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Chapter 3 Connected Topological Spaces 3.1 Connected Spaces Definition 3.1.1. A topological space (X, J ) is said to be a disconnected topological space if there exist nonempty open sets A and B of X such that (i) A ∩ B = φ, (ii) X = A ∪ B. In such a case B = Ac and A = B c and hence A and B are closed sets. Also X contains a nonempty proper subset A (that is A 6= φ, X which is both open and closed in X. A topological space (X, J ) is said to be connected if there cannot exist nonempty closed (open) subsets A and B of X such that (i) A∩B = φ, (ii) X = A∪B. Equivalently, (X, J ) is connected if and only if φ and X are the only subsets of X which are both open and closed in X. Examples 3.1.2. (i) Let X be a set containing at least two elements and A ⊆ X, A 6= φ, X. Then J = {φ, X, A, Ac } is a topological space. Here A is such that A 6= φ, A 6= X and A is both open and closed. Hence (X, J ) is not a connected topological space. (ii) Let X be a nonempty set and Jf = {A ⊆ X, XKA = Ac finite or Ac = X} then (X, Jf ) is a topology on X, known as cofinite topology on X. 53 Note. If X is a finite set containing at least two elements then Jf is the discrete topology on X. In this case (X, Jf ) is not a connected topological space. What will happen when X is an infinite set and Jf is the cofinite topology on X. Is (X, Jf ) not connected? In other words can we find a subset A of X such that A 6= φ and A 6= X but A is both open and closed? A 6= X, A is closed implies A is a finite set. Also A 6= φ implies such a nonempty finite set cannot be an open set. Therefore φ and X are the only sets which are both open and closed. This implies (X, Jf ) is a connected topological space whenever X is an infinite set. ? 3.2 Connected Subsets of the Real Line Keeping our intuition alive, let us prove that intervals are connected subsets of R and they are the only connected subsets of R. Recall that a subset J of R is an interval if and only if whenever a, b ∈ J and a < c < b, we have c ∈ J. Note that the null set φ and singleton sets are also intervals. For example for x ∈ R, φ = (x, x) = (1, 1) and {x} = [x, x], the singleton set containing x. We say that a subset Y of a topological space X is connected if (Y, JY ) is connected. First let us prove that if A ⊆ R is not an interval then A is not connected. Here JA = {A ∩ U : U is an open set in R}. So, we will have to prove that the topological space (A, JA ) is not connected. The given set A is not an interval implies there exist x, y ∈ A and z ∈ R such that x < z < y and z ∈ / A. We know that (−∞, z), (z, ∞) are open sets in R. This implies that (−∞, z) ∩ A and (z, ∞) ∩ A are open sets in (A, JA ). Also x ∈ (−∞, z) ∩ A = C and y ∈ (z, ∞) ∩ A = D and A = C ∪ D. That is C, D are nonempty open sets in (A, JA ) such that C ∩ D = φ 54 and A = C ∪ D. Hence the topological space (A, JA ) cannot be connected. That is the given subset A of R (which is not an interval) is not connected. Now let us prove: Theorem 3.2.1. Every interval in R is connected. Proof. Let J be an interval and let us assume that J contains at least two elements. For if J = φ or a singleton set then the null set φ and the whole space J are the only sets which are both open and closed in J and hence J is connected. Now let us suppose that there exist nonempty closed sets A, B in J (that is A, B ⊆ J and A, B are closed sets in (J, JJ ), where JJ = {U ∩ J : U is open in R}) such that J = A ∪ B. Fix a ∈ A, b ∈ B and without loss of generality let us say a < b. Note that a ∈ A ⊆ J, b ∈ B ⊆ J. a z b Figure 3.1 Since J is an interval [a, b] ⊆ J. Let y = sup(A ∩ [a, b]). Now let us prove that y ∈ A ∩ B. First let us prove that y ∈ A. Let U be an open set in J containing y (y ∈ [a, b] ⊆ J). Then there exists an open set V in R such that V ∩ J = U. Hence there exists > 0 such that (y − , y + ) ∩ J ⊆ V ∩ J = U. (If y = a, we are through, otherwise take 0 < < y − a and < b − y when y 6= b.) Now y − is not an upper bound for [a, b] ∩ A implies there exists x0 ∈ [a, b] ∩ A such that y − < x0 . This implies x0 ∈ (y − , y + ) ∩ A ⊆ (y − , y + ) ∩ J ⊆ U. Hence x0 ∈ U ∩ A. That is, whenever U is an open set in J containing y, then U ∩ A 6= φ. This proves that y ∈ AJ = A, AJ is the closure of A in J. Also if y = b, then y ∈ B. Suppose y < b, (y is the supremum of C = A ∩ [a, b] and b is an upper bound of C then take 55 y0 ∈ (y, y + ). Now y0 ∈ U ∩ B. Hence y ∈ B J = B. Hence we have y ∈ A ∩ B. Therefore there cannot exist nonempty closed subsets of A, B in the subspace (J, JJ ) such that A ∩ B = φ, A ∪ B = J. This proves that J is connected. 