# Homework Help: Proving equivalence for a collection of subsets?

1. Dec 8, 2012

### SMA_01

1. The problem statement, all variables and given/known data
Let f:X→Y where X and Y are sets. Prove that if {S$\alpha$}$\alpha\in$I is a collection of subsets of Y, then f-1($\cup$$\alpha\in$IS$\alpha$)=$\cup$$\alpha\in$If-1(S$\alpha$)

Would I prove this by showing set inclusion both ways? And any hints on how to begin?

Thanks.

2. Dec 8, 2012

### SammyS

Staff Emeritus
OK. So if $\displaystyle x\in f^{-1}\left(\cup_{\alpha\in I}\,S_\alpha \right)$, what does that say?

3. Dec 8, 2012

### SMA_01

That f(x) is in the union of the subsets?

4. Dec 8, 2012

### SammyS

Staff Emeritus
and that means ...

5. Dec 8, 2012

### SMA_01

This is what i'm thinking:
If you take $\cup$S$\alpha$ in Y, then that's the collection of subsets of Y, and "map" it back to X, then it would include every x in X such that f(x) is in one of the subsets, correct?
I'm not sure if i'm interpreting the other side of the equivalence correctly, but here's my attempt: If I take a particular S$\alpha$ and I map it back to X, then that will encompass every x value such that f(x) is in that particular subset. Then, I do this for each S$\alpha$ and take the union of all the x values in the preimage...is this a correct line of thinking?

6. Dec 8, 2012

### SMA_01

That f(x) must be in at least one particular subset...?

7. Dec 8, 2012

### Michael Redei

How about giving that subset a name, say, $S_a$? Now, if $f(x)\in S_a$, where does $x$ lie?

8. Dec 8, 2012

### SMA_01

Then x would be in the preimage of S$\alpha$?

9. Dec 8, 2012

### SMA_01

Okay, I see how this shows the left-hand side of the equivalence is a subset of the right-hand side. But when writing my proof, will I have to generalize it so that f(x) can be in more than one S$\alpha$?

10. Dec 8, 2012

### Michael Redei

No, one is enough. If some element $e\in S_k$, then $e\in\bigcup_{k\in I}S_k$.

11. Dec 9, 2012

### SammyS

Staff Emeritus
I would be inclined to say, $\displaystyle \dots\text{ "then }f(x)\in S_{\alpha_0}\,, \text{ for some }\alpha_0\in I/ .\text{"}$

12. Dec 9, 2012

### SMA_01

Got it, thank you both for your help!