Topology Exam 1 - Answers

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1 Topology Exam 1 - Answers John Terilla Fall 2014 Problem 1: Prove that the quotient topology is characterized by its universal property Definition 1. Let X be a topological space, let S be a set, and let π : X S be surjective. The quotient topology on S is defined to be the finest topology for which π is continuous. Equivalently, a set U in S is open in the quotient topology if and only if π 1 (U) is open in X. Prove that the quotient topology on S is characterized by the universal property stated below. That is, prove (1) that the quotient topology has the universal property and (2) that the quotient topology is the only topology on S that has the universal property. Universal property for the quotient topology. For every topological space Z and every function f : S Z, f is continuous if and only if f π : X Z is continuous. Here is the picture: π X f π S f Z Solution Let τ Q denote the quotient topology on S. To see that τ Q satisfies the property, let Z be a space and f : S Z. Since π : X S is continuous in the quotient topology, if f is continuous, then the composition f π is continuous. To show f π is continuous implies f is continuous, assume f π is continuous and let U Z be open. So (f π) 1 (U) = π 1 (f 1 (U)) is open in X. Since π 1 (f 1 (U)) is open in X, the set f 1 (U) is open in the quotient topology. This proves that f is continuous. So τ Q satisfies the universal property. 1

2 Now suppose that τ is a topology on S satisfying the universal property. Consider id : (S, τ) (S, τ Q ). π X id π = π (S, τ) (S, τ Q ) id Since (S, τ) satisfies the property, and the diagonal map π : X (S, τ Q ) is continuous, the map id : (S, τ) (S, τ Q ) is continuous. This means that τ Q τ. Consider id : (S, τ) (S, τ). π X id π = π (S, τ) (S, τ) id Since (S, τ) satisfies the property, and the map along the bottom id : (S, τ) (S, τ) is continuous, the diagonal map π : X (S, τ) is continuous. Since τ Q is the finest topology on S for which π : X S is continuous, we have τ τ Q. Therefore, τ = τ Q. This proves that τ Q is the only topology satisfying the universal property. 2

3 Problem 2: Give an example, or prove that no such example exists (a) A continuous surjection (0, 1) [0, 1] Answer. Here s the graph of one: (b) A continuous surjection [0, 1] (0, 1) Answer. Impossible. [0, 1] is compact and (0, 1) is not. (c) A path p : [0, 1] X connecting a to b in the space (X, τ) where Answer. Define p by X = {a, b, c, d} and τ = {, {c}, {a, c}, {b, c, d}, X}. a 0 t 1 3 p(t) = 1 c 3 < t < b 3 t < 1 Note p 1 ({c}) = ( 1 3, 3) 2, p 1 ({a, c}) = [ 0, 2 3 p 1 (X) = [0, 1] are all open in [0, 1] so p is continuous. ), p 1 ({b, c, d}) = ( 1 3, 1], and 3

4 Problem 3: Fill in the missing proofs: Hausdorff spaces, KC spaces, US spaces Here are the first two pages of an article published in The American Mathematical Monthly, Vol. 74, No. 3 (Mar., 1967), pp Parts have been redacted. Your problem: Prove Theorem 1 and Theorem 2. This content downloaded from on Thu, 25 Sep :20:08 PM All use subject to JSTOR Terms and Conditions 4

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6 Solution Theorem. T 2 KC US T 1. Proof. T 2 KC. Suppose X is T 2 and C X is compact. If C = X then C is closed. If C X, let y X \ C. For each x C, there exist disjoint open sets U x and V x with x U x and y V x. The sets {U x } cover C, so there exists a finite subcover U x1,..., U xn of C. Then the set V = V x1 V xn is an open set around y that doesn t inerset C. KC US. Suppose that X is KC and let {x n } be a sequence in X with {x n } x and {x n } y. If y x, then {y} is compact, hence closed, so X \ {y} is an open set around x. Therefore, there exists a natural number N so that for n N, x n X \ {y} x n y. But the set K = {x, x n+1, x n+2,...} is a compact set that does not contain y, which is a limit of the sequence {x n+1, x n+2,...} in K, so K is not closed. Therefore, x = y and X, proving that X is US. US T 1. If X is not T 1, then there exist two distinct points x, y X so that every open set containing x contains y also. Therefore, the sequence y, y, y, y,... converges to x and converges to y. So, X is not US. To see that the implications T 2 KC US T 1 are strict, here are some examples. Example 1. N with the cofinite topology is T 1 but not US. The sequence {1, 2, 3,..., for example, converges to {1} and {2} (and to every n N). Example 2. R with the cocountable topology is KC but not T 2. It s not T 2 since every two open sets intersect. To see that it s KC, note that an infinite set A X is not compact since if {x 1, x 2,...} i N is a countably infinite set of distinct elements of A, the open cover {U i } i N where U i = X \ {x i, x i+1,...} is an open cover of A with no finite subcover. So every compact set is finite, hence closed. Example 3. It takes a bit of searching to find a US space that is not KC. I ll refer you to example 4 in the article. Theorem 1. If X is first countable, then T 2 = KC = US. Proof. Suppose X is a first countable space that is not T 2. Then there exist two distinct points x, y X so that every neighborhood of x intersects every neighborhood of y. Let {U n } be a neighborhood base of x and V n } be a neighborhood base of y. For every n N, the set U 1 U n is a neighborhood of x and V 1 V n is a neighborhood of y. So, there exists a point, call it z n, in the intersection z n U 1 U n V 1 V n. The sequene {z n } converges to both x and y, hence X is not US. 6