What is the Impact of Functions on Sets?

[number0]

Homework Statement

It is located in the pdf or the following link: http://www.scribd.com/doc/50331847/hw". There seems to be 2 typos:

"1. There should not be an initial "If" at the beginning.
2. The intersection on the right side of the equation should be a union, I believe."

Homework Equations

The Attempt at a Solution

I spent a large amount of time thinking about how to do this, and I cannot figure it out. Can anyone please help me? Thanks.

 

[Mark44]

Homework Statement

It is located in the pdf or the following link: http://www.scribd.com/doc/50331847/hw". There seems to be a typo; there is no "if" before the set relationship.

Homework Equations

The Attempt at a Solution

I spent a large amount of time thinking about how to do this, and I cannot figure it out. Can anyone please help me? Thanks.

I believe there are two typos.
As stated, the proposition is:
Let f: S -> T. If f(A $\cup$ B) = f(A) $\cap$ f(B) for all subsets A
and B of S if and only if f is an injection.

1. There should not be an initial "If" at the beginning.
2. The intersection on the right side of the equation should be a union, I believe. Here's a counterexample. Let f(x) = csc(x) with domain the sets A = [-1, 0) and B = (0, 1]. f is clearly one-to-one, so is an injection.

f(A $\cup$ B) = (-$\infty$, -1] $\cup$ [1, $\infty$), but f(A) $\cup$ f(B) is the empty set. This is a contradiction of one of the two conditions if the iff proposition.

With the foregoing in mind, I believe the proposition should be:
Let f: S -> T. f(A $\cup$ B) = f(A) $\cup$ f(B) for all subsets A
and B of S if and only if f is an injection.

 

[number0]

I believe there are two typos.
As stated, the proposition is:
Let f: S -> T. If f(A $\cup$ B) = f(A) $\cap$ f(B) for all subsets A
and B of S if and only if f is an injection.

1. There should not be an initial "If" at the beginning.
2. The intersection on the right side of the equation should be a union, I believe. Here's a counterexample. Let f(x) = csc(x) with domain the sets A = [-1, 0) and B = (0, 1]. f is clearly one-to-one, so is an injection.

f(A $\cup$ B) = (-$\infty$, -1] $\cup$ [1, $\infty$), but f(A) $\cup$ f(B) is the empty set. This is a contradiction of one of the two conditions if the iff proposition.

With the foregoing in mind, I believe the proposition should be:
Let f: S -> T. f(A $\cup$ B) = f(A) $\cup$ f(B) for all subsets A
and B of S if and only if f is an injection.

Yes, you are correct. I just checked the problem. Any ideas on how to approach this?

 

[Mark44]

Your book might have some similar examples.

 

[number0]

Your book might have some similar examples.

My book has examples about the functions acting on sets. But, it does not have any examples relating to injections, surjections, bijections relating to functions on sets.

 

[Mark44]

It doesn't have to have examples where the functions are injections, surjections, whatever. Look for examples or theorems about what a function does to a union of sets, and then look up the definition of injection (i.e. one-to-one) .