3.3 Some Properties of Connected Spaces Now let us prove that continuous image of a connected topological space is connected. Theorem 3.3.1. Let (X, J ) be a connected topological space and (Y, J 0 ) be any topological space. Suppose f : (X, J ) → (Y, J 0 ) is a surjective continuous map then the image f(X) = Y is a connected topological space. Proof. (By contradiction). Let A, B be nonempty open subsets of Y such that A ∩ B = φ, A ∪ B = Y . Now f is a continuous map, and A, B are open sets in Y implies that f −1 (A), f −1 (B) are open sets in X. Also f −1 (A) ∩ f −1 (B) = f −1 (A ∩ B) = φ (3.1) and X = f −1 (Y ) = f −1 (A ∪ B) (from 3.1) = f −1 (A) ∪ f −1 (B). (3.2) Since A, B 6= φ, let y ∈ A and y 0 ∈ B, and f is a surjective map implies that there exist x, x0 ∈ X such that f (x) = y and f (x0 ) = y 0 implies f (x) ∈ A and f (x0 ) ∈ B implies x ∈ f −1 (A) and x0 ∈ f −1 (B). Hence f −1 (A), f −1 (B) are nonempty open subsets of the connected topological space (X, J ) satisfying Eqs. (3.1) and (3.2). This means (X, J ) cannot be a connected topological space. That is we have proved: f (X) = Y is not connected implies X is also not connected and this gives a contradiction. Hence 56 our assumption that (Y , J 0 ) is not connected is not valid. Therefore (Y , J 0 ) is a connected topological space. It is to be noted that if X is a connected topological space and f : X → Y is a continuous function, where Y is any topological space, then the image f (X) is also a connected topological space. Here we will have to consider f (X) as a subspace of the given topological space Y . Also if f : X → Y is continuous then f : X → f (X) is also continuous. Hence this result will follow from the previous result. Definition 3.3.2. A subspace Y of a topological space (X, J ) is said to have a separation if and only if there exist nonempty subsets A, B of X such that (i) Y = A ∪ B, (ii) A ∩ B = φ = A ∩ B. Here A is the closure of A in X. Example 3.3.3. Let X = R and Y = [0, 2) ∩ (2, 5). Then Y has a separation. Take A = [0, 2), B = (2, 5), then (i) Y = A ∪ B is satisfied. Now (ii) A ∩ B = [0, 2] ∩ (2, 5) = φ, and A ∩ B = [0, 2) ∩ [2, 5] = φ. Hence from (i) and (ii) we see that Y has a separation. Theorem 3.3.4. A subspace Y of a topological space (X, J ) is connected if and only if there does not exist any separation for Y . Proof. First let us assume that the subspace Y is connected. This means the subspace (Y , JY ) is a connected topological space. So, we will have to prove that Y does not admit any separation. Suppose Y has a separation. Hence there exist nonempty subsets A, B of X such that Y = A ∪ B, A ∩ B = φ = A ∩ B. Now AY = A ∩ Y = A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B) = A ∪ φ = A. This implies that A is a closed subset of (Y , JY ). (B ∩ (A ∪ B)) = (B ∩ A) ∪ (B ∩ B) = B. Similarly B Y = B ∩ Y = This implies that B is a closed set in (Y, JY ). Also BYc = Y KB = Y ∩B c = (A∪B)∩B c = (A∩B c )∪(B c ∩B) = A∩B c = A 57 (since A ∩ B = φ). This means complement of B with respect to Y is A. Hence B is such that B 6= φ, B 6= Y (since Y = A ∩ B and A 6= φ) and B is both open and closed in Y . This implies that the subspace (Y , JY ) is a disconnected space. This is a contradiction and we arrived at this contradiction by assuming that Y has a separation. Hence there does not exist any separation for Y . Conversely, now assume that there does not exist any separation for Y . Now suppose that the subspace (Y , JY ) is a disconnected space. Then there exist nonempty closed subsets A, B in (Y , JY ) such that Y = A ∪ B and A ∩ B = φ. Now A ∩ B = A ∩ (Y ∩ B) = (A ∩ Y ) ∩ B = AY ∩ B = A ∩ B = φ. (A is closed in Y implies AY = A) Similarly, A ∩ B = A ∩ Y ∩ B = A ∩ B Y = A ∩ B = φ. (B is closed in Y implies B Y = B.) We have nonempty subsets A, B in X such that (i) Y = A ∪ B, (ii) A ∩ B = φ = A ∩ B. This means Y has a separation and this gives a contradiction. We arrived at this