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4. Connectedness 4.1 Connectedness Let d be the usual metric on R2 , i.e. d(x, y) = (x1 , x2 ), y = (y1 , y2 ). Let p (x1 − y1 )2 + (x2 − y2 )2 , for x = X = {x ∈ R2 | d(x, 0) ≤ 1 or d(x, (4, 1)) ≤ 2} and Y = {x = (x1 , x2 ) ∈ R2 | − 1 ≤ x1 ≤ 1, −1 ≤ x2 ≤ 1}. So X is X =A S B and Y is Are X and Y homeomorphic? Both are Hausdorff (subspaces of R2 ) and both are compact since closed and bounded. However, it seems unlikely that they are homeomorphic since X is in two bits and Y is not. Definition A topological space X is disconnected if there exist two disjoint nonempty S T open sets U, V such that X = U V (and U V = ∅). A topological space is connected if it is not disconnected. 55 S (i) X as in the above example. X = A B and A is a closed subset of R2 T (its a closed ball). Hence X = X A is a closed subset of X (in the subspace topology). S Hence B = CX (A) is open in X. Similarly A = CX (B) is open. Hence X = A B, for Examples nonempty disjoint open sets A, B. Hence X is disconnected. (ii) Y (above) is connected (see later). (ii) Let X be a discrete topological space with at least 2 elements. Pick x ∈ X. Then S X = {x} CX ({x}) and so X is disconnected. (4.1a) Lemma A topological space X is disconnected if and only if there exists a proper subset (i.e. neither ∅ nor X) M, say, such that M is both open and closed. Proof (⇒) Assume X is disconnected. Then X = A S B with A, B disjoint and nonempty open sets. So A is a proper subset. Now A is closed and A = CX (B) is closed. Hence A is open and closed. (⇐) Suppose X has a subset M which is both open and closed. Then X = M expresses X as the union of disjoint open sets. Hence X is disconnected. S CX (M) (4.1b) A topological space X is disconnected if and only if there exist non-null, disjoint S closed sets A, B such that X = A B. Proof Exercise. (4.1c) Lemma A topological space X is connected if and only if every continuous map from X into a discrete topological space is constant. Proof (⇒) Suppose X connected and let f : X → D be a continuous map into a discrete S space D. Let x0 ∈ X and let α = f(x0 ). Put A = {α} and B = D\{α}. Then D = A B and hence X = f −1 D = f −1 A [ f −1 B. Now A, B are open in D (every subset of D is open) and hence f −1 A, f −1 B are open in X. T S Note also that f −1 A f −1 B = f −1 ∅ = ∅. Hence X = f −1 A f −1 B expresses X as the disjoint union of two open sets. Since X is connected, one of these must be empty. But x0 ∈ f −1 A so f −1 A 6= ∅ and hence f −1 B = ∅ and we have X = f −1 A. Thus f(x) ∈ A for every x ∈ X, i.e. f (x) = α for every x ∈ X. Hence f is the constant function. 56 (⇐) We suppose that every continuous map from f into a discrete space is constant and prove X is connected. Assume for a contradiction that X is not connected. Thus we may S write X = A B, with A, B nonempty and open. Let D{α, β} be the discrete space with two elements α, β. Define f : X → D by f (x) = ½ α, if x ∈ A; β, if x ∈ B. Check that f is continuous and hence nonconstant. This is a contradiction, so X is connected. (4.1d) Proposition The image of a connected space is connected, i.e. if f : X → Y is continuous and X is connected then f(X) is connected. Proof (i) Let Z = f (X) = Im(f ) and let g : X → Z be the restriction of f , i.e. g(x) = f (x) for all x ∈ X (the only difference between f and g is that the codomain of f is Y and the codomain of g is Z). “Direct Proof” If Z is disconnected then we can write Z = U S V for disjoint nonempty S S open sets U, V but then we would have X = g −1 Z = g −1 (U V ) = g−1 U g −1 V , a disjoint union of nonempty