1. ## Sets/Functions

I have set A{1,2,3,4,5}, B{3,4,5,6,7}, C{5,6,7,8,9}

A union B = {1,2,3,4,,6,7}
A intersection B = {3,4,5}

Define an onto(surjective) function f: A union B -> A intersection B
Define an 1-1 function f: A union B -> A intersection B

2nd part--
Let U = A union B union C
Let D = {x|x,y,z in U) ^ (there exits y)(there exists z)(z = x * y)

U = {1,2,3,4,5,6,7,8,9}
I am not sure how to get set D.
Do I choose a value for y and z from set U and if there is an value for x inside Set U satstify the equation? and if it does, the value for x goes in set D?

2. Originally Posted by darken4life
2nd part--
Let U = A union B union C
Let D = {x|x,y,z in U) ^ (there exits y)(there exists z)(z = x * y)

U = {1,2,3,4,5,6,7,8,9}
I am not sure how to get set D.
Do I choose a value for y and z from set U and if there is an value for x inside Set U satstify the equation? and if it does, the value for x goes in set D?
Basically. I'd choose an x, say x = 1. Then try to find a y such that $\displaystyle z = x \cdot y$ such that $\displaystyle z \in U$.

For example: y = 1. Thus we ask the question is $\displaystyle z = 1 \cdot 1 \in U$? Yes. So $\displaystyle 1 \in D$.

etc for all values of y that create an appropriate z.

-Dan

3. Thanks! I got it clearly.
Can you help me on the first part?
Is not that I want you guys to answer it for me but simply the question.

Define an onto(surjective) function f: A union B -> A intersection B
-if A is union B then A is intersection B
How I define an function out of that?
Not sure on how to approach this.

4. Originally Posted by darken4life
Thanks! I got it clearly.
Can you help me on the first part?
Is not that I want you guys to answer it for me but simply the question.

Define an onto(surjective) function f: A union B -> A intersection B
-if A is union B then A is intersection B
How I define an function out of that?
Not sure on how to approach this.
The surjective function is easy: Just make up any function you like. For example, consider:
$\displaystyle f: \{1, 2, 3, 4, 5, 6, 7 \} \to \{3, 4, 5 \}$:
$\displaystyle f(1) = 3$
$\displaystyle f(2) = 3$
$\displaystyle f(3) = 4$
$\displaystyle f(4) = 4$
$\displaystyle f(5) = 5$
$\displaystyle f(6) = 5$
$\displaystyle f(7) = 4$

For each element in the codomain there is at least one element that exists in the domain.

The one to one function is rather harder: in fact it is impossible. In order to have a one to one function between sets they must have the same cardinality. These sets don't.

Though if we are allowed to restrict the domain to three elements, for example, we can do it:
$\displaystyle f: \{2, 3, 4 \} \to \{3, 4, 5 \}$:
$\displaystyle f(2) = 4$
$\displaystyle f(3) = 5$
$\displaystyle f(3) = 3$
is one example.

-Dan