Math 541 Fall 2008 Connectivity Transition from Math 453/503 to Math 541 Ross E. Staffeldt-August 2008 Closed sets We have been operating at a fundamental level at which a topological space is a set together with a distinguished family of subsets. This family of subsets is called a topology for the given set. Definition. Let X be a set. A collection of subsets U = {U} of X is called a topology for X, if it satisfies the following properties: 1. X ∈ U and ∅ ∈ U. 2. The intersection of finitely many members of U is again in U. 3. The union of arbitrarily many members of U is again in U. A topological space is a set together with a topology. The family U is generally called the family of open sets of the topological space. We have also observed that a topology is also determined by specifying the family of closed sets, namely, the family of sets whose complements are open. Proposition 1. Let X be a set with a topology U. Define a family of subsets C = {U c}, where U c denotes X − U, the complement of U in X. Then the family C has the following properties. 1. X ∈ C and ∅ ∈ C. 2. The union of finitely many members of C is again in C. 3. The intersection of arbitrarily many members of C is again in C. The family C is generally called the family of closed sets of the topological space X. In some respects it is a matter of convenience whether or not one chooses to work with closed sets or open sets. However, for connectivity, we need to develop a few more properties of closed sets. First we define an operation called closure. Definition. Let X be a topological space and let A ⊂ X. Define A = \ C C closed, C ⊃ A We call A the closure of A. 1 Thus, A is the smallest closed set in X containing A, and, if C is closed, C = C. Often, formal properties of the closure operation follow easily from this conceptual defintion. For the sake of calculation of set closures in examples, we characterize the elements of a closed set in terms of open sets. Proposition 2. Let X be a topological space and let A ⊂ X. A point x ∈ X is in A if, and only if, for every open set U with x ∈ U, U ∩ A =6 ∅. Proof. One way to see this result is to convert it to the contrapositive equivalent. A point x ∈ X is not in A if, and only if, there is an open set U, x ∈ U, U ∩ A = ∅. Now, if x∈ / A, then there is a closed set C, C ⊃ A, and x∈ / C. Then x ∈ U = X − C, which is an open set, and, since A ⊂ C, U ∩ A = ∅. On the other hand, if, given x, there is an open set U such that U ∩A = ∅, then C = X−U is a closed set, A ⊂ C, and x∈ / C. So x∈ / A. Here is a characterization of continuous functions in terms of the closure operation that is occasionally useful. In fact, Hatcher uses this on page 35 in his calculation of the fundamental group of high-dimensional spheres. Proposition 3. Let X and Y be topological spaces and let f : X → Y be a function. The following conditions are equivalent. 1. The function f is continuous. 2. For any B ⊂ X, f(B) ⊂ f(B). Proof. If f is continuous, we know that the preimage of a closed subset of Y is closed. In particular, for any B ⊂ X, f −1(f(B)) is closed. Obviously, B ⊂ f −1(f(B)). Since B is the smallest closed set containing B, we have B ⊂ f −1(f(B)), which is a restatement of condition 2. On the other hand, assume condition 2 holds, and let V ⊂ Y be open. Let B = X − f −1(V )= f −1(Y − V ). By condition 2, we have f(B) ⊂ f(f −1(Y − V )) ⊂ Y − V = Y − V, since V is open. Equivalently, B ⊂ f −1(Y − V )= B. Since we always have B ⊂ B, we have, in fact, B = B. That is, B = f −1(Y − V )= X − f −1(V ) is closed. In turn, f −1(V ) is open, which means that f is continuous. This characterization of continuity reflects the intuitive idea that a continuous function is one that preserves limits. Here we have that B and its limit points are carried into f(B) and its limit points, if f is continuous. There is a dual notion, called the (set-theoretic) interior of a set, defined as ◦ IntA =A= [ U U open, U ⊂ A and described as the largest open subset of A. We won’t use this very often. 