open sets. But X is connected so this can’t happen. So Z is connected. “Another Proof” Let h : Z → D be a continuous map, with D discrete. Then h ◦ g : X → D is constant (since X is connected, see (4.1c)). But then if z1 , z2 ∈ Z we can write z1 = g(x1 ), z2 = g(x2 ) for some x1 , x2 ∈ X. So h(z1 ) = h(g(x1 )) = h(g(x2 )) = h(z2 ) (since h ◦ g is constant). But then h(z1 ) = h(z2 ) so h is constant and Z is connected, by (4.1c). (4.1e) Corollary Proof Connectedness is a topological property. We must show that if X is connected and X is homeomorphic to Y then Y is connected. Assume X is connected and X is homeomorphic to Y . Thus there is a homeomorphism f : X → Y . The map f is in particular a surjective (onto) continuous map. By (4.1e), Y = f(X) is connected. Example Let X be an infinite set regarded as a topological space with the cofinite topology. Then X is connected. 57 S T If not we can write X = A B with A, B nonempty and open and with A B = ∅. But S then X = CX (A) CX (B), a union of two finite sets and hence finite. But X is infinite so this is impossible. Hence X is connected. We shall need the following general result. (4.1f ) Lemma Let X be a topological space and let Y be a subspace of X. Let A be a subset of Y (so A ⊂ Y ⊂ X). (i) A is closed in Y if and only if it has the form K T Y for some closed set K of X. (ii) If A is open in Y (in the subspace topology on Y ) and Y is an open subset of X then S is an open subset of X. (iii) If A is closed in Y (in the subspace topology on Y ) and Y is a closed subset of X then A is a closed subset of X. T (i) Suppose A is closed in Y (in the subspace topology). Then CY (A) = Y U , S for some open set U in X. Let K be the closed set CX (U ). Then X = U K and so T S T T T Y = (Y U ) (Y K). Thus Y K is the complement in Y of CY (A) = Y U . But T the complement of CY (A) in Y is A, i.e. we have A = Y K. T (ii) We have A = U Y , with U open in X. So A is a union of two open sets in X hence Proof open in X. (iii) We have A = Y T K for some closed set K in X, by (i). Thus A is the intersection of two closed sets in X and hence A is closed in X. (4.1g) Theorem Proof The closed interval [a, b] is connected (for a < b). [a, b] is homeomorphic to [0, 1] so by (4.1e) it suffices to prove [0, 1] is connected. Let A be a nonempty subset of [0, 1] which is open and closed (in the subspace topology on [0, 1]). Let α be the greatest lower bound of A. Now A is closed in R (by (4.1f)(iii) since A is closed in [0, 1] and [0, 1] is closed in R). So α ∈ A by (3.2l). We claim that in fact T α = 0. Suppose not. Since A is open in [0, 1] we have A = [0, 1] U for some open set U in R. So we have (α − r, α + r) ⊂ U for some r > 0. Choose s > 0 such that s < min{r, α}. T Then (α − s, α] ⊂ U [0, 1] = A. But then α − s/2 ∈ A, contradicting the fact that α is a lower bound of A. Hence α = 0. We have shown that if A is any nonempty subset of [0, 1] which is open and closed then 0 ∈ A. But now if [0, 1] is disconnected then we can 58 write [0, 1] = A S B for nonempty disjoint subsets A, B which are open and closed. But T then 0 ∈ A and 0 ∈ B contradicting the fact that A B = ∅. Hence [0, 1] is connected. (4.1h) Corollary (The Intermediate Value Theorem) Let f : [a, b] → R be con- tinuous and suppose that f (a) < k < f (b). Then there exists some c ∈ (a, b) such that f (c) = k. Proof Suppose for a contradiction that there is no such c. Let U = (∞, k), V = (k, ∞). Then we have [a, b] = f −1 U [ f −1 V and this is a disjoint union of nonempty open subsets of [a, b]. But [a, b] is connected so this is impossible. (4.1i) Corollary (a, b) is