Generell Topologi. Richard Williamson. May 6, 2013

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1 Generell Topologi Richard Williamson May 6,

2 8 Thursday 7th February 8.1 Using connectedness to distinguish between topological spaces I Proposition 8.1. Let (, O ) and (Y, O Y ) be topological spaces. Suppose that f Y defines a homeomorphism between (, O ) and (Y, O Y ). Let A be a subset of, equipped with the subspace topology O A with respect to. Let f(a) be equipped with the subspace topology O f(a) with respect to Y. Let \ A be equipped with the subspace topology O \A with respect to, and let Y \ f(a) be equipped with the subspace topology with respect to Y. Then (A, O A ) is homeomorphic to ( f(a), O f(a) ), and ( \A, O\A ) is homeomorphic to ( Y \ f(a), O Y \f(a) ). Proof. See Exercise Sheet 4. Corollary 8.2. Let [a, b], (a, b), [a, b), and (a, b], for a, b R, be equipped with their respective subspace topologies with respect to (R, O R ). Then: (1) [a, b] is not homeomorphic to (a, b), [a, b), or (a, b]. (2) (a, b) is not homeormorphic to (a, b] or [a, b). (3) [a, b) is homeomorphic to (a, b]. Proof. We have that [a, b] \ {a, b} = (a, b). By Proposition 7.9, (a, b) equipped with the subspace topology with respect to (R, O R ) is connected. However, by removing any two points from (a, b), [a, b), or (a, b], and equipping the resulting set with the subspace topology with respect to (R, O R ), we will obtain a topological space which is not connected. By Proposition 8.1 and Corollary 7.3, we deduce that [a, b] is not homeomorphic to any of (a, b), [a, b), or (a, b]. An example of removing two points from (a, b] is depicted below. In this case, we obtain a disjoint union of two open intervals. ( ] ( ) ( ) A second example of removing two points from (a, b] is depicted below. In this case, we obtain a disjoint union of two open intervals and a half open interval. 68

3 ( ] ( ) ( ) ( ] We have that [a, b) \ {a} = (a, b). Again, by Proposition 7.9, (a, b) equipped with the subspace topology with respect to (R, O R ) is connected. However, removing any one point from (a, b) and equipping the resulting set with the subspace topology with respect to (R, O R ), we will obtain a topological space which is not connected. By Proposition 8.1 and Corollary 7.3, we deduce that [a, b) is not homeomorphic to (a, b). ( ) ( ) ( ) We also have that (a, b] \ {b} = (a, b). By the same argument as above, we deduce that (a, b] is not homeomorphic to (a, b). We will prove that [a, b) = (a, b] on Exercise Sheet Connected components Terminology 8.3. Let (, O ) be a topological space, and let A be a subset of equipped with the subspace topology O A with respect to (, O ). Then A is a connected subset of if (A, O A ) is a connected topological space. Definition 8.4. Let (, O ) be a topological space, and let x. The connected component of x in (, O ) is the union of all connected subsets A of such that x A. Proposition 8.5. Let (, O ) be a topological space, and let {A j } j J be a set of connected subsets of such that j J A j =. Suppose that j J A j. Then (, O ) is connected. Proof. Let {0, 1} be equipped with the discrete topology, and let be a continuous map. Given j J let f {0, 1} A j i j 69

4 denote the inclusion map. By Proposition 2.15 we have that i j is continuous. We deduce by Proposition 2.16 that f i j A j {0, 1} is continuous. By assumption, A j is a connected subset of. We deduce by Proposition 6.5 that f i j is constant. This holds for all j J. Since j J A j, we deduce that f is constant. Corollary 8.6. Let (, O ) be a topological space, and let x. The connected component of x in is a connected subset of. Proof. Follows immediately from Proposition 8.5. Remark 8.7. Thus the connected component of x in a topological space (, O ) is the largest connected subset of which contains x. Terminology 8.8. Let be a set. A partition of is a set { j } j J of subsets of such that = j J j. Proposition 8.9. Let (, O ) be a topological space, and let {A x } x denote the set of connected components of. Then {A x } x defines a partition of. Proof. Since x A x, we have that x A x =. Suppose that x, x and that A x A x. We must prove that A x = A x. By Proposition 8.5 we then have that A x A x is connected. By definition of A x, we deduce that A x A x A x. Since we also have that A x A x A x, we conclude that A x A x = A x as required. Examples (1) A connected topological space (, O) has exactly one connected component, namely itself. (2) At the other extreme, let be a set and let O disc denote the discrete topology on. The connected components of (, O disc ) are the singleton sets {x} for x. Let us prove this. Suppose that A has more than one element. Let O A denote the subspace topology on A with respect to (, O disc ). For any a A, we have A = {a} \ {a}, and both {a} and A \ {a} = A ( \ {a}) are open in (A, O A ). Thus A is not a connected subset of (, O ). (3) Let (, O ) and (Y, O Y ) be connected topological spaces. Then ( Y, O Y ) has two connected components, namely and Y. For instance, by Example 7.16 we have that (T 2, O T 2) is connected, and hence that (T 2 T 2, O T 2 T 2) consists of two connected components. 70

