and is, therefore, path connected, so there exists a continuous map g : [0, 1] → x2 × Y such that g(0) = x2 × y1 g(1) = x2 × y2. TOPOLOGY HW 4 3 Now, define x 1 f 2 0 ≤ x ≤ 2 h(x) = x 1 1 g 2 + 2 2 ≤ x ≤ 1. By the pasting lemma, then, h is continuous. Furthermore, 0 h(0) = f = f(0) = x × y 2 1 1 and 1 1 h(1) = g + = g(1) = x × y . 2 2 2 2 Hence, h is a path from x1 × y1 to x2 × y2. Since our choice of x1 × y1 and x2 × y2 was arbitrary, we see that X × Y is path-connected. ♣ (b) If A ⊂ X and A is path-connected, is A necessarily path connected? Answer: No. In Example 7, we saw that, if A = {x×sin(1/x)|0 < x ≤ 1}, then A, the topologist’s sine curve, is not path connected. To see that A is path-connected, let s, y ∈ A. Then x = a × sin(1/a) for some a ∈ (0, 1] and y = b × sin(1/b) for some b ∈ (0, 1]. Define the map f :[a, b] → A by f(z) = z × sin(1/z). Then f is continuous since its coordinate functions are continuous and f(a) = a × sin(1/a) = x and f(b) = b × sin(1/b) = y, so f is a path from x to y. Since our choice of x and y was arbitrary, we see that A is path-connected. ♣ (c) If f : X → Y is continuous and X is path-connected, is f(X) neces- sarily path connected? Answer: Yes. Let f be continuous and let y1, y2 ∈ f(X). Let x1 ∈ −1 −1 f (y1) and x2 ∈ f (y2). Then, since X is path-connected, there exists a continuous map g :[a, b] → X such that g(a) = x1 and g(b) = x2. Define h := f ◦ g. Then h(a) = (f ◦ g)(a) = f(g(a)) = f(x1) = y1 and h(b) = (f ◦ g)(b) = f(g(b)) = f(x2) = y2. Furthermore, h is continuous, since f and g are, so h is a path from y1 o y2. Since our choice of y1 and y2 was arbitrary, we conclude that f(X) is path-connected. ♣ T (d) If {Aα} is a collection of path-connected subspaces of X and if Aα 6= S ∅, is Aα necessarily path-connected? S T Answer: Yes. Let x, y ∈ Xα and let z ∈ Aα. Then x ∈ Aβ and y ∈ Aγ for some β and γ. Furthermore, z ∈ Aβ, z ∈ Aγ. Since Aβ is path connected, there exists a path f from x to z. Since Aγ is path connected, 4 CLAY SHONKWILER there exists a path g from z to y. Using the pasting lemma, we can glue these two paths together to make a path h from x to y (much as we did in part (a) above). ♣ 25.1 What are the components and path components of R`? What are the continuous maps f : R → R`? Answer: Each component and path component of R` consists of a single point. To see this, suppose not. Then there exists a component containing two points, x and y. This means there exists a connected subspace A ⊂ R` containing x and y. However, x ∈ A ∩ (−∞, y) and y ∈ A ∩ [y, ∞), both of these subsets are open in A, and A is equal to their union, so A can be separated by A ∩ (−∞, y) and A ∩ [y, ∞), a contradiction. Hence, we conclude that each component (and, therefore, path component) of R` is a point. Furthermore, since the continuous image of a connected space is con- nected, we can conclude that the continuous maps f : R → R` are just the constant maps. This is because R is connected, so it’s continuous image in R` must be connected. The only connected subspaces of R` are single points, so such a continuous map must map all of R to a single point. ♣ 26.1 (a) Let τ and τ 0 be two topologies on the set X; suppose that τ 0 ⊃ τ. What does compactness of X under one of these topologies imply about compactness under the other? Answer: If X is compact under τ 0, then it must be compact under τ. To see this, suppose A is an open cover of X in τ. Then A is also an open cover of X in τ 0, since every open set in τ is open in τ 0. Since X is compact under τ 0, A contains a finite subcover of X. Since our choice of A was arbitrary, we conclude that X is compact under τ. 0 On the other hand, if X = R, τ is the trivial topology and τ is the discrete topology, then τ 0 ⊃ τ and X is compact under τ, but not under 0 τ . To see this last, we merely construct the open cover {{x}|x ∈ R}, which certainly contains no finite subcover. ♣ (b) Show that if X is compact Hausdorff under both τ and τ 0, then either τ and τ 0 are equal or they are not comparable.