Problems - Section 17-2, 4, 6c, 9, 10, 13, 14; Section 18-1, 3, 4, 6, 8, 10; Section 19-1, 3, 5, 7, 8, 9;

Math 553 - Topology Todd Riggs Assignment 2 Sept 17, 2014 Problems - Section 17-2, 4, 6c, 9, 10, 13, 14; Section 18-1, 3, 4, 6, 8, 10; Section 19-1, 3, 5, 7, 8, 9; 17.2) Show that if A is closed in Y and Y is closed in X, then A is closed in X. If we can show that X A is open in X, then A will be closed in X. Since Y is closed in X, we have X Y open in X. Similarly, we have Y A open in Y. By the subspace topology of Y, there exist an open subset U of of X such that U Y = Y A and hence (Y A) U. So X A = (X Y ) U and since unions of open sets are open, X A is open in X as desired. 17.4) Show that if U is open in X and A is closed in X, then U A is open in X, and A U is closed in X. To show U A is open, we must show X (U A) is closed. Notice that X (U A) = X U A which is a finite union of closed sets and thus closed. Similarly, for A U closed, we will show X (A U) is open. As above, we have X (A U) = X A U which is a union of open sets and thus open. 17.6c) Let A, B, and A α denote subsets of a space X. Prove the following: (c) A α A α ; give an example where equality fails. Let x A α. Then x A αo for some α o and, by Theorem 17.5, for every open set U, containing x, we have U A αo. So U { A α } = and as such x A α. Thus A α A α as desired. Example where equality fails: Let A n = { 1 } where n N. Then n 1 A n = { 1 n N}. So A n α = { 1 } and n A α = { 1 n N}, but A n α = { 1 n N} {0}. n Page 1 of 8

17.9) Let A X and B Y. Show that in the space X Y, A B = A B. Let (a, b) A B. So, for all open neighborhoods U that contain (a, b), U (A B). Now (a, b) will be in A B iff a A and b B. This will be true iff for all open W containing a and for all open V containing b we have W A and V B. Let W be an open set containing a and V an open set containing b. Then W V is open in X Y and contains (a, b). So (W V ) (A B) = (W A) (V B) and as such W A and V B. Now since W and V are arbitrary open sets containing a and b respectively, a A and b B. Thus (a, b) A B and A B A B. Conversely, let (a, b) A B. Then a A and b B. So for all open U containing a and all open V containing b, we have U A and V B. Now (a, b) will be contained in A B iff for all open W in X Y containing (a, b), W (A B). Let W be open in X Y and contain (a, b). Since W is open, there exists U V,where U is open in X and V is open in Y, such that (a, b) U V W. Then U open in X and a U implies U A and similarly V B. So, W (A B) (U V ) (A B) = (U A) (V B). Hence, (a, b) A B and thus A B = A B. 17.10) Show that every order topology is Hausdorff. Let X be a non-empty set, with more than one element, with the order topology. By definition of Hausdorff we must show for each pair x 1, x 2 of distinct points of X that there exists neighborhoods U 1 and U 2 of x 1 and x 2, respectively, that are disjoint. We will look at three cases: (1) X is finite (or bounded), (2) X is unbounded. Case (1): X is finite (or bounded). Let X = {x 1, x 2,...} where x 1 is the least, (or similarly greatest), and x i < x i+1. Then by definition of the order topology there exists disjoint neighborhoods nbr(x 1 ) = [x 1, x i ) and nbr(x i ) = (x i 1, x n ], where 1 < i n, and X is Hausdorff. Case (2): X is unbounded. Let X be an unbounded ordered set. If X is unbounded than X must have an infinite number of elements. So for all element x i, x j an ordering is possible, say... < x i < x j <.... Then neighborhoods (, x j ) of x i and (x i, ) of x j are disjoint and X is again Hausdorff. Page 2 of 8