contradiction by assuming that the subspace (Y , JY ) is a disconnected space. Hence the assumption is wrong. That means (Y , JY ) is connected. Theorem 3.3.5. Let (X, J ) be a disconnected topological space and A be a subset of X such that (i) A 6= φ, X (ii) A is both open and closed in X. Suppose Y is a nonempty connected subspace of X. Then either Y ⊆ A or Y ⊆ Ac . Proof. X = A∪B, where B = Ac implies Y = X∩Y = (A∪B)∩Y = (A∩Y )∪(B∩Y ). Also (A ∩ Y ) ∩ (B ∩ Y ) ⊆ A ∩ B = A ∩ B = φ (since B = Ac is a closed set in X) implies (A ∩ Y ) ∩ (B ∩ Y ) = φ. Similarly, (A ∩ Y ) ∩ (B ∩ Y ) ⊆ A ∩ B = A ∩ B = φ implies (A ∩ Y ) ∩ (B ∩ Y ) = φ. It is given that Y is a connected subspace of X. Hence it cannot happen that A ∩ Y 6= φ and B ∩ Y 6= φ. Since Y is a connected 58 subspace of X, Y cannot admit any separation. Hence A ∩ Y = φ or B ∩ Y = φ implies Y ⊆ Ac or Y ⊆ B c = A. Alternate Proof. Now A ∩ Y, B ∩ Y (where B = Ac ) are open sets in the subspace Y such that (i) Y = (A ∩ Y ) ∪ (B ∩ Y ) and (ii) (A ∩ Y ) ∩ (B ∩ Y ) ⊆ A ∩ B = φ implies A ∩ Y = φ or B ∩ Y = φ implies Y ⊆ Ac or Y ⊆ A. Let us prove the following: Theorem 3.3.6. Let (X, J ) be a topological space and Y1 , Y2 be connected subspaces of X. Further suppose Y1 ∩ Y2 6= φ. Then Y1 ∪ Y2 is a connected subspace of X. Proof. We will have to prove that the subspace Y1 ∪ Y2 cannot admit any separation. Suppose A, B are subsets of X such that (i) Y1 ∪ Y2 = A ∪ B, (ii) A ∩ B = φ = A ∩ B (here A is the closure of A in X). From (i) Y1 = (Y1 ∩ A) ∪ (Y1 ∩ B), (3.3) and from (ii) (Y1 ∩ A) ∩ (Y1 ∩ B) = φ and (Y1 ∩ A) ∩ (Y ∩ B) = φ. (3.4) It is given that Y1 is a connected subspace. Hence Y1 cannot admit a separation. Therefore from Eqs. (3.3) and (3.4) we conclude that Y1 ∩ A = φ or Y1 ∩ B = φ. Let us say that Y1 ∩ A = φ. This implies that Y1 ⊆ Ac . (3.5) Similarly, using the fact that Y2 is a connected subspace we conclude that Y2 ∩ A = φ or Y2 ∩ B = φ. If Y2 ∩ B = φ then Y2 ⊆ B c . 59 (3.6) From Eqs. (3.5) and (3.6) we have Y1 ∩ Y2 ⊆ Ac ∩ B c = (A ∪ B)c . (3.7) Also Y1 ∩ Y2 6= φ. Hence there exists an element say x0 ∈ Y1 ∩ Y2 . Then from Eq. (3.7), x0 ∈ / A ∪ B, a contradiction to Y1 ∪ Y2 = A ∪ B. Therefore both Y1 ⊆ Ac and Y2 ⊆ Ac implies Y1 ∪ Y2 ⊆ Ac implies A = φ. (If there exists x ∈ A ⊆ Y1 ∪ Y2 then x ∈ Y1 ∪ Y2 ⊆ Ac .) What we have proved is the following: If (i) and (ii) happens then that leads to A = φ or B = φ. (Y1 ∩ A = φ or Y1 ∩ B = φ implies Y1 ⊆ Ac or Y1 ⊆ B c . So if we assume Y1 ⊆ B c then we would have arrived at B = φ.) That is we have proved that Y1 ∪ Y2 cannot admit any separation and this implies Y1 ∪ Y2 is a connected subspace of X. Remark 3.3.7. If X is a topological space and Y1 , Y2 , . . . , Yn for some n ∈ N is a n finite collection of connected topological spaces such that ∩ Yi 6= φ then by using i=1 induction we can prove that Y1 ∪ Y2 ∪ · · · ∪ Yn is a connected subset of X. In fact using exactly the same idea of proving that the subspace Y1 ∪ Y2 is connected whenever Y1 , Y2 are connected subspaces with the added condition that Y1 ∩ Y2 6= φ, we prove the following theorem. Theorem 3.3.8. If {Yα }α∈J is a collection of connected subspaces of a topological space X and further there exists an x0 ∈ X such that x0 ∈ Yα , for each α ∈ J, then ∪ Yα is a connected subspace of X. α∈J Proof. Here again we will prove that ∪ Yα cannot admit any separation. Suppose α∈J A, B are subsets of X such that (i) ∪ Yα = A ∪ B, (ii) A ∩ B = A ∩ B = φ. Then our α∈J aim is to prove A = φ or B = φ. For each fixed α0 ∈ J, Yα0 = (Yα0 ∩ A) ∪ (Yα0 ∩ B) 60 and (Yα0 ∩ A) ∩ (Yα0 ∩ B) = φ = (Yα0 ∩ A) ∩ (Yα0 ∩ B). Hence Yα0 is a connected subspace of X implies Yα0 ∩ A = φ or Yα0 ∩ B = φ. Let us say Yα0 ∩ A = φ. This will imply that Yα0 ⊆ Ac . Note that α0 ∈ J is an arbitrary element. Hence for each α ∈ J, Yα ⊆ Ac or Yα ⊆ B c . But if Yα1 ⊆ Ac for some α1 ∈ J and Yα2 ⊆ B c for c c c c some α2 ∈ J then x0 ∈ Yα1 ∩ Yα2 ⊆ A ∩ B = (A ∪ B) = ∪ Yα . This means α∈J x0 ∈ / Yα for some α ∈ J and that gives a contradiction. Therefore Yα ⊆ Ac for all α ∈ J or Yα ⊆ B c for all α ∈ J implies ∪ Yα ⊆ Ac or ∪ Yα ⊆ B c implies A = φ or α∈J α∈J B = φ. This means ∪ Yα cannot admit any separation and that is what we wanted α∈J to prove. Theorem 3.3.9. Let X be a topological space and Y1 , Y2 , . . . , Yn , . . . be