connected and R is connected. (a, b) is homeomorphic to R so it is enough to show that R is connected. Suppose S not. Then we have R = U V , for U, V open, nonempty disjoint subsets of R. Pick x ∈ U Proof and y ∈ V . We may suppose that x < y (otherwise relabel U and V ). Hence we have \ \ [ \ [x, y] = [x, y] R = ([x, y] U) ([x, y] V ) and this disconnects [x, y] contrary to (4.1g). (4.1j) Corollary Proof [a, b) is connected (for real numbers a < b). [a, b) is homeomorphic to [0, 1) so it is enough to prove [0, 1) connected. Define f : (−1, 1) → [0, 1) by f(x) = ½ x, if x ≥ 0; 0, if x ≤ 0. Then f is continuous and onto so [0, 1) = f(−1, 1) is connected by (4.1i) and (4.1e). We still haven’t resolved the question : Is [0, 1) homeomorphic to (0, 1)? Both spaces are Hausdorff, neither is compact, both are connected. However, removing {0} from [0, 1) leaves a connected space whereas removing any point from (0, 1) disconnects the space. (4.1k) Suppose that X and Y are homeomorphic spaces and that x1 , . . . , xn are distinct points of X. Then there exist distinct points y1 , . . . , yn of Y such that X\{x1 , . . . , xn } is homeomorphic to Y \{y1 , . . . , yn }. 59 Proof Let f : X → Y be a homeomorphism with inverse g : Y → X. Let yi = f(xi ), 1 ≤ i ≤ n. Let X0 = X\{x1 , . . . , xn } and let Y0 = Y \{y1 , . . . , yn }. Then f0 : X0 → Y0 , the restriction of f , is continuous, (1.4b), and is a bijection with continuous inverse g0 : Y0 → X0 , the restriction of g. Hence f0 is a homeomorphism (and X0 and Y0 are homeomorphic spaces). (4.1l) [0, 1) is not homeomorphic to (0, 1). Proof Well suppose these spaces are homeomorphic. In the Lemma take X = [0, 1), Y = (0, 1), take n = 1 and x1 = 0. Then X0 = X\{0} = (0, 1) is connected. But S Y0 = Y \{y1 } = (0, y1 ) (y1 , 1) is disconnected. These spaces are not homeomorphic and so X is not homeomorphic to Y . We are now going to show that Rn is connected. We do this by proving generally that if X and Y are connected then X × Y is connected. (4.1m) Let X, Y be topological spaces and let y0 ∈ Y . The map α : X → X × Y , given by α(x) = (x, y0 ) is continuous. Proof By (3.3e), α is continuous provided that p ◦ α : X → X and q ◦ α : X → Y are continuous (where p : X × Y → X and q : X × Y → Y are the projections). But we p ◦ α : X → X is the identity, which is continuous, and q ◦ α : X → Y is the constant map with value y0 , and this map is continuous. Hence α is continuous. (4.1n) Let X, Y, Z be topological spaces and let φ : X × Y → Z be a continuous map. For each y ∈ Y the map φy : X → Z, given by φy (x) = φ(x, y) is continuous. Proof We have φy (x) = φ(x, y) = φ(α(x)), where α : X → X × Y is the map α(x) = (x, y). Thus φy = φ ◦ α, a composite of continuous maps and hence continuous. (4.1o) Proposition Proof If X, Y are connected then X × Y is connected. By (4.1c) it suffices to show that every continuous map φ : X ×Y → D is constant, where D is a discrete space. For each y ∈ Y the map φy : X → D, given by φy (x) = φ(x, y) is constant, by (4.1c), since X is connected. In other words, we have φ(x, y) = φ(x0 , y) 60 (1) for all x, x0 ∈ X and y ∈ Y . Similarly the function φx : Y → D, given by φx (y) = φ(x, y), is constant, in other words φ(x, y) = (x, y 0 ) (2) for all x ∈ X, y, y 0 ∈ Y . Now by (1) and (2), for (x, y), (x0 , y 0 ) ∈ X × Y we have φ(x, y) = φ(x0 , y) = φ(x0 , y 0 ) so that φ is constant. Hence X × Y is connected. (4.1p) Rn is connected, for n > 0. Proof We have already shown that R = R1 is connected. For n > 1 we have that Rn is homeomorphic to R × Rn−1 . So the result follows by induction on n, and (4.1o). 61