There is a bijection between 0 1 and 0 1 2 Why There exists an injection 0 1 0

There is a bijection between 0 1 and 0 1 2 why there

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There is a bijection between [0 , 1] and [0,1]2. Why? There exists an injection [0,1][0,1]2 1] 2 [0 , 1]. The details of this second map are as follows. Write x = k =0 x k 10 - k with 0 x k 9 and the decimal expansion not terminating in an infinite sequence of 9s. If x = 1 then x 0 = 1 and x k = 0 for k > 0. y = k =0 y k 10 - k with the same properties as x above. Define the function h : [0 , 1] 2 [0 , 1] by h ( x, y ) = 0 .x 0 y 0 x 1 y 1 . . . = k =0 x k 10 - (2 k +1) + k =0 y k 10 - (2 k +2) . h : [0 , 1] 2 [0 , 1] is not onto, but it can be easily be verified that h is one-to-one. Then by the Schroeder-Bernstein Theorem, there exists a bijection [0 , 1] [0 , 1] 2 . The Peano Curve is a function f : [0 , 1] [0 , 1] 2 which is onto and continuous. (Note: There is no homeomorphism [0 , 1] [0 , 1] 2 .) Lemma 6.5. Suppose X is compact and f : X Y is continuous, onto and one-to-one. Then f is a homeomorphism. Proof. We can define g : Y X with g ( y ) = x if f ( x ) = y . Given an open V X , we need to check that g - 1 ( V ) is open in Y . K = X \ V is closed in X , hence it is compact. Therefore, f ( K ) is also compact. f ( K ) = g - 1 ( K ). Then g - 1 ( V ) = g - 1 ( X \ K ) = g - 1 ( X ) \ g - 1 ( K ) which is open in Y . Definition 6.6. A (topological or) metric space X is path-connected if for any x, y X there exists a path connecting them, i.e., γ : [0 , 1] X , which is continuous with γ (0) = x and γ (1) = y . Clearly, R k and [ a, b ] are connected. More generally, if K R is convex , it is path- connected. K is convex if for all x, y K and for all t [0 , 1], γ ( t ) = ty + (1 - t ) x K . Definition 6.7. A set K X is called clopen if it is closed and open in X . Definition 6.8. X is connected if the only clopen sets in X are and X . Equivalently, X is connected if and only if for any partition X = V 1 V 2 of X into two open and disjoint sets, one of them is . For instance, Q = { x Q : x < 2 } ∪ { x Q : x > 2 } and thus Q is not connected. 29
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30 Lecture 12: October 06 Fact 6.9. For a < b R , [ a, b ] is connected. Proof. Suppose K [ a, b ] is clopen in [ a, b ] and K = , K = [ a, b ]. If a K , let s = sup K [ a, b ]. Then s K because K is closed. We have three cases: (1) Suppose s = a . This is impossible because we know, since K is open, that there exists > 0 such that [ a, a + ) K . (2) Suppose s < b . This is impossible because we know, since K is open, that there exists > 0 such that [ s, s + ) K . (3) Suppose s = b . This implies that b K . To get a contradiction, we examine sup K c . (a) Suppose sup K c < b . We apply the argument in (2) and reach a contradiction. (b) Suppose sup K c = b . Then b K c . But b K , so we have a contradiction. Theorem 6.10. Suppose that X is path-connected. Then X is connected. Proof. Assume X is not connected. Suppose X = V 1 V 2 with V 1 , V 2 open, disjoint and nonempty. Let x V 1 , y V 2 . We find γ : [0 , 1] X with γ continuous and γ (0) = x and γ (1) = y . Then [0 , 1] = γ - 1 ( V 1 ) γ - 1 ( V 2 ) is a disjoint union of open sets and nonempty since 0 γ - 1 ( V 1 ) and 1 γ - 1 ( V 2 ). This is a contradiction since [0 , 1] is connected (meaning one of γ - 1 ( V 1 ) and γ - 1 ( V 2 ) has to be empty.) Fact 6.11. [0 , 1] and the circle C = { ( x, y ) R 2 | x 2 + y 2 = 1 } are not homeomorphic.
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  • Spring '14
  • KLASS,MJ
  • Topology, Integers, Metric space, Lecturer, Xn, Yuval Peres

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