a collection of connected topological spaces. Further suppose Yk ∩ Yk+1 6= φ for all k ∈ N. Then ∞ ∪ Yk is also a connected space. k=1 Proof. Now Y1 , Y2 are connected subspaces of X and Y1 ∩Y2 6= φ implies E2 = Y1 ∪Y2 is a connected subspace. Now E2 is a connected subspace of X and Y3 is a connected subspace of X. Further Y2 ∩ Y3 ⊆ (Y1 ∪ Y2 ) ∩ Y3 = E2 ∩ Y3 . Hence Y2 ∩ Y3 6= φ implies E2 ∩ Y3 6= φ implies E3 = E2 ∪ Y3 = Y1 ∪ Y2 ∪ Y3 is a connected subspace of X. Now use induction to prove that, for each k ∈ N, Ek = Y1 ∪ Y2 ∪ · · · ∪ Yk is a connected subspace of X. We have a collection {Ek }∞ k=1 of connected subspaces of X ∞ ∞ ∞ ∞ k=1 k=1 k=1 k=1 such that Y1 = E1 ⊆ ∩ Ek . Also ∩ Ek 6= φ. Since Y1 6= φ. Hence ∪ Ek = ∪ Yk is a connected subspace. Now let us prove that if Y is a nonempty connected subspace of a topological space X and then it remains connected after adding some (or all) of its limits points to E. 61 Theorem 3.3.10. Let Y be a nonempty connected subspace of a topological space X and E be a subset of X such that Y ⊆ E ⊆ Y . Then E is also a connected subspace of X. Proof. Let A, B be subsets of X such that (i) E = A ∪ B (ii) A ∩ B = A ∩ B = φ then Y = Y ∩E = (A∩Y )∪(B ∩Y ). Also (A ∩ Y )∩(B ∩Y ) = φ = (A∩Y )∩(B ∩ Y ). Hence Y is a connected subspace of X implies A ∩ Y = φ or B ∩ Y = φ. Let us say A ∩ Y = φ. This implies Y = B ∩ Y ⇒ Y ⊆ B ⇒ Y ⊆ B. Since, E ⊆ Y , E = A ∪ B and A ∩ B = φ we get A ⊆ E ⊆ B and A = A ∩ B = φ. That is we have proved that for subsets A, B of X, E = A ∪ B, A ∩ B = A ∩ B = φ. This implies A = φ. We arrived at this conclusion by assuming that A ∩ Y = φ. If we had assumed B ∩ Y = φ, then we would have proved that B = φ. Therefore E does not admit any separation and hence E is a connected subspace of X. Theorem 3.3.11. If (X1 , J1 ), (X2 , J2 ) are connected topological spaces, then the product space X1 × X2 is also a connected space. Proof. Fix a1 ∈ X1 , a2 ∈ X2 . (Note. X1 = φ or X2 = φ ⇒ X1 × X2 = φ and in this case X1 × X2 is a connected space.) For each x1 ∈ X1 , fx1 : X2 → X1 × X2 defined as fx1 (x2 ) = (x1 , x2 ) is a continuous function. Also we know that continuous image of a connected space is connected. In this case, the continuous image is fx1 (X2 ) = x1 ×X2 . That is x1 × X2 = {(x1 , x2 ) : x2 ∈ X2 } is a connected subspace of the product space X1 × X2 (refer the vertical line passing through x1 ). Similarly, X1 × a2 is a connected subspace of the product space. Also (x1 , a2 ) ∈ (x1 × X2 ) ∩ (X1 × a2 ). Hence (x1 × X2 ) ∪ (X1 × a2 ) is a connected subspace of the product space. Let Tx1 = (x1 × X2 ) ∪ (X1 × a2 ). Also note that (a1 , a2 ) ∈ Tx1 , for each x1 ∈ X1 . That is 62 {Tx1 } is a collection of connected subspaces of the product space X1 × X2 . Further (a1 , a2 ) ∈ Tx1 for all x1 ∈ X1 . Hence X2 ∪ Tx1 = X1 × X2 is a connected space. x1 ∈X1 X1 X X2 a2 (a1 , a 2 ) x1 a1 X 1 X a2 X1 Figure 3.2 Now we use mathematical induction to prove: if (X1 , J1 ), (X2 , J2 ), . . . , (Xn , Jn ) are a finite collection of connected topological spaces then the product space X1 × X2 × · · · × Xn is also a connected space. But it is to be noted that we cannot use (say why?) mathematical induction to prove: If (Xn , Jn ), n ∈ N is a collection ∞ of connected topological spaces then the product space X1 × X2 × · · · = Π Xn is n=1 also a connected space. However, we prove the following theorem when we have a collection (Xα , Jα ), α ∈ J (where J is a nonempty index set) of connected topological spaces. Theorem 3.3.12. Let (Xα , Jα ), α ∈ J be a collection of connected topological spaces. Then the product space X = Π Xα is also a connected space. α∈J Proof. When J is a finite set we have already proved this result. So let us assume that J is an infinite set. Also let us assume that each Xα 6= φ. From each Xα , fix an 63 element say aα ∈ Xα . That is we have a function f : J → ∪ Xα such that α∈J f (α) = aα ∈ Xα for all α ∈ J. That is f ∈ Π Xα . We normally write f = (aα )α∈J α∈J and just say that (aα )α∈J ∈ Π Xα . If α1 , α2 , . . . , αk ∈ J then X(α1 , α2 , . . . , αk ) = α∈J {x = (xα )α∈J : xα = aα , when α 6= α1 , α2 , . . . , αk }. If our J = N and αj = j, j = 1, 2, . . . , k then X(1, 2, . . . , k) = {x1 , x2 , . . . , xk , ak+1 , ak+2 , . . .