17.13) Show that X is Hausdorff iff the diagonal = {x x x X} is closed in X X. Let X be a Hausdorff Space. To show the diagonal is closed we will show (X X) is open. Let (x, y) (X X). Since (x, y) we have x y. Furthermore, since X is Haudorff, there exists open neighborhoods U and V of x and y, respectively, such that U V =. Now (U V ) = since if not, there exists (x 1, x 1 ) (U V ) implying x 1 U V and contradicting U V =. So (x, y) (U V ) (X X) and U V is open in X X. Hence, (X X) is open as desired. Conversely, Suppose X X is closed so that (X X) is open. Let x, y X be such that x y. So (x, y) (X X) and since (X X) is open, there exists open neighborhoods U and V, of x and y respectively, such that (U V ) (X X). Now U V = since if x 1 U V then (x 1, x 1 ) U V and (x 1, x 1 ) would be in. Thus we have found neighborhoods U and V, of x and y respectively, such that U V =. That is, X is Hausdorff. 17.14) In the finite complement topology on R, to what point or points does the sequence x n = 1 n converge? Claim: The sequence x n = 1 converges to all x X. n In the finite complement topology, U is open in a topological space X if U = X or X U is finite (pg 77). Let U be an open neighborhood of x X. Then if X U is finite, x n U for all but finitely many n and by definition, x n converges to x. 18.1) Prove that for functions f : R R, the ɛ δ definition of continuity implies the open set definition. Let f : R R be a continuous function based on the ɛ δ definition and V be an open set in the range space. We must show that f 1 (V ) is open. Let x o f 1 (V ) so that f(x o ) V. Since V is open and f(x o ) V, there exists an ɛ > 0 such that (f(x o ) ɛ, f(x o ) + ɛ) V. From the ɛ δ definition, choose δ > 0, depending on x o and ɛ, such that if x R satisfies x x o < δ, then f(x) f(x o ) < ɛ. Now let x (x o δ, x o + δ) so that x x o < δ. Then f(x) f(x o ) < ɛ and f(x) (f(x o ) ɛ, f(x o ) + ɛ) V. So f(x) V implies x f 1 (V ) and f 1 (V ) is open as desired. Page 3 of 8

18.3) Let X and X denote a singleset in the two topologies T and T, respectively. Let i : X X be the idenity function. (a) Show that i is continuous iff T is finer than T. (b) Show that i is a homeomorphism iff T = T. (a) Suppose i is continuous and let U T. Since i : X X is continuous and U is open in X, we have i 1 (U) = U is open in X. That is U T and T is finer than T. Conversely, assume T is finer than T. Then for all U T we have U T. So given an open set U in X we have i 1 (U) = U is open in X. Hence i is continuous. (b) Suppose i is a homeomorphism. Then i : X X and i 1 : X X are continuous. So for U open in X we have i 1 (U) = U is open in X. That is T T. Likewise, for W open in X we have (i 1 ) 1 (W ) = i(w ) = W is open in X and T T. Thus T = T. Conversely, let T = T. Since i is the identity, it is bijective. So if U is open in X we have i 1 (U) = U is open in X. That is, i : X X is continuous. Likewise, if W is open in X we have (i 1 ) 1 (W ) = i(w ) = W is open in X and i 1 : X X is continuous. Thus i is a homeomorphism. 18.4) Given x 0 X and y 0 Y, show that the maps f : X X Y and g : Y X Y defined by f(x) = x y 0 and g(y) = x 0 y are imbeddings. We need to show that the restriction of f and g to their ranges leads to homeomorphisms. That is, f : X X {y o }, f (x) = f(x) and g : Y {x o } Y, g (y) = g(y) are homeomorphisms. Since f and g are bijective with their range, it needs to be shown that they and their inverses are continuous. For the continuity of f let W be open in X {y o }. By definition of subspace topology, W = (X {y 0 }) (U V ), where U V is open in X Y. So U is open in X and V is open in Y. Now (f ) 1 (W ) = (f ) 1 ((X U) ({y o } V ) = X U = U which is open in X. Hence f is continuous. Similarly, for the continuity of g, let W be open in{x o } Y. By definition of subspace topology, W = ({x 0 } Y ) (U V ) = ({x o } U) (Y V ), where U V is open in X Y so that U is open in X and V is open in Y. Now (g ) 1 (W ) = (g ) 1 (({x o } U) (Y V ) = Y V = V which is open in Y. Hence g is continuous. Now we need to show (f ) 1 : X {y o } X and (g ) 1 : {x o } Y Y are continuous. For (f ) 1 if U is open in X, then ((f ) 1 ) 1 (U) = f (U) = U {y o } = (X {y o }) (U Y ) which is open in X {y o }. Similarly, for (g ) 1, if V is open in Y, then ((g ) 1 ) 1 (V ) = g (V ) = {x o } V = ({x o } Y ) (X V ) which is open in {x o } Y. Since f and g are continuous bijective mappings with continuous inverses, they are homeomorphisms. Thus f and g imbeddings as desired. Page 4 of 8

