week24 - > 1, then | C | > 2 | s |...

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Cardinality Aside (Short Example) Original Notes adopted from March 26, 2002 (Week 24) c ± P. Rosenthal , MAT246Y1, University of Toronto, Department of Mathematics NOTE: Tests were handed back today, and solutions to Term Test 3 were discussed. Everything taken up in this lecture is material from previous lecture notes. However, one extra note which was added for proving/finding Cardinality. Let C = { f : S T } . If | T | = 1, then | C | = 1. If | T |
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Unformatted text preview: > 1, then | C | > 2 | s | Aside: ' f : S { t 1 ,t 2 } = Characteristic Functions = 2 | S | . Each function from S to T is a set of ordered pairs of the form ( s,t ), with s S,t T . That is, every function from S to T is a subset of S T . Therefore | C | 6 2 | S T | . In particular: If | S | = | S T | and | T | > 1, then | C | = 2 | S | . 1...
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