} = X1 × X2 × · · · × Xk × ak+1 × ak+2 × · · · Note that g(x1 , x2 , . . . , xk ) = (x1 , x2 , . . . , xk , ak+1 , ak+2 . . .) ∞ is a continuous function from the connected space X1 × X2 × · · · × Xk → Π Xn n=1 and its image is X1 × X2 × · · · × Xk × ak+1 × ak+2 × · · · . Hence X(1, 2, . . . , k) = X1 × X2 × · · · × Xk × ak+1 × ak+2 × · · · is a connected subspace of Π Xn ) and n∈N X(α1 , α2 , . . . , αk ) is a connected subspace of the product space. Also our fixed a = (aα )α∈J ∈ X(α1 , α2 , . . . , αk ). Therefore Y = ∪ {α1 ,α2 ,...,αk }⊆J X(α1 , α2 , . . . , αk ) (that is x ∈ Y if and only if there exists k ∈ N and α1 , α2 . . . , αk ∈ J such that x ∈ X(α1 , α2 , . . . , αk )) is a connected subspace of the product space. Again Y is a connected subspace of the product space implies Y is also a connected space. Now let us prove that Y = Π Xα = X. So take an element say α∈J x = (xα )α∈J ∈ X our aim is to prove that x ∈ Y . Start with a basic open set say U = Π Uα containing the point x = (xα )α∈J . U is a basic open set in the product α∈J space implies there exists k ∈ N and α1 , α2 , . . . , αk ∈ J such that Uα = Xα for α ∈ J and α 6= α1 , α2 , . . . , αk and Uαj ∈ Jαj for all j = 1, 2, . . . , k. Let y = (yα )α∈J be such that yαj = xαj for all j = 1, 2, . . . , k and yα = aα when α 6= α1 , α2 , . . . , αk . Then y ∈ U ∩ X(α1 , α2 , . . . , αk ) ⊆ U ∩ Y and hence U ∩ Y 6= φ for every basic open set U containing x. This implies x ∈ Y . Hence X ⊆ Y . This implies X = Y . Now Y is a connected subspace of X implies Y is also a connected subspace of X and hence X = Y is a connected space. 64 3.4 Connected Components Now let us define a relation say R on a given topological space (X, J ). We know that a relation on X is a subset of X × X. Here R = {(x, y) ∈ X × X : x, y ∈ Exy for a connected subset Exy of X}. If (x, y) ∈ R then we write xRy (read as x is related to y). So we say that xRy if and only if there is a connected set containing x, y. Now it is easy to see that this relation is an equivalence relation. For each x ∈ X, singleton set {x} is a connected subset of X. That is Exx = {x} is a connected set containing x, x. Hence xRx for all x ∈ X. Therefore R is reflexive. Now suppose for x, y ∈ X, xRy this implies there is a connected subset say Exy of X such that x, y ∈ Exy . But x, y ∈ Exy implies y, x ∈ Exy this implies yRx. That is for x, y ∈ X, xRy ⇒ yRx and hence R is symmetric. Now suppose for x, y, z ∈ X xRy and yRz. Now xRy implies there exists a connected subset say Exy of X such that x, y ∈ Exy . Similarly, yRz implies there exists a connected subset say Eyz of X such that y, z ∈ Eyz . Now Exy , Eyz are connected subsets of X such that y ∈ Exy ∩ Eyz . This gives us Exy ∪ Eyz is a connected set and further y, z ∈ Exy ∪ Eyz . Hence there exists a connected set containing y, z. This implies yRz. That is we have proved that xRy and yRz implies xRz that is R is transitive. Thus R is an equivalence relation on X. Hence this relation will partition the set X into disjoint equivalence classes. It is to be noted that for each x ∈ X there is exactly one equivalence class containing x. Let us denote the equivalence class containing x by [x]. That is for x ∈ X, [x] = {y ∈ X : yRx} is the equivalence class containing x. Now suppose 65 x ∈ [x], which is true since xRx and also x ∈ [y], for some y ∈ X. What is [y]? Note that [y] = {z ∈ X : zRy}. If such a situation arises then we aim to prove that [x] = [y]. So let z ∈ [x]. This implies zRx. Also we have xRy. Now our relation R is a transitive relation. Hence zRy and xRy together implies that zRy this implies that z ∈ [y]. That is z ∈ [x] implies z ∈ [y] implies [x] ⊆ [y]. Exactly in the same way we can prove that [y] ⊆ [x]. Hence [x] = [y], whenever x ∈ [y]. For x, y ∈ X, either [x] ∩ [y] = φ or [x] = [y]. For x ∈ X there is exactly one equivalence class containing x. Such an equivalence class which we denoted by [x], is called a connected component (also known as a component or a maximal connected set containing x. Why should we call [x] as a maximal connected set containing x ? So we will have to prove that if E is a connected subset of X containing x ∈ E, then E ⊆ [x]. That is [x] is the largest connected set containing x. Have we proved that [x] is a connected subset of X? Note that for each y ∈ [x], there is