2 Topological connectedness Essentially, we start this section by defining what we mean for a subset of a topological space to be disconnected. Roughly speaking, we know how points are separated by open sets in Hausdorff space, so we try to extend the separation idea to subsets. Definition. Let X be a topological space, A ⊂ X a subset, possibly X itself. A separation of A is a pair U, V of open subsets of X, such that 1. A ⊂ U ∪ V . 2. U ∩ V = ∅. 3. neither A ∩ U nor A ∩ V is empty. Once we know when parts of a subset are separated, then, by definition, we also “know” when a subset is connected. This may be true in a strict logical sense, but it may take some effort to accept some of the examples. Definition. Let X be a topological space and let A be a subset. The set A is called connected if it has no separation. That is, A is connected, if, for every pair of open subsets U, V such that 1. A ⊂ U ∪ V . 2. U ∩ V = ∅. 3. either A ∩ U or A ∩ V is empty. The empty subset is always connected. Take A = X. If X has a separation X = U ∪ V , then U and V are both open and closed in X. On the other hand, if X has no separation, then ∅ and X are the only subsets of X that are both open and closed. Sometimes we exploit this fact of connected spaces to prove that a property holds everywhere in a connected topological space by proving that set of points having the property is both open and closed. We will see examples later. The first part of the following proposition says connectedness is a topological property, preserved by continuous functions. The second part is a characterization of connected sets in terms of maps to discrete spaces. Proposition 4. Let f : X → Y be a map and let A be a connected subset of X. Then f(A) is a connected subset of Y . Let f : X → {0, 1} be a map, where {0, 1} has the discrete topology. If A ⊂ X is connected, then the map f|A is constant. Proposition 5. If A is a connected subset of X and B satisfies A ⊂ B ⊂ A, then B is also connected. Proof. Let U, V be open, suppose B ⊂ U ∪ V and U ∩ V = ∅. If x ∈ B, then x ∈ A, and for every open subset W containing x, W ∩ A =6 ∅. Now, if x ∈ B, either x ∈ U or x ∈ V . If x ∈ U, then U ∩ A =6 ∅. But A is connected, and has no separation, so we must have A ∩ V = ∅. That is, A ⊂ X − V , which implies B ⊂ A ⊂ X − V . Thus, B ∩ V = ∅. We conclude that B has no separation, and, so, B is connected. 3 Proposition 6. Let X be a topological space and let {Ai : i ∈ I} be a family of connected subsets. If ∩i∈I Ai =6 ∅, then ∪i∈I Ai is connected as well. Proof. Write ∪i∈I Ai = U ∪ V , where U and V are open in X. Choose an index i0. We have Ai0 ⊂ U ∪ V , and Ai0 is connected. Then all of Ai0 is in one or the other of the open sets, so we may assume Ai0 ⊂ U. Now let x ∈ ∩i∈I Ai. We have x ∈ Ai0 ⊂ U, so x ∈ U. Then, for all i ∈ I, Ai ∩ U =6 ∅. Since each Ai is connected, we must have Ai ⊂ U and Ai ∩ V = ∅. Then ∪i∈I Ai ⊂ U and ∪i∈I Ai ∩ V = ∅. In other words, ∪i∈I Ai has no separation, so it’s connected. So far we don’t know any examples of topologically connected spaces, except for trivial ones. The following proposition cures this. The essential ingredients of the proof are that the real numbers have the least upper bound property and that between any two real numbers there is another. Proposition 7. Let I = [a, b] be an interval in R. Then I is connected. Proof. Let U, V be open subsets in R that provide a separation of I. That is, suppose also I ⊂ U ∪ V and U ∩ V = ∅. Without loss of generality suppose a ∈ U. Let S = {x ∈ I : The interval [a, x] is in U.} The set S is non-empty, because a ∈ S, and S is clearly bounded above by b. Therefore, S has a least upper bound c. Now we want to show two things: First, that c ∈ S and, second, that c = b. Once we have these facts, I ⊂ U and and I ∩ V = ∅, so our “separation” was not real. Toward showing these facts, we first show c ∈ U. If not, then c ∈ V , and we obtain a contradiction, as follows.