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cardinality - Cardinality The"cardinality of a set provides...

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Cardinality The “cardinality” of a set provides a precise meaning to “the number of elements” in the set, even when the set contains infinitely many elements. We start with the following very reasonable definitions of “the set S contains the same number of elements as the set T ”, “the set S contains fewer elements than the set T ” and “the set S contains more elements then the set T ”. Definition 1 Let S and T be nonempty sets. Then card ( S ) = card ( T ) ⇐⇒ there exists an f : S T which is one–to–one and onto card ( S ) card ( T ) ⇐⇒ there exists an f : S T which is one–to–one card ( S ) card ( T ) ⇐⇒ there exists an f : S T which is onto card ( S ) < card ( T ) ⇐⇒ card ( S ) card ( T ) but card ( S ) 6 = card ( T ) card ( S ) > card ( T ) ⇐⇒ card ( S ) card ( T ) but card ( S ) 6 = card ( T ) Definition 2 Let S be a nonempty set. Then (a) The set S has card ( S ) = n IN if card ( S ) = card ( { 1 , 2 , 3 , · · · , n } ) . (b) The set S is countable if card ( S ) card (IN). The set S is countably infinite if card ( S ) = card (IN). (c) The set S is uncountable if it is not countable. (d) The set S has the cardinality of the continuum if card ( S ) = card (IR). We now verify that and , in the sense of cardinality, work as expected. Proposition 3 Let S , T and U be nonempty sets. Then (a) card ( S ) card ( T ) if and only if card ( T ) card ( S ) . (b) Either card ( S ) card ( T ) or card ( S ) card ( T ) . (c) If card ( S ) card ( T ) and card ( S ) card ( T ) , then card ( S ) = card( T ) . (d) If card ( S ) card ( T ) and card ( T ) card ( U ) , then card ( S ) card ( U ) . Proof: (a) If card ( S ) card ( T ), then there is a one–to–one function f : S T . Fix any s 0 S . Then g ( t ) = f - 1 ( t ) if t is in the range of f s 0 otherwise maps T onto S , so that card ( T ) card ( S ). If card ( T ) card ( S ), then there is a function g that maps T onto S . So the sets g - 1 ( s ) = t T g ( t ) = s , s S are disjoint and nonempty. For each s S select one t s g - 1 ( s ). Then f ( s ) =
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