a connected subset say Ey (Ey is just a notation) of X containing x, y. That is {Ey }y∈[x] is a collection of connected sets and x ∈ Ey for all y ∈ [x]. This implies that ∪ Ey is also connected set. It is simple y∈[x] exercise to see that [x] = ∪ Ey . Hence [x] is a connected set containing x. If E is y∈[x] a connected set containing x then for each y ∈ E, yRx. Hence E ⊆ [x]. Also we know that if A is a connected subset of a topological space X then A is also a connected subset of X. Hence [x], the closure of the set [x], is also connected set containing x. But we have just proved that [x] is the maximal connected set containing x. Hence [x] ⊆ [x] implies [x] = [x] and [x] is a closed set. Therefore our maximal connected set containing x is a closed set. 66 Theorem 3.4.1. Intermediate value theorem. Let (X, J ) be a connected topological space and f : (X, J ) → R be a continuous function. If x, y are points of X and α is a real number such that α lies between f (x) and f (y). So let us say that f (x) < α < f (y). Then there exists z ∈ X such that f (z) = α. Proof. Now f is a continuous function and X is a connected topological space implies f (X) is a connected subset of R. But we know that every connected subset of R is an interval. Hence f (X) is an interval in R. Now f (x), f (y) ∈ f (X) such that f (x) < f (y), f (X) is an interval implies [f (x), f (y)] ⊆ f (X). Suppose α ∈ R such that f (x) < α < f (y). f (x) α f (y) Figure 3.3 In particular α ∈ [f (x), f (y)] ⊆ f (X). That is α ∈ f (X). This implies that there exists z ∈ X such that f (z) = x. Using the fact that [a, b](a, b ∈ R, a < b) is a connected subspace of R, we prove the following result: Result 3.4.2. Let f : [a, b] → [a, b] be a continuous function, then there exists x0 in [a, b] such that f (x0 ) = x0 . Proof. (Proof by contradiction:) Suppose f (x) 6= x for each x ∈ [a, b]. Let A = {x ∈ [a, b] : f (x) < x }, B = {x ∈ [a, b] : f (x) > x}. Since f is a continuous function implies that both A 67 and B are open subsets of [a, b]. Also A ∩ B = φ and [a, b] = A ∪ B. Hence [a, b] is a connected topological space implies either A = φ or B = φ. Suppose A = φ then we get that B = [a, b]. That is [a, b] = {x ∈ [a, b] : f (x) > x}. In particular b ∈ [a, b] implies f (b) > b and this gives a contradiction to our assumption that f : [a, b] → [a, b]. (Note: If B = φ then A = [a, b] and a ∈ [a, b] implies f (a) < a again we will get a contradiction.) We get a contradiction if we assume that f (x) 6= x, for every x ∈ [a, b]. Hence there exists at least one x0 ∈ [a, b] such that f (x0 ) = x0 . Note. In the above result such an x0 ∈ [a, b] (satisfying f (x0 ) = x0 ) is called a fixed point of f. ? Remark 3.4.3. In the proof of the above result we have used the fact that [a, b] is a connected topological subspace (in addition to the fact that f : [a, b] → [a, b] is a continuous map). What is to be noted here is we have not used the intermediate value theorem to prove the above result. Now let us use the intermediate value theorem to observe the following: Define g : [a, b] → R as g(x) = f (x) − x then g is a continuous map. Also g(a) = f (a) − a ≥ 0 and g(b) = f (b) − b ≤ 0. If f (a) = a or f (b) = b then we are through. If not g(b) = f (b) − b < 0 < f (a) − a = g(a). That is g(b) < 0 < g(a). Hence by intermediate value theorem there exists x0 ∈ [a, b] such that f (x0 ) = x0 . Definition 3.4.4. A topological space (X, J ) is said to be totally disconnected if and only if the connected components of X are singletons. That is if A is a nonempty connected subset of X then A is a singleton set. Exercises 3.4.5. Prove that Q and RKQ = Qc are totally disconnected topological 2 spaces. 