triclino Posted August 2, 2009 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) }
the tree Posted August 2, 2009 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.
triclino Posted August 2, 2009 Author 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??
triclino Posted August 27, 2009 Author 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
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