18.6) Find a function f : R R that is continuous at precisely one point. Claim: The piecewise function f : R R defined by { x : x Q f(x) = 0 : x Q is continuous only at 0. First we will show f is continuous at 0 and than that it is not continuous at any other point. Let U be a neighborhood of f(0) = 0. (We will be utilizing Theorem 18.1(4) pg 104.) We need to show there is a neighborhood V of 0 such that f(v ) U. Since U is an open set containing 0, there exists ɛ such that 0 ( ɛ, ɛ) U. Let V = ( ɛ, ɛ). Then V is a neighborhood of 0 such that f(v ) ( ɛ, ɛ) U. So, f is continuous at 0. For the second part, let x R such that x 0. If x Q, set ɛ = x, and let U = 2 (x ɛ, x+ɛ). So U is a neighborhood of f(x) = x that does not contain 0. Now given any neighborhood V of x, it contains irrational numbers, so 0 f(v ). We cannot find a neighborhood V of x such that f(v ) U. Hence f is not continuous at x. If x R Q then U = ( x, x ) is a neighborhood of f(x) = 0. Given any neighborhood 2 2 V of x, we can choose ɛ > 0 sufficiently small so that x (x ɛ, x + ɛ) V and ɛ < x. Between x and x + ɛ we can find q Q. So f(q) = q f(v ) but q U, so 2 f(v ) is not contained in U. Hence f is not continuous at x and our claim is proved. 18.8) Let Y be an ordered set in the order topology. Let f, g : X Y be continuous. (a) Show that the set {x f(x) g(x)} is closed in X. (b) Let h : X Y be the function h(x) = min{f(x), g(x)}. Show that h is continuous. (Hint: Use pasting lemma) PART A By problem 17.10 we have that Y is Hausdorff. To show {x f(x) g(x)} is closed we will show that the complement is open. That is, X {x f(x) g(x)} = {x f(x) > g(x)} = A is open. Let x A so that f(x) > g(x). Since Y is Hausdorff and f and g are continuous, we have disjoint neighborhoods U and V of f(x) and g(x) in Y, respectfully, such that f 1 (U) and g 1 (V ) are open in X. Now we have f 1 (U) g 1 (V ) is open in X and since x f 1 (U) g 1 (V ) there exists a neighborhood N such that x N (f 1 (U) g 1 (V )). Moreover, since U and V are disjoint, f(n) g(n) =. That is, for all x N we have f(x) > g(x) implying N A and we have A open as desired. PART B Using the pasting lemma, with f continuous on A = {x f(x) g(x)}, and g continuous on B = {x f(x) g(x)}, where B is closed from part (a) as well as A by a similar argument to (a), we have X = A B and A B = {x f(x) = g(x)}. Page 5 of 8

