Group Theory Sets and Mappings

[ZTV]

Homework Statement

Prove that f: S -> T is one-to-one if and only if f(AnB) = f(A) n f(B)
for every pair of subsets A and B of S

Homework Equations

See above

The Attempt at a Solution

Part 1:

Starting with the assumption f(AnB) = f(A) n f(B)

Let f(a) = f(b) [I'm going to try and prove that a=b for one-to-one]

f(a) is an element of f(A)
f(b) is an element of f(B)
f(a) = f (b)
So
f(a) is an element of f(A) n (B)
From assumption
f(a) and f(b) are elements of f(A) n f(B) so also elements of f (AnB)

so a, b are elements of AnB

Can't work out how to show they are equal though..

2. Assume f is one-to-one

So f(a)=f(b) and a=b

If a=b, a is an element of AnB
so f(a) is an element of f(AnB)

Likewise, if f(a)=f(b) then f(a) is an element of f(A) n f(B)

Don't know where to go from here.

 

[CompuChip]

For the first one, note that the assumption holds for any sets A and B. So you can choose them conveniently (hint: sets can consist of just one point).

For the second one, I suggest showing two inclusions f(A n B) c f(A) n f(B) and f(A) n f(B) c f(A n B).
One of them is trivial (if y is the image of a point x in A n B, then x is in A so it's the image of a point in A and x is in B so it's the image of a point in B, so it's definitely in f(A) and f(B)). For the other one you will need the injectivity, just write out definitions.

 

[ZTV]

Thanks... for the second one I'd considered looking at inclusions but couldn't work out how to implement it.

Don't understand what you mean about choosing convenient sets though, can you explain?

 

[CompuChip]

Well, you have assumed (I hope, otherwise do it now ) that f(AnB) = f(A) n f(B) for any subsets A and B of S, because that's what it said in the statement.

So to conclude that a = b, you can choose them as you like: if the assumption holds for any A and B, then in particular they hold for some A and B that you choose. For example, you are allowed to take A = B, A empty, B = S, A = S \ B, or whatever you think you need to prove that a = b.
My hint to you is: a single point x of S is also a subset {x} c S.

 

[ZTV]

Is for every A and B the same as any A and B?

 

[CompuChip]

Yep, they both mean $\forall A \subset S, \, \forall B \subset S$.