TOPOLOGY HW 2 CLAY SHONKWILER 20.9 Show that the euclidean metric d on R n is a metric, as follows: If x, y R n and c R, define x + y = (x 1 + y 1,..., x n + y n ), cx = (cx 1,..., cx n ), x y = x 1 y 1 +... x n y n. (a) Show that x (y + z) = (x y) + (x z). Proof. Let x, y, z R n. Then x (y + z) = x (y 1 + z 1,..., y n, z n ) = x 1 (y 1 + z 1 ) +... x n (y n + z n ) = x 1 y 1 + x 1 z 1 +... x n y n + x n z n = x y + x z (b) Show that x y x y. Proof. Let x, y R n. Then. Then x x ± y y 2 = n i=1 Then, since x y 0, x x ± y y 0 ( xi x ± y ) 2 i = y n ( ) xi y ± y i x 2 0. x y i=1 n (x i y ± y i x ) 2 0. i=1 This in turn implies that 0 n ( i=1 x 2 i y 2 + yi 2 x 2 ± 2x i y i x y ) = y n i=1 x2 i + x n i=1 y2 i ± 2 x y n i=1 x iy i = 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
2 CLAY SHONKWILER or ±(x y) x 2 y 2 x y Multiplying both sides by 1, we see that = x y. (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 R n, 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 R n. Then d(x, y) = x y = ((x 1 y 1 ) 2 +... + (x n y n ) 2 ) 1/2 0 and d(x, y) = x y = 0 if and only if x i = y i for all i = 1,... n, which is to say precisely when x = y. Now, d(x, y) = x y = ((x 1 y 1 ) 2 +... + (x n y n ) 2 ) 1/2 = ((y 1 x 1 ) 2 +... + (y n x n ) 2 ) 1/2 = y x = d(y, x). Finally, if z R n 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 R n.
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 R 2 given by the equation ρ((x, y), (x 0, y 0 )) = max{ x x 0, y y 0 }. (a) Show that addition is continuous. Proof. Let ɛ > 0 and let δ < ɛ/2. If then, ρ((x, y), (x 0, y 0 )) = max{ x x 0, y y 0 } < δ, d(x + y, x 0 + y 0 ) = (x + y) (x 0 + y 0 ) (x x 0 ) + (y y 0 ) x x 0 + y y 0 < δ + δ < 2(ɛ/2) = ɛ, so addition is continuous. (b) Show that multiplication is continuous. Proof. Given (x 0, y 0 ) R 2, let 1 > ɛ > 0 and let δ < then ɛ x 0 + y o +1. If ρ((x, y), (x 0, y 0 )) = max{ x x 0, y y 0 } < δ d(xy, x 0 y 0 ) = xy x 0 y 0 = xy + (xy 0 xy 0 ) + (yx 0 yx 0 ) + (x 0 y 0 x 0 y 0 ) x 0 y 0 = x 0 (y y 0 ) + y 0 (x x 0 ) + (x x 0 )(y y 0 ) x 0 (y y 0 ) + y 0 (x x 0 ) + (x x 0 )(y y 0 ) = x 0 y y 0 + y 0 x x 0 + x x 0 y y 0 < x 0 δ + y 0 δ + δ 2 = δ( x 0 + y 0 + δ) < δ( x 0 + y 0 + 1) < ( x 0 + y 0 + 1) ɛ x 0 + y 0 +1 < ɛ. 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 ɛ > 0 and let δ < ɛ/2. If ρ((x, y), (x 0, y 0 )) = max{ x x 0, y y 0 } < δ, then d(x y, x 0 y 0 ) = (x y) (x 0 y 0 ) = (x x 0 ) + (y 0 y) x x 0 + y 0 y < δ + δ < 2(ɛ/2) = ɛ, 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 y = x. y Thus, if q is the quotient map, p the product map, r the reciprocal map, and r : R R R R {0} is defined by r (x, y) = (x, r(y)) then q = p r. Note that r 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 p(x) = b if x < 0 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 which is open in R. Also, p 1 ({b}) = (, 0), 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 and p 1 ({a, c}) = [0, ) 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 id Y, 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 id A. 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 = R 2 as follows: x 0 y 0 x 1 y 1 if x 0 + y 2 0 = x 1 + y 2 1. 