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TOPOLOGY HW 2 CLAY SHONKWILER 20.9 Show that the euclidean metric d on Rn is a metric, as follows: If x, y ∈ Rn and c ∈ R, define x + y = (x1 + y1 , . . . , xn + yn ), cx = (cx1 , . . . , cxn ), x · y = x1 y1 + . . . xn yn . (a) Show that x · (y + z) = (x · y) + (x · z). Proof. Let x, y, z ∈ Rn . Then x · (y + z) = x · (y1 + z1 , . . . , yn , zn ) = x1 (y1 + z1 ) + . . . xn (yn + zn ) = x1 y1 + x1 z1 + . . . xn yn + xn zn =x·y+x·z (b) Show that |x · y| ≤ ||x||||y||. Proof. Let x, y ∈ Rn . Then || y x ± || ≥ 0 ||x|| ||y|| . Then n || x y 2 X ± || = ||x|| ||y|| i=1 xi yi ± ||x|| ||y|| 2 = n X xi ||y|| ± yi ||x|| 2 i=1 ||x||||y|| Then, since ||x||||y|| ≥ 0, n X (xi ||y|| ± yi ||x||)2 ≥ 0. i=1 This in turn implies that P 2 ||y||2 + y 2 ||x||2 ± 2x y ||x||||y|| 0 ≤ ni=1 x i i i i P P P = ||y|| ni=1 x2i + ||x|| ni=1 yi2 ± 2||x||||y|| ni=1 xi yi = ||y||2 ||x||2 + ||x||2 ||y||2 ± 2||x||||y||(x · y) = 2||x||2 ||y||2 ± 2||x||||y||(x · y). In other words, ||x||2 ||y||2 ± ||x||||y||(x · y) ≥ 0, 1 ≥ 0. 2 CLAY SHONKWILER or ±(x · y) ≥ −||x||2 ||y||2 = −||x||||y||. ||x||||y|| Multiplying both sides by −1, we see that ∓(x · y) ≤ ||x||||y||, which means that |x · y| ≤ ||x||||y||. (c) Show that ||x + y|| ≤ ||x|| + ||y||. Proof. To see this, we note that, for any z ∈ Rn , ||z||2 = z · z. Hence, ||x + y||2 = (x + y) · (x + y) = (x + y) · x + (x + y) · y =x·x+y·x+x·y+y·y = ||x||2 + 2(x · y) + ||y||2 ≤ ||x||2 + 2||x||||y|| + ||y||2 = (||x|| + ||y||)2 since x · y ≤ ||x||||y|| by part (b) above. Hence, we can conclude that ||x + y|| ≤ ||x|| + ||y||. (d) Verify that d is a metric. Proof. Let x, y ∈ Rn . Then d(x, y) = ||x − y|| = ((x1 − y1 )2 + . . . + (xn − yn )2 )1/2 ≥ 0 and d(x, y) = ||x − y|| = 0 if and only if xi = yi for all i = 1, . . . n, which is to say precisely when x = y. Now, d(x, y) = ||x − y|| = ((x1 − y1 )2 + . . . + (xn − yn )2 )1/2 = ((y1 − x1 )2 + . . . + (yn − xn )2 )1/2 = ||y − x|| = d(y, x). Finally, if z ∈ Rn as well, then d(x, z) = ||x−z|| = ||(x−y)+(y −z)|| ≤ ||x−y||+||y −z|| = d(x, y)+d(y, z) by part (c) above. Hence, d is a metric on Rn . TOPOLOGY HW 2 3 21.12 Prove continuity of the algebraic operations on R as follows: Use the metric d(a, b) = |a − b| on R and the metric on R2 given by the equation ρ((x, y), (x0 , y0 )) = max{|x − x0 |, |y − y0 |}. (a) Show that addition is continuous. Proof. Let > 0 and let δ < /2. If ρ((x, y), (x0 , y0 )) = max{|x − x0 |, |y − y0 |} < δ, then, d(x + y, x0 + y0 ) = |(x + y) − (x0 + y0 )||(x − x0 ) + (y − y0 )| ≤ |x − x0 | + |y − y0 | <δ+δ < 2(/2) = , so addition is continuous. (b) Show that multiplication is continuous. Proof. Given (x0 , y0 ) ∈ R2 , let 1 > > 0 and let δ < |x0 |+|yo |+1 . If ρ((x, y), (x0 , y0 )) = max{|x − x0 |, |y − y0 |} < δ then d(xy, x0 y0 ) = |xy − x0 y0 | = |xy + (xy0 − xy0 ) + (yx0 − yx0 ) + (x0 y0 − x0 y0 ) − x0 y0 | = |x0 (y − y0 ) + y0 (x − x0 ) + (x − x0 )(y − y0 )| ≤ |x0 (y − y0 )| + |y0 (x − x0 )| + |(x − x0 )(y − y0 )| = |x0 ||y − y0 | + |y0 ||x − x0 | + |x − x0 ||y − y0 | < |x0 |δ + |y0 |δ + δ 2 = δ(|x0 | + |y0 | + δ) < δ(|x0 | + |y0 | + 1) < (|x0 | + |y0 | + 1) |x0 |+|y 0 |+1 < . Hence, multiplication is continuous. (c) Show that the operation of taking reciprocals is a continuous map from R − {0} to R. Proof. Let r : R − {0} → R be the reciprocal map. Let (a, b) be an open interval in R. If both a and b are positive, then, r−1 (a, b) = (1/b, 1/a) which is open in R − {0}. If both a and b are negative, then r−1 (a, b) = (1/b, 1/a), 4 CLAY SHONKWILER which is again open in R − {0}. If a is