Proving Function Properties with Subsets

[Enjoicube]

Alright, 2+1/2 problems here:

First:

36. Let F be a function from the set A to the set B. Let S and T be subsets of A. Show that:

a) f(S\cupT)=f(S)\cupf(T)
b)f(S\capT)\subseteq f(S)\capf(T)

Note: This must be done using a membership proof. i.e. prove that f(S\cupT)\subseteqf(S)\cupf(T) AND f(S)\cupf(T)\subseteqf(S\cupT)


40. Let f be a function from A to B. Let S be a subset of B. We define the inverse image of S to be the subset of A whose elements are precisely all pre images of all elements of S. We denote the inverse image of S by f^{-1}(S) so f^{-1}(S)={a\inA| f(a)\inS}

Prove parts a and b of 36 substituting the inverse image for f, and an equality in both parts (rather than subset)

1) (not a question from the text). This was not covered in the lecture or in the book, but how do most people go about proving that a function is surjective from the definition?

Homework Equations

f^{-1}(S)={a\inA| f(a)\inS}
f(S)={f(s)|s\inS}

The Attempt at a Solution

Alright, I only have an idea of how to do 36, 40 I am lost. The professor said to use proof by case for 36, so here is how I thought it would go:

a) prove:f(S\cupT)=f(S)\cupf(T)

Proof:

1) assume x\in f(SUT), then, by definition x\in{f(s)|s\inSUT}

2) If x\in{f(s)|s\inSUT}, then x\in{f(s)|s\inS or T} by definition of union.

3) If x\in{f(s)|s\inS}, then x\inf(S), and since f(S)\subseteqf(S)Uf(T), thus if x\in{f(s)|s\inS} then x\in{f(s)|s\inS}\cup{f(s)|s\inT}.

4)If x\in{f(s)|s\inT}, then x\inf(T), and since f(T)\subseteqf(S)Uf(T), thus if x\in{f(s)|s\inT} then x\in{f(s)|s\inS}\cup{f(s)|s\inT}.

4)Therefore if x\inf(SUT) then x\inf(S)Uf(T)

5 Now we must prove that f(S)\cupf(T)\subseteqf(S\cupT, to do this, just invert the steps of this proof (obviously I wouldn't put this on a test but to cut length, this is reasonable)

b)f(S\capT)\subseteq f(S)\capf(T)

proof:

1) assume x\in f(S\capT), then, by definition x\in{f(s)|s\inS\capT}

2) Then x\in{f(s)|s\inS and T}.

3) Thus s\inS and s\inT and so, through simplification, we can say:

4) s\inS, if s\inS then x\in{f(s)|s\inS}

5) We can also say s\inT, and if s\inT then x\in{f(s)|s\inT}.

6) Therefore, if s\inS and s\inT, which are true, we have x\in{f(s)|s\inS} and x\in{f(s)|s\inT} (we can say through addition).

7) Note that by the definition of intersection, x\in{f(s)|s\inS}\cap{f(s)|s\inT}.

8) Therefore, if x\in{f(s)|s\inS\capT} then x\in{f(s)|s\inS}\cap{f(s)|s\inT}.

9)if x\in{f(s)|s\inS\capT} then x\in{f(s)|s\inS}\cap{f(s)|s\inT}.


10)This means f(S\capT)\subseteqf(S)\capf(T).

For Problem 40, I am wondering whether the proofs follow the same steps, or is there something else I have to worry about?

 

[Enjoicube]

To make this easier, forget problem 36, and 40, I can go in for help for that. Now my main question is on proving surjectivity: How is this done. I understand, you must pick a random member y0 of the codomain and prove that there is some value x0 from the domain such that f(x0)=y0, in practice however, this gives me trouble. For example, f(x)=x^2 isn't surjective on R-->R, however, it is surjective on R-->R+, how would one guarantee that the y0 value picked from the codomain reflects the restrictions put on it?

 

[HallsofIvy]

To make this easier, forget problem 36, and 40, I can go in for help for that. Now my main question is on proving surjectivity: How is this done. I understand, you must pick a random member y0 of the codomain and prove that there is some value x0 from the domain such that f(x0)=y0, in practice however, this gives me trouble. For example, f(x)=x^2 isn't surjective on R-->R, however, it is surjective on R-->R+, how would one guarantee that the y0 value picked from the codomain reflects the restrictions put on it?

By saying so? If you want to prove that f(x)= x²:R-->R⁺ is surjective, you would start by saying "Let y be in R⁺." Then you are looking for x in R such that x²= y. Since y is in R⁺, x= y^1/2 exists.

 

[Enjoicube]

Aha, thank you, why it couldn't be explained as clearly as that in my text is beyond me.