Proving equivalence for a collection of subsets?

[SMA_01]

Homework Statement

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.

 

[SammyS]

Homework Statement

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.

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

 

[SMA_01]

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

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

 

[SammyS]

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

and that means ...

 

[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?

 

[SMA_01]

and that means ...

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

 

[Michael Redei]

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

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

 

[SMA_01]

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

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

 

[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_$\alpha$?

 

[Michael Redei]

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

 

[SammyS]

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

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

 

[SMA_01]

Got it, thank you both for your help!