negative and b is positive, then r−1 (a, b) = (1/a, 1/b), which is open in R − {0}. If a is zero, then r−1 (a, b) = (1/b, ∞) which is open in R − {0}. Finally, if b is zero, then r−1 (a, b) = (−∞, 1/a) which is open in R − {0}. Hence, we can conclude that taking reciprocals is a continuous map. (d) Show that the subtraction and quotient operations are continuous. Proof. To show that subtraction is continuous, let > 0 and let δ < /2. If ρ((x, y), (x0 , y0 )) = max{|x − x0 |, |y − y0 |} < δ, then d(x − y, x0 − y0 ) = |(x − y) − (x0 − y0 )| = |(x − x0 ) + (y0 − y)| ≤ |x − x0 | + |y0 − y| <δ+δ < 2(/2) = , so subtraction is continuous. Finally, to show that the quotient is continuous from R × R − {0} to R, we note that, for (x, y) ∈ R × R − {0}, x 1 =x . y y Thus, if q is the quotient map, p the product map, r the reciprocal map, and r0 : R × R → R × R − {0} is defined by r0 (x, y) = (x, r(y)) then q = p ◦ r0 . Note that r0 is continuous, since its coordinate functions (the identity and the reciprocal maps) are, so q is continuous, since it is the composition of two continuous functions. 22.1 Check the details of Example 3 Check: We want to show that, if p is the map from R to A = {a, b, c} where a if x > 0 b if x < 0 p(x) = c ifx = 0 the topology on A induced by p consists of the following open sets: {a}, {b}, {a, b}, {a, b, c}. TOPOLOGY HW 2 5 In the quotient topology, we know that {a} is open, since p−1 ({a}) = (0, ∞), which is open in R. Similarly, {b} is open, since p−1 ({b}) = (−∞, 0), which is open in R. Also, p−1 ({a, b}) = (−∞, 0) ∪ (0, ∞), which is open in R, so {a, b} is open. Finally, {a, b, c} is open, since p−1 ({a, b, c}) = R which is open in R. Now, we want to show that none of the other subsets of A (other than the empty set, which is obviously open) are open in the quotient topology. p−1 ({c}) = {0} which is not open in R, so {c} is not open. Also, {a, c} and {b, c} are not open, since p−1 ({a, c}) = [0, ∞) and p−1 ({b, c}) = (−∞, 0], neither of which is open in R. We’ve exhausted all subsets of A, so we see that, indeed, the topology on A is precisely that depicted in Example 3. ♣ 22.2 (a) Let p : X → Y be a continuous map. Show that if there is a continuous map f : Y → X such that p ◦ f equals the identity map of Y , then p is a quotient map. Proof. If there exists a continuous map f : Y → X such that p ◦ f ≡ idY , then we want to show that p is a quotient map. p is clearly surjective since, if it were not, p ◦ f could not be equal to the identity map. Now, let U ⊂ Y . If p−1 (U ) is open in X, then U = (p ◦ f )−1 (U ) = f −1 (p−1 (U )) is open in Y since f is continuous. Hence, p is a surjective, continuous open map, so it is necessarily a quotient map. (b) If A ⊂ X, a retraction of X onto A is a continuous map r : X → A such that r(a) = a for each a ∈ A. Show that a retraction is a quotient map. 6 CLAY SHONKWILER Proof. Let i : A → X be the inclusion map. Then i is continuous and, for all a ∈ A, (r ◦ i)(a) = r(i(a)) = r(a) = a so r ◦ i ≡ idA . Hence, since r is continuous, we can conclude, using part (a) above, that r is a quotient map. 22.4 (a) Define an equivalence relation on the plane X = R2 as follows: x0 × y0 ∼ x1 × y1 if x0 + y02 = x1 + y12 . Let X ∗ be the corresponding quotient space. It is homeomorphic to a familiar space; what is it? Answer: X ∗ is homeomorphic to R. We see this as follows. Define g : R2 → R by g(x × y) = x + y 2 . Then g is surjective since, for any r ∈ R, g(r × 0) = r + 02 = r. Also, if A = (a, b) × (c, d) ⊂ R2 , then A = [a, b] × [c, d]. Now g(A) = (a + inf x2 , b + sup