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) }

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.

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??

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