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4. Connectedness 4.1 Connectedness Let D Be the Usual Metric on R 2, Ie 4. Connectedness 4.1 Connectedness Let d be the usual metric on 2, i.e. d(x, y) = (x y )2 + (x y )2, for x = R 1 − 1 2 − 2 (x1, x2), y = (y1, y2). Let p X = x 2 d(x, 0) 1 or d(x, (4, 1)) 2 { ∈ R | ≤ ≤ } and Y = x = (x , x ) 2 1 x 1, 1 x 1 . { 1 2 ∈ R | − ≤ 1 ≤ − ≤ 2 ≤ } So X is X = A B and Y is S 2 Are X and Y homeomorphic? Both are Hausdorff (subspaces of R ) 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 open sets U, V such that X = U V (and U V = ). A topological space is connected if ∅ it is not disconnected. S T 55 2 Examples (i) X as in the above example. X = A B and A is a closed subset of R (its a closed ball). Hence X = X A is a closed subsetS of X (in the subspace topology). Hence B = CX (A) is open in X.T Similarly A = CX (B) is open. Hence X = A B, for nonempty disjoint open sets A, B. Hence X is disconnected. S (ii) Y (above) is connected (see later). (ii) Let X be a discrete topological space with at least 2 elements. Pick x X. Then ∈ X = x C ( x ) and so X is disconnected. { } X { } S (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 B with A, B disjoint and ⇒ nonempty open sets. So A is a proper subset. Now A isS 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 C (M) ⇐ X expresses X as the union of disjoint open sets. Hence X is disconnected. S (4.1b) A topological space X is disconnected if and only if there exist non-null, disjoint closed sets A, B such that X = A B. S 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 ⇒ → space D. Let x X and let α = f(x ). Put A = α and B = D α . Then D = A B 0 ∈ 0 { } \{ } and hence S 1 1 1 X = f − D = f − A f − B. [ 1 1 Now A, B are open in D (every subset of D is open) and hence f − A, f − B are open in X. 1 1 1 1 1 Note also that f − A f − B = f − = . Hence X = f − A f − B expresses X as the ∅ ∅ disjoint union of twoT open sets. Since X is connected, one ofS these must be empty. But 1 1 1 1 x f − A so f − A = and hence f − B = and we have X = f − A. Thus f(x) A for 0 ∈ 6 ∅ ∅ ∈ 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 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 S → α, if x A; f(x) = β, if x ∈ B. ½ ∈ 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 V for disjoint nonempty 1 1 1 1 open sets U, V but then we would have X = g− Z = g− (US V ) = g− U g− V , a disjoint union of nonempty open sets. But X is connected so thisS can’t happen.S 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 z , z Z we can write → 1 2 ∈ z = g(x ), z = g(x ) for some x , x X. So h(z ) = h(g(x )) = h(g(x )) = h(z ) 1 1 2 2 1 2 ∈ 1 1 2 2 (since h g is constant). But then h(z ) = h(z ) so h is constant and Z is connected, by ◦ 1 2 (4.1c). (4.1e) Corollary Connectedness is a topological property. Proof 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 If not we can write X = A B with A, B nonempty and open and with A B = . But ∅ then X = CX (A) CX (B), aS union of two finite sets and hence finite. ButTX is infinite so this is impossible.S 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 Y for some closed set K of X. (ii) If A is open in Y (in the subspace topology on YT) 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. Proof (i) Suppose A is closed in Y (in the subspace topology). Then CY (A) = Y U, for some open set U in X. Let K be the closed set CX (U). Then X = U K andT so Y = (Y U) (Y K). Thus Y K is the complement in Y of CY (A) = YS U. But the complementT S ofTCY (A) in Y is TA, i.e. we have A = Y K. T (ii) We have A = U Y , with U open in X. So A is a unionT of two open sets in X hence open in X. T (iii) We have A = Y K for some closed set K in X, by (i). Thus A is the intersection of two closed sets in X Tand hence A is closed in X. (4.1g) Theorem The closed interval [a, b] is connected (for a < b). Proof [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 ). So α A by (3.2l). We claim that in fact R ∈ α = 0. Suppose not. Since A is open in [0, 1] we have A = [0, 1] U for some open set U in . So we have (α r, α + r) U for some r > 0. Choose s > 0 such that s < min r, α . R − ⊂ 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.T 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 B for nonempty disjoint subsets A, B which are open and closed. But then 0 A and 0 B contradicting the fact that A B = . Hence [0, 1] is connected. ∈ S∈ ∅ T (4.1h) Corollary (The Intermediate Value Theorem) Let f :[a, b] be con- → R 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 1 1 [a, b] = f − U f − V and this is a disjoint union of nonempty open subsets[ of [a, b]. But [a, b] is connected so this is impossible.
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