y 2 ) x∈(c,d) y∈(c,d) and g(A) = [a + inf x2 , b + sup y 2 ] = g(A), x∈(c,d) y∈(c,d) so g is continuous. Also, we see that g(A) is open, so g is a surjective, continuous, open map. In other words, a quotient map. Now, for each z ∈ R, g −1 (z) = {x × y ∈ R2 |x + y 2 = z} Hence, X ∗ = {g −1 (z)|z ∈ R}. Thus, by Corollary 22.3, g induces a bijective continuous map f : R2 → R which is a homeomorphism. ♣ (b) Repeat (a) for the equivalence relation x0 × y0 ∼ x1 × y1 if x20 + y02 = x21 + y12 . Answer: X ∗ is homeomorphic to R+ ∪{0}. We see this as follows. Define g : R2 → R by g(x × y) = x2 + y 2 . √ Then g is surjective since, for any r ∈ R+ ∪ {0}, g( r × 0) = r2 + 02 = r. Also, if A = (a, b) × (c, d) ⊂ R2 , a basis element of the topology on R2 , then A = [a, b] × [c, d]. Now g(A) = ( inf x0 ∈(a,b) x20 + inf y0 ∈(c,d) y02 , sup x21 + sup y12 ), x1 ∈(a,b) y1 ∈(c,d) and g(A) = [ inf x0 ∈(a,b) x20 + inf y0 ∈(c,d) y02 , sup x21 + sup y12 ] = g(A), x1 ∈(a,b) y1 ∈(c,d) TOPOLOGY HW 2 7 so g is continuous. Also, we see that g(A) is open, so g is a surjective, continuous, open map. In other words, a quotient map. Now, for each z ∈ R, g −1 (z) = {x × y ∈ R2 |x2 + y 2 = z} Hence, X ∗ = {g −1 (z)|z ∈ R}. Thus, by Corollary 22.3, g induces a bijective continuous map f : R2 → R which is a homeomorphism. ♣ 23.1 Let T and T 0 be two topologies on X. If T 0 ⊃ T , what does connnectedness of X in one topology imply about connectedness in the other? Answer: If X is connected in T 0 , then there exist no two elements of T 0 that separate X. Since every element of T is an element of T 0 , this means no two elements of T that separate X. In other words, if X is connected in T 0 , then X is connected in T . On the other hand, if X is connected in T , then it is not necessarily connected in T 0 . For example, if we consider the three point set A from 22.1 above under the quotient topology described above and under the discrete topology, then the discrete topology is finer than the quotient topology, A is connected under the quotient topology, but A is not connected in the discrete topology. This last is clearly true, since {a, b} and {c} are disjoint open sets in the discrete topology that cover A. To see that A is connected in the quotient topology, we note that the only open set containing the point c is A, so it will be impossible to separate A by open sets. ♣ 23.2 Let {An } be a sequence of S connected subspaces of X, such that An ∩ An+1 6= ∅ for all n. Show that An is connected. S Proof. Suppose An is not connected. S S Then there exist disjoint open sets C and D in An such that C ∪ D = An . Then, to arrive at a contradiction, we aim to demonstrate that each An lies in C or D. Since A1 is connected, it must be contained entirely in either C or D, by Lemma 23.2. Suppose, without loss of generality, that it lies in C. Fix k ∈ Z+ and suppose that Ak ⊂ C. By Lemma 23.2, since Ak+1 is connected, it must lie entirely within either C or D. Furthermore, ∅= 6 Ak+1 ∩ Ak ⊂ Ak ⊂ C, so Ak+1 ∩ C 6= ∅. Therefore, it must be true that Ak+1 ⊂ C. 8 CLAY SHONKWILER S Hence, by induction, every AS An ⊂ C, contradicting the n ⊂ C, so assertion thatSC and D separate An . From this contradiction we conclude that, in fact, An is connected. 