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Elements of Point Set Topology

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Elements of Point Set Topology Charpter 3 Elements of Point set Topology Open and closed sets in R1 and R2 3.1 Prove that an open interval in R1 is an open set and that a closed interval is a closed set. proof: 1. Let a,b be an open interval in R1, and let x a,b. Consider minx a,b x : L. Then we have Bx,L x L,x L a,b.Thatis,x is an interior point of a,b.Sincex is arbitrary, we have every point of a,b is interior. So, a,b is open in R1. 2. Let a,b be a closed interval in R1, and let x be an adherent point of a,b.Wewant to show x a,b.Ifx a,b, then we have x a or x b. Consider x a, then Bx, a x a,b 3x a , x a a,b 2 2 2 which contradicts the definition of an adherent point. Similarly for x b. Therefore, we have x a,b if x is an adherent point of a,b.Thatis,a,b contains its all adherent points. It implies that a,b is closed in R1. 3.2 Determine all the accumulation points of the following sets in R1 and decide whether the sets are open or closed (or neither). (a) All integers. Solution: Denote the set of all integers by Z.Letx Z, and consider x1 Bx, 2 x S .So,Z has no accumulation points. x1 However, Bx, 2 S x .SoZ contains its all adherent points. It means that Z is closed. Trivially, Z is not open since Bx,r is not contained in Z for all r 0. Remark: 1. Definition of an adherent point: Let S be a subset of Rn,andx a point in Rn, x is not necessarily in S. Then x is said to be adherent to S if every n ball Bx contains at least one point of S. To be roughly, Bx S . 2. Definition of an accumulation point: Let S be a subset of Rn,andx a point in Rn, then x is called an accumulation point of S if every n ball Bx contains at least one point of S distinct from x. To be roughly, Bx x S .Thatis,x is an accumulation point if, and only if, x adheres to S x. Note that in this sense, Bx x S Bx S x. 3. Definition of an isolated point: If x S, but x is not an accumulation point of S, then x is called an isolated point. 4. Another solution for Z is closed: Since R Z nZ n,n 1, we know that R Z is open. So, Z is closed. 5. In logics, if there does not exist any accumulation point of a set S, then S is automatically a closed set. (b) The interval a,b. solution: In order to find all accumulation points of a,b, we consider 2 cases as follows. 1. a,b :Letx a,b, then Bx,r x a,b for any r 0. So, every point of a,b is an accumulation point. 2. R1 a,b ,a b, : For points in b, and ,a, it is easy to know that these points cannot be accumulation points since x b, or x ,a,there exists an n ball Bx,rx such that Bx,rx x a,b . For the point a, it is easy to know that Ba,r a a,b . That is, in this case, there is only one accumulation point a of a,b. So, from 1 and 2, we know that the set of the accumulation points of a,b is a,b. Since a a,b, we know that a,b cannot contain its all accumulation points. So, a,b is not closed. Since an n ball Bb,r is not contained in a,b for any r 0, we know that the point b is not interior to a,b.So,a,b is not open. (c) All numbers of the form 1/n, (n 1,2,3,.... Solution: Write the set 1/n : n 1,2,... 1,1/2,1/3,...,1/n,... : S. Obviously, 0 is the only one accumulation point of S.So,S is not closed since S does not contain the accumulation point 0. Since 1 S,andB1,r is not contained in S for any r 0, S is not open. Remark: Every point of 1/n : n 1,2,3,... is isolated. (d) All rational numbers. Solutions: Denote all rational numbers by Q. It is trivially seen that the set of accumulation points is R1. So, Q is not closed. Consider x Q,anyn ball Bx is not contained in Q.Thatis,x is not an interior point of Q. In fact, every point of Q is not an interior point of Q. So, Q is not open. (e) All numbers of the form 2n 5m, (m,n 1,2,.... Solution: Write the set 2n 5m : m,n 1,2,.. m 1 5m, 1 5m,..., 1 5m,... : S m1 2 4 2n 1 1 , 1 1 ,..., 1 1 ,... 1 2 5 2 52 2 5m 1 1 , 1 1 ,..., 1 1 ,... 2 4 5 4 52 4 5m ........................................................ 1 1 , 1 1 ,..., 1 1 ... 3 2n 5 2n 52 2n 5m ..................................................... 1 1 So,wefindthatS 2n : n 1,2,... 5m : m 1,2,... 0.So,S is not 1 1 closed since it does not contain 0. Since 2 S,andB 2 ,r is not contained in S for any r 0, S is not open. Remark: By (1)-(3), we can regard them as three sequences m m m 1 m 1 m 1 m 5 , 5 and n 5 , respectively. 2 m1 4 m1 2 m1 m m 1 And it means that for (1), the sequence 5 m 1 moves . Similarly for others. So, it is 2m easy to see why 1 is an accumulation point of 1 5m . And thus get the set of all 2 2 m1 accumulation points of 2n 5m : m,n 1,2,.. (f) All numbers of the form 1n 1/m, (m,n 1,2,.... Solution: Write the set of all numbers 1n 1/m,(m,n 1,2,... as m m 1 1 1 1 : S. m m1 m m1 And thus by the remark in (e), it is easy to know that S 1,1.So,S is not closed since S S.Since2 S,andB2,r is not contained in S for any r 0, S is not open. (g) All numbers of the form 1/n 1/m, (m,n 1,2,.... Solution: Write the set of all numbers 1/n 1/m,(m,n 1,2,... as m m m 1 1/mm1 1/2 1/mm1 ...1/n 1/mm1 ...: S. We find that S 1/n : n N 1/m : m N 0 1/n : n N 0.So,S is not closed since S S.Since1 S,andB1,r is not contained in S for any r 0, S is not open. (h) All numbers of the form 1n/1 1/n, (n 1,2,.... Soluton: Write the set of all numbers 1n/1 1/n,(n 1,2,... as k k 1 1 : S. 1 1 1 1 2k k1 2k1 k1 1 1 We find that S 1,1.So,S is not closed since S S.Since 2 S,andB 2 ,r is not contained in S for any r 0, S is not open. 3.3 The same as Exercise 3.2 for the following sets in R2. (a) All complex z such that |z| 1. Solution: Denote z C :|z| 1 by S. It is easy to know that S z C :|z| 1. |z|1 So, S is not closed since S S.Letz S, then |z| 1. Consider Bz, 2 S,soevery point of S is interior. That is, S is open. (b) All complex z such that |z| 1. Solution: Denote z C :|z| 1 by S. It is easy to know that S z C :|z| 1. So, S is closed since S S.Since1 S,andB1,r is not contained in S for any r 0, S is not open. (c) All complex numbers of the form 1/n i/m, (m,n 1,2,.... Solution: Write the set of all complex numbers of the form 1/n i/m, (m,n 1,2,... as m m m 1 i 1 i ... 1 i ...: S. m m1 2 m m1 n m m1 We know that S 1/n : n 1,2,... i/m : m 1,2,... 0.So,S is not closed since S S.Since1 i S,andB1 i,r is not contained in S for any r 0, S is not open. (d) All points x,y such that x2 y2 1. Solution: Denote x,y : x2 y2 1 by S. We know that S x,y : x2 y2 1.So,S is not closed since S S.Letp x,y S, then x2 y2 1.Itiseasytofindthatr 0 such that Bp,r S.So,S is open. (e) All points x,y such that x 0. Solution: Write all points x,y such that x 0asx,y : x 0 : S. It is easy to know that S x,y : x 0.So,S is not closed since S S.Letx S, then it is easy to find rx 0 such that Bx,rx S.So,S is open.
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