# Troubled with Definition

1. May 16, 2010

### annoymage

1. The problem statement, all variables and given/known data

For sets A and B, the Cartesianproduct A x B is
{<a,b>: a $$\in$$ A and b $$\in$$ B}. A subset f of A x B is said to be a function from A to B (in symbols,f: A->B) just in case, for each element a of A there is exactly one element b of B with <a, b> $$\in$$ f:

Given B be the set {a,b}

and

There are four functions from B to B, namely:
{<a,a>, <b,a>}
{<a,a>, <b,b>}
{<a,b>, <b,a>}
{<a,b>, <b,b>}

2. Relevant equations

3. The attempt at a solution

by definition, i know how BxB

but i don understand B->B how to get that 4 function?
{<a,a>, <b,a>} meaning two output? or <a,a> is input, and <b,a> the output? or that is the subset of BxB?

this bring me to next question

"A subset f of A x B" or let make it simpler "A subset M of N" is this mean N is the subset of M?

help me clear my confusion

2. May 16, 2010

### tiny-tim

Hi annoymage!
Let's call those four subsets f g h and i.

Then f = {<a,a>, <b,a>} is a subset of AxB and is the function such that f(a) = a, f(b) = a.
Yes, every function is a subset of the Cartesian product of its range and domain spaces.

3. May 16, 2010

### HallsofIvy

Staff Emeritus
Since B contains only two members, a and b, and a "function from B" must have each of those exactly once, any function from B to another set can have only two members: <a, > and <b, >. Now, fill in those blank spots with elements of the other set. Since her the "target set" is also B, we have two things, a and b, we can put into those blank spots. since we can choose either a or b for each of those two blank spots, there are $2^2= 4$ choices:
1) a, a
2) a, b
3) b, a
4) b, b.

That gives the 4 possible functions: {<a, a>, <b, a>}, {<a, a>, <b, a>}, {<a, b>, <b, a>}, and {<a, b>, <b, b>}.

4. May 16, 2010

### annoymage

i'm still trying to understand what HallsofIvy trying to say, sorry but im no good in english ;P.

anyway, maybe i understand tiny-tim, ok, check my understanding

if B={a,b}
C={c,d}

possible fucntion of B->C are:

{<a,c>,<b,c>} is a subset of BxC and is the function such that f(a)=c , f(b)=c
{<a,d>,<b,d>} is a subset of BxC and is the function such that f(a)=d , f(b)=d
{<a,c>,<b,d>} is a subset of BxC and is the function such that f(a)=c , f(b)=d
{<a,d>,<b,c>} is a subset of BxC and is the function such that f(a)=d , f(b)=c

and now suddenly i get what HallsofIvy saying

but now, this bothers me instead

M={<a,c>,<b,c>} is a subset of BxC and is the function such that f(a)=c , f(b)=c

is it f(B)=c, B={a,b}???

{<a,c>,<b,d>} is a subset of BxC and is the function such that f(a)=c , f(b)=d

is it

f(B)={ c , B=a
d , B=b }

(pice wise function)

is it correct? sorry if i over-complicating things

5. May 17, 2010

### HallsofIvy

Staff Emeritus
Be careful here. f is defined on members of b, not B itself. We can give a separate definition: If f is a function from A to B and C is a subset of A, rather than a member of A, f(C) is defined as the set $\{y | y= f(x)\}$ for some x in A.
Since B contains a and b and both f(a)= c and f(b)= c, f(B)= {c}. That is, f(B) is NOT "c", it is the set containing only c.

If f(a)= c and f(b)= d then f(B)= f({a, b))= {c, d}.

Again, you have to distinguish between a function, defined on a set, applied to elements of that set and to subsets of that set.

If we are looking at sets of numbers, and $f(x)= x^2$, then f([-2, 2])= [0, 4] because any number in the interval from -1 to 1, squared, gives a number from 0 to 4.

6. May 17, 2010

### annoymage

understood, thank you thank you very much