23.6 Let A ⊂ X. Show that if C is a connected subspace of X that intersects both A and X − A, then C intersects Bd A. Proof. By contrapositive. Suppose C is connected and does not intersect Bd A. Then (1) ∅ = C ∩ (A ∩ X − A) = (C ∩ A) ∩ (C ∩ X − A). Now, by definition of the subspace topology, C ∩A and C ∩X − A are closed in C, so C − (C ∩ A) and C − (C ∩ X − A) are open in C. Now, C − (C ∩ A) ∩ (C ∩ X − A) = C − ∅ = C. However, by DeMorgan’s, C = C − (C ∩ A) ∩ (C ∩ X − A) = C − (C ∩ A) ∪ C − (C ∩ X − A) . Since C is connected, it must be true that either C − (C ∩ A) or C − (C ∩ X − A) is empty. If C − (C ∩ A) = ∅, then C ∩ A = C. By (1), then, C ∩ X − A = ∅. Since X − A ⊂ X − A, we can conclude that C ∩ (X − A) = ∅. On the other hand, if C − (C ∩ X − A) = ∅, then C ∩ X − A = C. By (1), C ∩ A = ∅. Since A ⊂ A, we see that C ∩ A = ∅. In either case, we see that, if C ∩ Bd A = ∅, then it is not the case that C intersects both A and X − A. Hence, by contrapositive, we conclude that if C intersects both A and X − A, then C intersects Bd A. A Let X and Y be connected topological spaces and endow X × Y with the product topology. Now suppose that A ( X and B ( Y are proper subsets. Show that (X × Y ) − (A × B) is connected. Give an example showing that this fails if one of A and B is not a proper subset. Proof. Fix a point a×b ∈ (X ×Y )−(A×B). Then the set X ×b is connected, as it is homeomorphic to the connected set X and, for all x ∈ X − A, x × Y is connected, since it is homeomorphic to the connected set Y . Now, X × b and x × Y have the point a × b in common, so Tx = (x × Y ) ∪ (X × b) TOPOLOGY HW 2 9 is connected for all x ∈ X − A. Furthermore, the union [ Cx = Tx x∈X−A is connected, since each Tx is connected and contains the point a × b. Now, also, a×Y is connected, since it is homeomorphic to Y and, for all y ∈ Y −B, X ×y is connected, since it is homeomorphic to X, and contains a×b. Hence, Ty = (a × Y ) ∪ (X × y) is connected. Furthermore Cy = [ Ty y∈Y −B is connected, since each Ty is connected and contains the point a × b. Now, finally, we can conclude that Cx ∪ Cy is connected, since both Cx and Cy are connected and contain a × b. To see that Cx ∪ Cy = (X × Y ) − (A × B), let c × d ∈ (X × Y ) − (A × B). Then, either c ∈ X − A or d ∈ Y − B (or both). In the first case, c × d ∈ Cx and in the second, c × d ∈ Cy . Either way, c × d ∈ Cx ∪ Cy , so (X × Y ) − (A × B) ⊆ Cx ∪ Cy . Clearly, Cx ∪ Cy ⊆ (X × Y ) − (A × B), so, in fact, Cx ∪ Cy = (X × Y ) − (A × B), meaning (X × Y ) − (A × B) is connected. Counter-Example: Let X = Y = A = R and let B = [0, 1]. Then (X × Y ) − (A × B) = (R × (−∞, 0)) ∪ (R × (1, ∞)) . Now, R × (−∞, 0) and R × (1, ∞) are both clearly open in (R × (−∞, 0)) ∪ (R × (1, ∞)) and disjoint, so they form a separation. Hence, we see an example where A is not a proper subset of X and, thereby, (X ×Y )−(A×B) is not connected, even though X and Y are. ♣ B Let X = [0, 1] ⊂ R have the product topology. Now suppose f : X → X is continuous. Prove (without using the intermediate value theorem) that f has a fixed point. That is show that there exists an x0 ∈ X such that f (x0 ) = x0 . For the purposes of this exercise you may assume that X is connected. 10 CLAY SHONKWILER Proof. Define g : [0, 1] → [−1, 1] by g(x) = x − f (x). Then g is continuous, g(0) = −f (0) ≤ 0 and g(1) = 1 − f (b) ≥ 0. Let A = g([0, 1]) ∩ (−∞, 0) and B = g([0, 1]) ∩ (0, ∞). Both A and B are open in g([0, 1]). Furthermore, if there is no point c ∈ [0, 1] such that g(c) = 0, then f ([0, 1]) = A ∪ B. Since A and B are disjoint, this constitutes a separation of g([0, 1]). However, the image of a connected set like [0, 1] under a continuous map like g is connected by Thm. 23.5. Hence, there must be some c ∈ [0, 1] such that g(c) = 0. Now, f (c) = c − g(c) = c − 0 = c, so f has a fixed point. DRL 3E3A, University of Pennsylvania E-mail address: [email protected]