triclino Posted August 2, 2009 Share Posted August 2, 2009 Given that the definition of a function f, from A to B (sets) is: 1) f is a subset of AxB............................................................................... [math]f\subset AxB[/math] ,and 2)for all xεΧ ,there exists a unique yεΥ SUCH that (x,y)εf .......................................... [math]\forall x [x\in X\Longrightarrow\exists! y(y\in Y\wedge (x,y)\in f)][/math] prove that the following is a function from A to B,where A = {1,2} B = { a,b } f = { (1,a),(2,b) } Link to comment Share on other sites More sharing options...
the tree Posted August 2, 2009 Share Posted August 2, 2009 Since the sets are finite, you can just go through A, checking that each element appears in f once and only once. Not elegant - granted - but a proof. Link to comment Share on other sites More sharing options...
triclino Posted August 2, 2009 Author Share Posted August 2, 2009 Thanks. but it is the details of the proof that i got confused. For example in proving that :[math] f\subset A\times B[/math], i started: Let [math](x,y)\in f\Longrightarrow (x,y) = (1,a)\vee (x,y)=(2,b)[/math],then?? Link to comment Share on other sites More sharing options...
triclino Posted August 27, 2009 Author Share Posted August 27, 2009 Some body sujested the following proof which i do not understand: [math](x,y)\in f\Longrightarrow (x,y)=(1,a)\vee(x,y)=(2,b)[/math][math]\Longrightarrow (x,y)=(1,a)\vee (x,y)=(2,b)\vee (x,y)=(1,b)\vee (x,y)=(2,a)[/math] [math]\Longrightarrow (x,y)\in A\times B[/math] I do not know how we go from the 1st implication to the 2nd Link to comment Share on other sites More sharing options...
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