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5. Connectedness We Begin Our Introduction to Topology with the Study of Connectedness—Traditionally the Only Topic Studied in Both Analytic and Algebraic Topology

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5. Connectedness We Begin Our Introduction to Topology with the Study of Connectedness—Traditionally the Only Topic Studied in Both Analytic and Algebraic Topology 5. Connectedness We begin our introduction to topology with the study of connectedness—traditionally the only topic studied in both analytic and algebraic topology. C. T. C. Wall, 1972 The property at the heart of certain key results in analysis is connectedness. The definition, however, applies to any topological space. Definition 5.1. A space X is disconnected by a separation U, V if U and V are open, non-empty, and disjoint (U V = ) subsets of X with X = {U V}. If no separation of the space X exists, then X is connected∩ ∅ . ∪ Notice that V = X U is closed and likewise U is closed. A subset that is both open and closed is sometimes− called clopen. Closure leads to an equivalent condition. Theorem 5.2. A space X is connected if and only if whenever X = A B with A, B, non-empty, then A (cls B) = or (cls A) B = . ∪ ∩ ∅ ∩ ∅ Proof: If A (cls B)= and (cls A) B = , then, since A B = X, it will follow that X cls A, X∩ cls B is a∅ separation of∩X. To∅ see this, consider∪x (X cls A) (X cls B); then{ −x/cls A−and x/} cls B. But then x/cls A cls B = X, a∈ contradiction.− ∩ Therefore− (X cls∈ A) (X cls∈ B)= . Thus we have∈ a separation.∪ − ∩ − ∅ Conversely, if U, V is a separation of X, let A = X V = U and B = X U = V . Since U and V are{ open,} A and B are closed. Then X −= U V = A B.− However, A cls B = A B = U V = . ∪ ∪ ∩ ∩ ∩ ∅ ♦ Example: The canonical connected space is the unit interval [0, 1] (R, usual). To see this, suppose U, V is a separation of [0, 1]. Suppose that 0 U.⊂ Let c = sup 0 t 1 [0,t] U .{ If c }= 1, then V = , so suppose c<1. Since∈c [0, 1], c U or{ c≤ V≤. If |c U⊂, then} there exists an >0,∅ such that (c , c + ) U∈ and there∈ is a natural∈ number∈ N>1 such that c<c+(/N) < 1. But− this contradicts⊂ c being a supremum since c +(/N) [0, 1]. If c V , then there exists a δ>0, such that (c δ, c + δ) V . ∈ ∈ − ⊂ For some N > 1, c +(δ/N) < 1 and so (c (δ/N),c+(δ/N)) does not meet U so c could not be a supremum. Since the set 0 −t 1 [0,t] U is nonempty and bounded, it has a supremum. It follows that c = 1{ and≤ so≤ [0,|1] is connected.⊂ } ♦ Is connectedness a topological property? In fact more is true: Theorem 5.3. If f: X Y is continuous and X is connected, then f(X), the image of X in Y , is connected. → Proof: Suppose f(X) has a separation. It would be of the form U f(X),V f(X) 1 1 { ∩ ∩ } with U and V open in Y . Consider the open sets f − (U),f− (V ) . Since U f(X) = , 1 1 { } ∩ ∅ we have f − (U) = and similarly f − (V ) = . Since U f(X) V f(X)=f(X), we 1 1 ∅ ∅1 1∩ ∪ ∩ have f − (U) f − (V )=X. Finally, if x f − (U) f − (V ), then f(x) U f(X) and ∪ ∈ ∩ 1 1∈ ∩ f(x) V f(X). But (U f(X)) (V f(X)) = . Thus f − (U) f − (V )= and X is disconnected.∈ ∩ ∩ ∩ ∩ ∅ ∩ ∅ ♦ Corollary 5.4. Connectedness is a topological property. 1 Example: Suppose a<b, then there is a homeomorphism h: [0, 1] [a, b] given by h(t)= a +(b a)t. Thus, every [a, b] is connected. → − A subspace A of a space X is disconnected when there are open sets U and V in X for which A U = = A V , and A U V , and A U V = . Notice that U V can be nonempty∩ in X ,∅ but A∩ U V =⊂. ∪ ∩ ∩ ∅ ∩ ∩ ∩ ∅ Lemma 5.5. If A i J is a collection of connected subspaces of a space X with { i | ∈ } Ai = , then Ai is connected. i J i J ∈ ∅ ∈ Proof: Suppose U and V are open subsets of X with Ai U V and Ai U V = i J i J ∈ ⊂ ∪ ∈ ∩ ∩ . Let p Aj, then p Aj for all j J. Suppose that p U. Since U and V are j J ∅ ∈ ∈ ∈ ∈ ∈ open, U Aj,V Aj would separate Aj if they were both non-empty. Since Aj is a connected{ ∩ subspace,∩ this} cannot happen, and so A U. Since j J was arbitrary, we j ⊂ ∈ can argue in this way to show A U and hence, U, V is not a separation. i ⊂ { } ♦ Example: Given an open interval (a, b) R, let N>2/(b a). Then we can write ⊂ − (a, b)= [a + 1 ,b 1 ], a union with nonempty intersection. It follows from the n N n n ≥ − lemma that (a, b) is connected. Also R = [ n, n] and so R is connected. n>0 − Let us review our constructions to see how they respect connectedness. A subset A of a space X is connected if it is connected in the subspace topology. Subspaces do not generally inherit connectedness; for example, R is connected but [0, 1] (2, 3) R is disconnected. A quotient of a connected space, however, is connected since∪ it is⊂ the continuous image of the connected space. How about products? Proposition 5.6. If X and Y are connected spaces, then X Y is connected. × Proof: Let x0 and y0 be points in X and Y , respectively. In the exercises of Chapter 4 we can prove that the inclusions jx0 : Y X Y , given by jx0 (y)=(x0,y) and iy0 : X X Y , given by i (x)=(x, y ) are continuous;→ × hence j (Y ) and i (X) are connected in→X ×Y . y0 0 x0 y0 × Furthermore, jx0 (Y ) iy0 (X)=(x0,y0) so iy0 (X) jx0 (Y ) is connected. We express X Y as a union of similar∩ connected subsets: ∪ × X Y = iy (X) jx(Y ), x X 0 × ∈ ∪ a union with intersection given by iy (X) jx(Y )=iy (X), which is connected. By x X 0 ∪ 0 Lemma 5.5, X Y is connected. ∈ × ♦ n Example: By induction, R is connected for all n. Wrapping R onto S1 by w: R S1, given by w(γ) = (cos(2πγ), sin(2πγ)), shows that S1 is connected and so is the→ torus S1 S1. We can also prove this by arguing that [0, 1] [0, 1] is connected and the torus is × 2 × 2 1 a quotient of [0, 1] [0, 1]. It also follows that S is connected—S ∼= ΣS , a quotient of S1 [0, 1]. By induction× and Theorem 4.19, Sn is connected for all n 1. × ≥ A characterization of the connected subspaces of R has some interesting corollaries. Proposition 5.7. If W (R, usual) is connected, then W =(a, b), [a, b), (a, b], or [a, b] for a b . ⊂ −∞ ≤ ≤ ≤∞ 2 Proof: Suppose c, d W with c<d. We show [c, d] W , that is, that W is convex. (In other words, if c, d are∈ both in W , then (1 t)c+td W⊂ for all 0 t 1.) Otherwise there exists a value r , with c<r <dand r −/ W . Then∈ U =( ≤,r) ≤W , V = W (r, ) 0 0 0 ∈ −∞ ∩ ∩ ∞ is a separation of W . We leave it to the reader to show that a convex subset of R must be an open, closed, or half-open interval. ♦ Intermediate Value Theorem. If f:[a, b] R is a continuous function and f(a) < c<f(b) or f(a) >c>f(b), then there is a value→ x [a, b] with f(x )=c. 0 ∈ 0 Proof: Since f is continuous, f([a, b]) is a connected subset of R. Furthermore, this subset contains f(a) and f(b). By Proposition 5.7, the interval between f(a) and f(b), which includes c, lies in the image of [a, b], and so there is a value x [a, b] with f(x )=c. 0 ∈ 0 ♦ 1 1 Corollary 5.8. Suppose g: S R is continuous. Then there is a point x0 S with g(x )=g( x ). → ∈ 0 − 0 Proof: Defineg ˜ : S1 R byg ˜(x)=g(x) g( x). Wrap [0, 1] onto S1 by w(t)= (cos(2πt), sin(2πt)). Then→ w(0) = w(1/2). − − Let F =˜g w. It follows that− ◦ F (0) =g ˜(w(0)) = g(w(0)) g( w(0)) − − = [g( w(0)) g(w(0))] − − − = [g(w(1/2)) g( w(1/2))] − − − = F (1/2). − If F (0) > 0, then F (1/2) < 0 and since F is continuous, it must take the value 0 for some t between 0 and 1/2. Similarly for F (0) < 0. If F (t) = 0, then let x0 = w(t) and g(x )=g( x ). 0 − 0 ♦ Here is a whimsical interpretation of this result: There are two antipodal points on the equator at which the temperatures are exactly the same. in later chapters we will n generalize this result to continuous functions Sn R . It is the connectedness of the domain of a continuous→ real-valued function that leads to the Intermediate Value Theorem (IVT). Furthermore, the IVT can be used to prove that an odd-degree real polynomial has a real root (see the Exercises). Toward a proof of the Fundamental Theorem of Algebra, that every polynomial with complex coefficients has a complex root (see [Uspensky] and [Fine-Rosenberger]), we present an argument given by Gauss, in which connectedness plays a key role.
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