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 : R 2 R by g(x y) = x + y 2. Then g is surjective since, for any r R, g(r 0) = r + 0 2 = r. Also, if A = (a, b) (c, d) R 2, then A = [a, b] [c, d]. Now and g(a) = (a + inf x (c,d) x2, b + sup g(a) = [a + inf x (c,d) x2, b + sup y (c,d) y (c,d) y 2 ) y 2 ] = g(a), 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 R 2 x + y 2 = z} Hence, X = {g 1 (z) z R}. Thus, by Corollary 22.3, g induces a bijective continuous map f : R 2 R which is a homeomorphism. (b) Repeat (a) for the equivalence relation x 0 y 0 x 1 y 1 if x 2 0 + y 2 0 = x 2 1 + y 2 1. Answer: X is homeomorphic to R + {0}. We see this as follows. Define g : R 2 R by g(x y) = x 2 + y 2. Then g is surjective since, for any r R + {0}, g( r 0) = r 2 + 0 2 = r. Also, if A = (a, b) (c, d) R 2, a basis element of the topology on R 2, then A = [a, b] [c, d]. Now and g(a) = ( inf x 0 (a,b) x2 0 + g(a) = [ inf x 0 (a,b) x2 0 + inf y 0 (c,d) y2 0, inf y 0 (c,d) y2 0, sup x 1 (a,b) sup x 1 (a,b) x 2 1 + x 2 1 + sup y 1 (c,d) sup y 1 (c,d) y 2 1), y 2 1] = g(a),
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 R 2 x 2 + y 2 = z} Hence, X = {g 1 (z) z R}. Thus, by Corollary 22.3, g induces a bijective continuous map f : R 2 R which is a homeomorphism. 23.1 Let T and T be two topologies on X. If T T, what does connnectedness of X in one topology imply about connectedness in the other? Answer: If X is connected in T, then there exist no two elements of T that separate X. Since every element of T is an element of T, this means no two elements of T that separate X. In other words, if X is connected in T, then X is connected in T. On the other hand, if X is connected in T, then it is not necessarily connected in T. 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 {A n } be a sequence of connected subspaces of X, such that A n A n+1 for all n. Show that A n is connected. Proof. Suppose A n is not connected. Then there exist disjoint open sets C and D in A n such that C D = A n. Then, to arrive at a contradiction, we aim to demonstrate that each A n lies in C or D. Since A 1 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 A k C. By Lemma 23.2, since A k+1 is connected, it must lie entirely within either C or D. Furthermore, so A k+1 A k A k C, A k+1 C. Therefore, it must be true that A k+1 C.
8 CLAY SHONKWILER Hence, by induction, every A n C, so A n C, contradicting the assertion that C and D separate A n. From this contradiction we conclude that, in fact, A n 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, However, by DeMorgan s, C ( (C A) (C X A) ) = C = C. 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 T x = (x Y ) (X b)
TOPOLOGY HW 2 9 is connected for all x X A. Furthermore, the union C x = x X A is connected, since each T x 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, is connected. Furthermore T x T y = (a Y ) (X y) C y = y Y B is connected, since each T y is connected and contains the point a b. Now, finally, we can conclude that C x C y is connected, since both C x and C y are connected and contain a b. To see that C x C y = (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 C x and in the second, c d C y. Either way, c d C x C y, so Clearly, T y (X Y ) (A B) C x C y. C x C y (X Y ) (A B), so, in fact, C x C y = (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 x 0 X such that f(x 0 ) = x 0. 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, so f has a fixed point. f(c) = c g(c) = c 0 = c, DRL 3E3A, University of Pennsylvania E-mail address: shonkwil@math.upenn.edu