Math 541 Fall 2008 Connectivity Transition from Math 453/503 to Math 541 Ross E

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. Since V is open, there is an interval (c−ǫ, c+ǫ) ⊂ V . Then the ǫ ǫ interval [a, c− 2 ] 6⊂ U, and no interval [a, y] with c− 2 ≤ yconnected space.

Examples Proposition 8. The set R is connected. So are open intervals (c,d), half-lines (c, ∞) and (−∞,c) and rays [c, ∞) and (−∞,c]. ∞ Proof. For each n ≥ 1, the interval [−n, n] is connected, by proposition 7. Now ∩n=1[−n, n] is non-empty, containing the point 0, for example. Therefore, by proposition 6, ∞ [[−n, n]= R n=1 is connected. The other examples are handled similarly.

4 Proposition 9. Let X be a connected space, and let f : X → R be a map. If f(x) < 0 and f(y) > 0, then there is z ∈ X such that f(z)=0. Proof. Consider U = f −1((−∞, 0)) and V = f −1((0, ∞)). We have x ∈ U and y ∈ V , so neither is empty. Both are open in X, because f is continuous. If X = U ∪ V , then X has a separation, contradicting the hypothesis that X is connected. Consequently, there must be z ∈ X such that f(z)=0. Definition. Define a relation ∼ on X by x ∼ y if and only if there is a connected A ⊂ X such that x, y ∈ A. Proposition 10. This relation is an equivalence relation. Therefore X is partitioned into disjoint subsets, each of which is called a connected component of X. Proof. Proof of transitivity of ∼ requires proposition 6, but is otherwise straightforward. A somewhat more intuitive definition of connectedness, based on the idea of mapping connected test spaces into the space under investigation is called path-connectedness Definition. Let X be a topological space. We say X is path-connected, if, for any pair of points x, y ∈ X, there is a map f : I → X such that f(0) = x and f(1) = y. We call f a path in X from x to y. Alternatively, given a topological space X, define an relation ∼p on X by x ∼p y if there is a path f in X from x to y. The proof the following proposition is an exercise.

Proposition 11. For any space X, the relation ∼p is an equivalence relation. The space X is now partitioned into equivalence classes, called the path-components. Here are two examples that show connectedness and path-connectedness are diﬀerent notions. Example 1. Let I = [0, 1], as usual, and let A = {1, 1/2, 1/3, 1/4,...} ⊂ I. Deﬁne X = I × {0}∪A×I∪{(0, 1)}. X is called the deleted comb space. The space Y = I×{0}∪A×I∪{0}×I is called the comb space. Make a sketch showing both Y and X. Clearly Y is connected, by multiple applications of proposition 6. We see that X is connected, because X ⊂ I ×{0} ∪ A × I, and I ×{0} ∪ A × I is connected. Now we see that X is not path-connected. To show that X is not path-connected, let f : I → X be a path with f(0) = P = (0, 1). Obviously, f −1(P ) is closed in I, since f is continuous. We will also show that f −1(P ) is open in I. Once we have this additional fact we argue as follows. Since I is connected, its only subsets that are both open and closed are the empty set and I itself. Since f −1(P ) is not empty, f −1(P ) = I and the path f starting at P is constant. To see that f −1(P ) is open, let V be an open ball around P that does not touch the x- axis. f −1(V ) is open in I. Then, for any t ∈ f −1(V ) there is a small interval J =(t − ǫ, t + ǫ) such that f(J) ⊂ V ∩ X. Now let (1/n,y) ∈ V ∩ X. Choose r, 1/(n + 1)

5 define HL =(−∞,r) × R and HR =(r, ∞) × R. We note that P ∈ HL and (1/n,y) ∈ HR. Since the ball V does not touch the x-axis, we must have that f(J) ⊂ HL. This is true because J is connected, its image under f is connected, and the only way to reach HR in X is by taking a path through (r, 0) on the x-axis, which is specifically excluded from V . Thus, for any point (1/n,y) ∈ V , ( (1/n,y) =6 P !), (1/n,y) is not in the image f(J). We conclude that f −1(V )= f −1(P ) for an open ball V as above. Thus, f −1(P ) is also open. Example 2. Let Γ = {(x, sin(1/x): 0 < x ≤ 1} be a part of the graph of the function f(x) = sin(1/x), defined for x =06 . Let W =Γ ∪{0} × [−1, 1] ⊂ R2. It is easy to verify that W = Γ, so W is connected by proposition 5. An argument similar to the one proving the deleted comb space is not path-connected shows that W is not path-connected. This example is sometimes called the “Warsaw sine curve” or the “Polish sine curve”.