68 Now let us introduce a concept known as pathwise connected. Definition 3.4.6. A topological space (X, J ) is said to be pathwise connected space if for x, y ∈ X there exists a continuous function say f : [0, 1] → X such that f (0) = x, f (1) = y. That is, X is pathwise connected if and only if for x, y ∈ X there exists a curve joining x and y. Note. For a, b ∈ R, a < b, [0, 1] is homeomorphic to [a, b]. That is there exists a bijective continuous function f : [0, 1] → [a, b] such that f −1 : [a, b] → [0, 1] is also continuous. For let f (t) = (1 − t)a + tb = a + (b − a)t, 0 ≤ t ≤ 1. Then f is bijective and f, f −1 are continuous functions. ? Now let us prove the following: Theorem 3.4.7. A topological space (X, J ) is pathwise connected if and only if for x, y ∈ X and a, b ∈ R, a < b there exists a continuous function g : [a, b] → X such that g(a) = x and g(b) = y. Proof. Given x, y ∈ X and a, b ∈ R, a < b. Now X is a pathwise connected space implies there exists a continuous function say f1 : [0, 1] → X such that f1 (0) = x and f1 (1) = y. Now let f2 (x) = x−a , b−a x ∈ [a, b]. Now f2 is a continuous function such that f2 (a) = 0 and f2 (b) = 1. Hence g : f1 ◦ f2 : [a, b] → X is a continuous function such that g(a) = f1 (f2 (a)) = f1 (0) = x and g(b) = f1 (f2 (b)) = f1 (1) = y. Theorem 3.4.8. Every pathwise connected topological space X is a connected space. Proof. Suppose X is disconnected. Then there exist nonempty closed subsets A, B of X such that (i) X = A ∪ B and (ii) A ∩ B = φ. Take x ∈ A, y ∈ B. Now X is a pathwise connected space implies there exists 69 a continuous function say f : [0, 1] → X such that f (0) = x, f (1) = y. Hence X = A ∪ B implies f −1 (X) = f −1 (A ∪ B) = f −1 (A) ∪ f −1 (B). Also x = f (0) ∈ A implies 0 ∈ f −1 (A), y = f (1) ∈ B implies 1 ∈ f −1 (B) and f −1 (A), f −1 (B) are closed subsets of [0, 1] (f : [0, 1] → X is continuous A, B are closed sets in X implies f −1 (A), f −1 (B) are closed sets in [0, 1]). Now f −1 (A), f −1 (B) are nonempty closed subsets of [0,1] such that [0, 1] = f −1 (X) = f −1 (A) ∪ f −1 (B) and f −1 (A) ∩ f −1 (B) = f −1 (A ∩ B) = φ implies [0, 1] is not connected and this gives a contradiction. Hence our initial assumption namely X is disconnected is not valid. Therefore X is a connected space. It is easy to prove that the continuous image of a pathwise connected space is pathwise connected. Result 3.4.9. If X is a nonempty convex subset of Rn (n ∈ N) then X is pathwise connected and hence X is a connected space. Proof. For x, y ∈ X, define f : [0, 1] → X as f (t) = (1 − t)x + ty for all t ∈ [0, 1]. Then for tn ∈ [0, 1], tn → t ∈ [0, 1] implies f (tn ) → f (t). Hence f is a continuous function such that f (0) = x and f (1) = y. Therefore X is pathwise connected. The following result is known as pasting lemma is very useful in the study of connected spaces. Lemma 3.4.10. (Pasting lemma) Let A, B be nonempty closed subsets of a topological space X. Suppose f : A → Y , g : B → Y are continuous functions (where Y is a topological space) such that f (x) = g(x), whenever x ∈ A ∩ B. Then 70 h : A ∪ B → Y is defined as f (x) if x ∈ A h(x) = g(x) if x ∈ B is also a continuous function. Proof. Left as an exercise to the reader. (Hint. For a nonempty closed subset C of Y , prove that h−1 (C) = f −1 (C) ∪ g −1 (C).) Exercises 3.4.11. (i) In the above lemma assume that A, B are nonempty open sets and arrive at the conclusion that h defined as above is a continuous function. (ii) Given that (X, J ) is a topological space. For x, y ∈ X define xRy (read as x is related to y) if there is a path (or say curve) joining x and y. That is there exists a continuous function f : [0, 1] → X such that f (0) = x and f (1) = y. 2 Hint. Now xRy implies there exists a continuous function f : 0, 21 → X such that f (0) = x, f 12 = y. Also yRz implies there exists a continuous function g : 21 , 1 → X such that g 12 = y g(1) = z. Now use pasting lemma to prove xRz. That is it is easy to prove that R is an equivalence relation on X. Definition 3.4.12. A topological space (X, J ) is locally connected at a point x in X if and only if for each open set U containing x, there is a connected open set V such that x ∈ V and V ⊆ U . If (X, J ) is locally connected at each x in X then we say that (X, J ) is a locally connected topological space. Note that neither every connected topological space is a locally connected space nor every locally connected topological space is a connected space. 