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Unformatted text preview: to end; (2) that the arguments are sufﬁcient to prove the theorem; and (3) that the arguments are all necessary to prove the statement. Proof of proposition 2.3 Assume that A, B and C are arbitrary sets and that A ⊆ B and B ⊆ C . We need to show that, for an arbitrary element x, if x is a member of A then it must also be a member of C . Assume x ∈ A. By assumption, we know that A ⊆ B , and hence by the deﬁnition of the subset relation that x ∈ B . We also know that B ⊆ C , and hence x ∈ C as required. 2.1.2 Set Equality Whenever we introduce a structure (in this case sets), an important part of knowing what we mean is to be able to identify when two such structures are equal, in other words when they denote the same thing. Two sets are equal if and only if they contain the same elements. D E FI N I T I O N 2 . 4 ( S E T E Q UA L I T Y ) Let A, B be any two sets. Then A equals B , written A = B , if and only if A ⊆ B and B ⊆ A: that is, A = B ⇔A ⊆ B∧B ⊆ A This deﬁnition of equality on sets means that the number of occurrences of elements and the order of the elements of a set do not matter: the sets {a, b, c} and {b, a, a, c} are equal. Contrast this property with the (perhaps more familiar) list data structure, where the order and number of elements is important. 2.2 Constructing Sets There are several ways of describing sets. So far, we have informally introduced two approaches: (1) either we just list the elements inside curly brackets: V = {a, e, i, o, u}, P N = {0, 1, 2, . . . }, {∅, {a}, {b}, {a, b}}; or (2) we deﬁne a set by stating the property that its elements must satisfy: R = {x : x is a real number} 6 = {x ∈ N : x is a prime number} ...
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This note was uploaded on 01/02/2010 for the course MATH Math2009 taught by Professor Koskesh during the Spring '09 term at SUNY Empire State.
 Spring '09
 Koskesh
 Math, Sets

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