Product spaces and connectedness We will see that it is straightforward to demonstrate that a ﬁnite product of connected spaces is connected. For inﬁnite products, the result is not so obvious. Proposition 12. Let X and Y be connected non-empty topological spaces. Then X × Y is also connected.

Proof. We make two applications of proposition 6. Choose x0 ∈ X and let y ∈ Y . The subsets Ay = X ×{y} and B = {x0} × Y are connected, since they are homeomorphic to X and Y , respectively. And Ay and B have a point in common. So, by proposition 6, Cy = Ay ∪ B is connected. Now consider the family of subsets {Cy : y ∈ Y }. Each of these is connected, and ∩Cy = B is not empty. Therefore, the union ∪Cy = X × Y is connected. Inductively, any ﬁnite product of connected spaces is connected.

Proposition 13. Let {Xα : α ∈ A} be a family of non empty connected spaces indexed by the inﬁnite set A. Then Qα∈A Xα is connected, where Qα∈A Xα is given the product topology. Proof. This result requires a few minutes of consideration of the deﬁnition of open sets in the product topology. Recall that the basic open sets, whose unions are the general open sets of the product topology, are sets

U = Y Uα where Uα is open in Xα and where Uα = Xα for all but ﬁnitely many α. α∈A

Back to the proof of the proposition, we know that α∈F Xα, is connected for any finite 0 Q subset F of A. Now, choose a point (xα) ∈ Qα∈A Xα. For each finite subset F of A we have a copy of α∈F Xα sitting in side of the full product α∈A Xα as the set of points (xα) Q 0 Q defined by the equations xα = xα, for α∈ / F . That is, we restrict the coordinates, if the coordinate index is not in F . All subsets constructed this way are connected, by the finite 0 case, and these subsets have the point (xα) in common. Therefore the union W is connected. However, we are not yet finished. The union W is not all of the full product. Indeed, it consists of all points of the product for which at most finitely many of the coordinates differ 0 from those of the “basepoint” (xα). At this point we really need the definition of open sets

6 in the product topology. Given the description of the basic open sets containing any point (xα), it is easy to see that any of these basic open sets has non-empty intersection with W . But this says that, in the product topology the closure W = Qα∈A Xα, the full product. By proposition 5, Qα∈A Xα is connected.

Local path-connectedness: We have seen by means of the deleted comb space example and the Warsaw sine curve that connectedness and path-connectedness are distinct notions. However, because the interval is connected, every path-connected space is connected. From the properties of equivalence classes in general, it follows that every connected component of a topological space is a union of path components. However, path components are not necessarily open or closed subsets of a connected component. Just identify the path components in the two examples to see this. There is a useful condition that we can impose to make the situation a little better. That is the condition of being “locally path connected”. Read the definition very carefully; it may not coincide with your first idea. Definition. We say a space X is locally path-connected, if for every point x, and for every open set V containing x, there is a path-connected open set U such that x ∈ U ⊂ V . This says that every point has path-connected neighborhoods that are arbitrarily small, which is more than saying every point has a path-connected neighborhood. This is not the only way to define “locally path-connected”; see Hatcher’s remarks at the top of page 62. Proposition 14. Let X be a connected space that is locally path-connected. Then X is path-connected.

Proof. Let x0 ∈ X be a point, and deﬁne

U0 = {x ∈ X : There is a path in X from x0 to x.}

We will show that U0, which is not empty, is both open and closed in X. Since X is connected, U0 must be all of X. Let x ∈ U0. The set X is an open set containing x, so there is a path-connected open set U, such that x ∈ U ⊂ X. Following a path from x0 to x by a path from x to any other point y ∈ U shows that, if x ∈ U0, then there is an open subset U such that x ∈ U ⊂ U0. Thus, U0 is open. Now let y ∈ U0. We show that y ∈ U0. Recall the characterization of points of a closure given in proposition 2. If y ∈ U0, then for any open set V containing y, V ∩ U0 =6 ∅. By local path-connectedness, any open set V containing y contains a path-connected open set U, which also contains y. And U ∩ U0 =6 ∅. So, if z ∈ U ∩ U0, we can connect x0 to z, and we can connect z to y by a path in U. Joining the paths together shows that y ∈ U0. Thus, U0 is also closed. As indicated, U0 is a non-empty subset of X that is both open and closed. Since X is connected, U0 = X. This argument can be extended to show that if a space is locally path-connected, then the path components coincide with the components. Moreover, each path component is an open subset of the whole space.