71 Example 3.4.13. Let B = {(x, y) : 0 < x ≤ 1, y = sin ( x1 )} and X = B. Define f : (0, 1] → R2 as f (x) = (x, sin ( x1 )) then f is continuous. Hence (0, 1] is a connected space implies the image f ((0, 1]) = B is a connected subspace of R2 . This implies that B = X is also a connected space and X = B ∪ (0 × [−1, 1]). This space X is called the topologist’s sine curve and X is neither locally connected nor pathwise connected. Now U = B (0, 12 ), 14 ∩ X is an open set containing (0, 12 ). Now we leave it as an exercise to prove that there cannot exist any connected open set V satisfying (0, 12 ) ∈ V and V ⊆ U . Example 3.4.14. X = (0, 1) ∪ (2, 3) is locally connected but it is not connected. Theorem 3.4.15. Let Y be a connected open subset of Rn (Rn with Euclidean metric). Then Y is pathwise connected. Proof. Fix x ∈ Y . Let us prove that for each y ∈ Y there exists a path joining x and y. So let A = {y ∈ Y : there is a path joining y and x }. As x ∈ A, A 6= φ. We aim to prove that A = Y . Let y ∈ A. This implies y ∈ Y, and hence there exists r > 0 such that B(y, r) ⊆ Y . For z ∈ B(y, r). Define f : [0, 1] → Y as f (t) = (1 − t)z + ty. Then (1 − t)z + ty ∈ B(y, r), f is a continuous function satisfying the condition that f (0) = z and f (1) = y. So f is a path joining z and y. Also y ∈ A implies there exists a path joining y and x. That is there exists a continuous function say g : [0, 1] → Y such that g(0) = y, g(1) = x. Now define h : [0, 1] → Y as h(t) = f (2t) if 0 ≤ t ≤ 1 2 1 2 and h(t) = g(2t − 1) if ⇔ 0 ≤ 2t ≤ 1 and 21 ≤ t ≤ 1 ⇔ 0 ≤ 2t − 1 ≤ 1.) Here t → f (2t) is a continuous function on 0, 21 and t → g(2t−1) is a continuous function on 12 , 1 and f (2t) = g(2t − 1) when t = 12 . Hence by pasting lemma h : [0, 1] → Y is ≤ t ≤ 1. (Note: 0 ≤ t ≤ 1 2 a continuous function such that h(0) = z and h(1) = x. So we have a path h joining x 72 and z and hence z ∈ A. This gives that B(y, r) ⊆ A and hence each point y of A is an interior point of A. This proves that A is an open set in the subspace Y . Now in the same way we can prove that Y KA is also an open set in Y . That is Y = A ∪ (Y KA), where A, Y KA are both open sets in Y . As x ∈ A, A 6= φ. It is given that Y is connected and hence Y KA = φ. Therefore Y = A. Now A is pathwise connected and hence Y is pathwise connected. (For y, z ∈ A = Y , there is a path joining y and x also there is a path joining z and x. If there is path joining z and x then there is also a path joining x and z. Again there is a path joining y and x also a path joining x and z implies there is a path joining y and z.) Example of a topological space which is connected but not pathwise connected. Let Y1 = {(0, y) ∈ R2 : −1 ≤ y ≤ 1}, Y2 = {(x, sin( πx )) ∈ R2 : 0 < x ≤ 1} and X = Y1 ∪ Y2 . Let f : (0, 1] → Y2 defined as f (x) = (x, sin( πx )) is a continuous function, f (0, 1] = Y2 (that is f is surjective) and (0,1] is a connected space implies the continuous image f (0, 1] = Y2 is a connected space. Also Y20 = Y1 and hence Y 2 = Y1 ∪ Y2 = X. Now Y2 is a connected space implies Y 2 = X is also a connected space. Now let us prove that X is not pathwise connected. Suppose there exists a continuous function say g : [0, 1] → X such that g(0) = (0, 1) (that is we want to prove that there is no path joining (1, 0) and a point of Y2 ), Y1 is a closed subset of R2 implies Y1 is also a closed subset of X (X is a subspace of R2 ) and hence g −1 (Y1 ) is a closed subset of [0, 1]. Also g(0) = (0, 1) ∈ Y1 implies 0 ∈ g −1 (Y1 ). Now let us prove that g −1 (Y1 ) is also an open subset of [0,1]. So, let t ∈ g −1 (Y1 ) then g(t) ∈ Y1 . Now B(g(t), 12 ) ∩ X is an open set in X and g is a continuous function implies g −1 (B(g(t), 21 ) ∩ X) is an open set in [0, 1]. Also t ∈ g −1 (B(g(t), 21 ) ∩ X) implies there 73 exists r > 0 such that (t−r, t+r)∩[0, 1] ⊆ g −1 (B(g(t), 21 )∩X). Hence g((t−r, t+r)∩ [0, 1]) ⊆ B(g(t), 21 ) ∩ X, B(g(t), 12 ) ∩ X consists of an interval on the y-axis, together with segments of the curve y = sin( πx ), each of which is homomorphic to an interval. Further any two of these sets are separated from one another in B(g(t), 12 ) ∩ Y1 . Hence B(g(t), 12 ) ∩ Y1 is a connected component of B(g(t), 12 ) ∩ X containing g(t). Also g(t − r, t + r) ⊆ B(g(t), 21 ) ∩ Y1 . That is (t − r, t + r) ⊆ g −1 (B(g(t), 21 ) ∩ Y1 ) ⊆ g −1 (Y1 ). This proves that g −1 (Y1 ) is also an open set. That is g −1 (Y1 ) is a nonempty set which is both open and closed in [0,1]. Hence [0,1] is a connected space implies g −1 (Y1 ) = [0, 1]. That is g([0, 1]) ⊆ g(g −1 (Y1 )) ⊆ Y1 . Therefore there cannot exist any continuous function g : [0, 1] → X such that g(0) = (0, 1). This proves that X is not pathwise connected. 74