5 By induction, we may similarly cook up examples of topological spaces with n connected components for any finite n. (4) Let the set of rational numbers Q be equipped with its subspace topology O Q with respect to (R, O R ). The connected components of (Q, O Q ) are exactly the singleton sets {q} for q Q. Let us prove this. For any A Q which contains at least two distinct rational numbers q and q, there is an irrational number r with q < r < q. Then ( ) ( ) A = A (, r) A (r, ). Since both A (, r) and A (r, ) are open in (Q, O Q ), we deduce that A is not a connected subset of Q. Terminology A topological space (, O) is totally disconnected if the connected components of (, O) are the singleton sets {x} for x. Remark By (2) of Examples 8.10, a set equipped with its discrete topology is totally disconnected. However, as (4) of Examples 8.10 demonstrates there are totally disconnected topological spaces (, O) for which O is not the discrete topology. Lemma Let (, O ) and (Y, O Y ) be topological spaces, and let f Y be a continuous map. Let x. Let A x denote the connected component of x in, and let B f(x) denote the connected component of f(x) in Y. Then f(a x ) B f(x). Proof. By Corollary 8.6, A x is a connected subset of (, O ). By Proposition 7.1 we deduce that f(a) is a connected subset of Y. Since x A, we have that f(x) f(a). By definition of B f(x), we conclude that f(a) B f(x). Proposition Let (, O ) and (Y, O Y ) be topological spaces. If = Y then there exists a bijection between the set of connected components of and the set of connected components of Y. 71

6 Proof. Given x, let A x denote the connected component of x in (, O ). Given y Y, let B y denote the connected component of y in (Y, O Y ). Let Γ denote the set of connected components of, and let Γ Y denote the set of connected components of Y. Let f be a homeomorphism. By definition of f as a homeomorphism, there is a continuous map Y g such that g f = id and f g = id Y. For any x we have by Lemma 8.13 that g(b f(x) ) A g f(x) = A x. Hence B f(x) = f ( g(b f(x) ) ) f(a x ). Moreover, by Lemma 8.13 we have that f(a x ) B(f x ). We deduce that f(a x ) = B f(x). Thus A x f(a x ) defines a map Y η Γ Γ Y. By an entirely analogous argument we have that g(b y ) = A g(y) for any y Y. Thus defines a map B y g(b y ) ζ Γ Y Γ. We have that ζ η = id Γ since for any x we have that g ( f(a x ) ) = g(b f(x) ) = A g f(x) = A x. Moreover we have that η ζ = id ΓY since for any y Y we have that f ( g(b y ) ) = f(a g(y) ) = B f g(y) = B y. 72

7 Observation Let (, O ) and (Y, O Y ) be topological spaces, and suppose that = Y. Then every connected component of, equipped with the subspace topology with respect to (, O ), is homeomorphic to a connected component of Y, equipped with the subspace topology with respect to (Y, O Y ). This follows from observations made during the proof of Proposition See the Exercise Sheet. 8.3 Using connectedness to distinguish between topological spaces II Examples (1) Let us regard the letter T as a subset of (R 2, O R 2). For example, we can let T = {(x, 1) R 2 x [ 1, 1]} {(0, y) R 2 y [0, 1]}. We equip T with the subspace topology O T. Let us also regard the letter I as a subset (R 2, O R 2). For example, we can let I = {(0, y) R 2 y [0, 1]}. We equip I with the subspace topology O I. By Examples 4.10 (5) we have that (I, O I ) is homeomorphic to the unit interval (I, O I ). Let us prove that (T, O T ) is not homeomorphic to (I, O I ). Let x be the point (0, 1) of T. 73

8 Then T \ {x} equipped with the subspace topology with respect to (T, O T ) has three connected components. However, the topological space obtained by removing a single point of I and equipping the resulting set with the subspace topology with respect to (I, O I ) has either one connected component or two connected components. We obtain one connected component if we remove one of the two end points of I. Removing the lower end point (0, 0) of I is depicted below. We obtain two connected components if we remove any point of I which is not an end point. In particular, it is not possible to remove a single point from (I, O I ) and obtain a topological space with three connected components. We deduce by Proposition 8.14 that T \ {x} is not homeomorphic to I \ {y} for any y I. We conclude that T is not homeomorphic to I by Proposition 8.1 (2) The circle S 1 is not homeomorphic to I. Indeed, equipping I \ {t} for 0 < t < 1 with the subspace topology with respect to (I, O I ) gives a topological space with two connected components. 74

9 Removing any point from S 1 and equipping the resulting set with the subspace topology with respect to (S 1, O S 1) gives a topological space with exactly one connected component. In particular, it is not possible to remove a single point from (I, O I ) and obtain a topological space with two connected components. We deduce by Proposition 8.14 that I \ {t} is not homeomorphic to S 1 \ {x} for any x S 1. We conclude that S 1 is not homeomorphic to I by Proposition 8.1. (3) Let us regard the letters Å and A as subsets of R 2, equipped with their respective subspace topologies with respect to (R 2, O R 2). By Proposition 8.14 we have that Å is not homeomorphic to A, since Å has two connected components, whilst A has one. 75

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