Now, by noticing that h(x) h (x) = { f(x) : x A g(x) : x B we have h continuous. 18.10) Let f : A B and g : C D be continuous functions. Let us define a map f g : A C B D by the equation (f g)(a c) = f(a) g(c). Show that f g is continuous. Let a A and c C. Additionally, let U = U 1 U 2 be an open neighborhood of f(a) g(c) in B D where f(a) U 1 B and g(c) U 2 D. Then since f and g are continuous, there exists basis elements B A A and B C C such that f 1 (f(a)) = a B A and g 1 (g(c)) = c B C. So B = B A B C is a basis element of A C and as such open. That is (f(a) g(c)) 1 = f 1 (f(a)) g 1 (g(c)) B is open and thus f g is continuous. 19.1) Prove Theorem 19.2 The given set, B α, is a basis element in the box topology. We must show this α J is open in the generated topology. Letting α J U α be a basis element in the product topology and x α J U α, then x α will be in U α for all alpha. Since B α is a basis for X α, there exists B α such that x α B α U α. That is, x B α U α and thus open in the generated topology. Conversely, the given set, α J B α, is a basis element in the product topology. Let x U α where U α is a basis element such that all but finitely many U α = X α. Then, for all U α X α we choose basis elements B α1,..., B α2 such that x αj B αj U αj. So, as above, x B α U α and thus open in the generated topology. 19.3) Prove Theorem 19.4 Since the box topology is finer than the product topology (pg 115), it will suffice to show the theorem holds under the box topology. Let x and y be distinct elements in X α such that each X α is Hausdorff. Since x and y are distinct, there is at least one coordinate term such that x i y i. So for each x i y i there are open neighborhoods U i and V i, where x i U i and y i V i, that are subsets of X i and U i V i =. Now define open neighborhoods U and V in X α by U = U α and V = V α. Then U V = (U α V α ) = (U 1 V 1 )... (U i V i )... = (U 1 V 1 )... ( )... =. Thus X α is Hausdorff as desired. Page 6 of 8

19.5) One of the implications stated in Theorem 19.6 holds for the box topology. Which one? By example 2 pg 117, if each f α is continuous then f is not necessarily continuous. So, If f : A X α is continuous than f α : A X α is continuous for all alpha. Let f : A X α be continuous and U = U α be open in X α where U α is open in X α. Then since f 1 (U) is open in A, we have fα 1 (U α ) open for all alpha and f α is continuous. 19.7) Let R be the subset R ω consisting of all sequences that are eventually zero, that is, all sequences (x 1, x 2,...) such that x i = 0 for only finitely many values of i. What is the closure of R in R ω in the box and product topologies? Justify. Claim: (a) R is closed in the box topology and (b) The closure of R in the product topology is R ω. (a) We will show R ω R is open. The box topology has open sets of the form U = U 1 U 2.... Let {x n } n N = (x 1, x 2,...) (R ω R ). Then, since there are infinitely many nonzero elements, only finitely many zero elements, we can find an open neighborhood U nj about x nj that does not contain zero. So V = U 1... U n1 U n2... will be an open neighborhood containing (x n ) n N such that V is a proper subset of R. (b) We will show that for any open neighborhood U R ω containing (x n ) n N that U R. Notice, for U = U 1 U 2... U j R R open in the product topology, that (x n ) n N = (x 1, x 2,..., x j, 0, 0,...), for some j is an element of U and an element of R. That is, U R as desired. 19.8) Given the sequences (a 1, a 2,...) and (b 1, b 2,...) of real numbers with a i > 0 for all i, define h : R ω R ω by the equation h((x 1, x 2,...)) = (a 1 x 1 + b 1, a 2 x 2 + b 2,...). Show that if R ω is given the product topology, h is a homeomorphism of R ω with itself. What happens if R ω is given the box topology? Clearly h is bijective. By theorem 19.6, h is continuous in the product topology since the composition with the projection onto the n-th factor is continuous. In the box topology let h n = h n (x) = a n x + b n be the n-th coordinate function of h. Then, for U = U 1 U 2... open in the box topology, we have h 1 (U) = h 1 1 (U 1 ) h 1 2 (U 2 )... open since h n is continuous for all n. Page 7 of 8

19.9) Show that the choice axiom is equivalent to the statement that for any indexed family {A α } α J of nonempty sets, with J 0, the cartesian product A α is not empty. α J The axiom of choice (pg 59) states that from any collection of nonempty sets A = {A α } α J that there exists a choice function c : A such that c(a α ) = a α A α. So, c(a α ) = (a α ) α J = A α. ADDITIONAL PROBLEM(S) A α α J 17.3) Show that if A is closed in X and B is closed in Y, then A B is closed in X Y. As in problem 17.2, we need to show that (X Y ) (A B) is open in X Y. Since A is close in X and B is closed in Y, X A and Y B are open in X and Y respectively. So by the product topology, (X A) (Y B) = (X Y ) (A B) is open in X Y as desired. Page 8 of 8