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Unformatted text preview: Some Basic Notations Of Set Theory References There are some good books about set theory; we write them down. We wish the reader can get more. 1. Set Theory and Related Topics by Seymour Lipschutz. 2. Set Theory by Charles C. Pinter. 3. Theory of sets by Kamke. 4. Naive set by Halmos. 2.1 Prove Theorem 2.2. Hint. ( a,b ) = ( c,d ) means {{ a } , { a,b }} = {{ c } , { c,d }} . Now appeal to the definition of set equality. Proof : ( ⇐ ) It is trivial. ( ⇒ ) Suppose that ( a,b ) = ( c,d ) , it means that {{ a } , { a,b }} = {{ c } , { c,d }} . It implies that { a } ∈ {{ c } , { c,d }} and { a,b } ∈ {{ c } , { c,d }} . So, if a 6 = c, then { a } = { c,d } . It implies that c ∈ { a } which is impossible. Hence, a = c. Similarly, we have b = d. 2.2 Let S be a relation and let D ( S ) be its domain. The relation S is said to be (i) reflexive if a ∈ D ( S ) implies ( a,a ) ∈ S, (ii) symmetric if ( a,b ) ∈ S implies ( b,a ) ∈ S, (iii) transitive if ( a,b ) ∈ S and ( b,c ) ∈ S implies ( a,c ) ∈ S. A relation which is symmetric, reflexive, and transitive is called an equiv alence relation. Determine which of these properties is possessed by S, if S is the set of all pairs of real numbers ( x,y ) such that (a) x ≤ y Proof : Write S = { ( x,y ) : x ≤ y } , then we check that (i) reflexive, (ii) symmetric, and (iii) transitive as follows. It is clear that D ( S ) = R. 1 (i) Since x ≤ x, ( x,x ) ∈ S. That is, S is reflexive. (ii) If ( x,y ) ∈ S, i.e., x ≤ y, then y ≤ x. So, ( y,x ) ∈ S. That is, S is symmetric. (iii) If ( x,y ) ∈ S and ( y,z ) ∈ S, i.e., x ≤ y and y ≤ z, then x ≤ z. So, ( x,z ) ∈ S. That is, S is transitive. (b) x < y Proof : Write S = { ( x,y ) : x < y } , then we check that (i) reflexive, (ii) symmetric, and (iii) transitive as follows. It is clear that D ( S ) = R. (i) It is clear that for any real x, we cannot have x < x. So, S is not reflexive. (ii) It is clear that for any real x, and y, we cannot have x < y and y < x at the same time. So, S is not symmetric. (iii) If ( x,y ) ∈ S and ( y,z ) ∈ S, then x < y and y < z. So, x < z wich implies ( x,z ) ∈ S. That is, S is transitive. (c) x <  y  Proof : Write S = { ( x,y ) : x <  y } , then we check that (i) reflexive, (ii) symmetric, and (iii) transitive as follows. It is clear that D ( S ) = R. (i) Since it is impossible for 0 <   , S is not reflexive. (ii) Since ( 1 , 2) ∈ S but (2 , 1) / ∈ S, S is not symmetric. (iii) Since (0 , 1) ∈ S and ( 1 , 0) ∈ S, but (0 , 0) / ∈ S, S is not transitive. (d) x 2 + y 2 = 1 Proof : Write S = { ( x,y ) : x 2 + y 2 = 1 } , then we check that (i) reflexive, (ii) symmetric, and (iii) transitive as follows. It is clear that D ( S ) = [ 1 , 1] , an closed interval with endpoints, 1 and 1 ....
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 Fall '09
 Frade
 Natural number, Countable set, Seymour Lipschutz

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