Basic proof about inverse images of sets

1. Jan 1, 2008

jacobrhcp

[SOLVED] basic proof about inverse images of sets

1. The problem statement, all variables and given/known data

let f map X onto Y, and let A be a subset of X, B subset of Y

1) Prove that if y is in B, then f^(-1) (y) is in f^(-1) (B)

3. The attempt at a solution

1) I think I'm missing some two-line proof, because I'm getting nowhere and this is supposed to be simple. I just don't know where to start, so just a hint would be appreciated.

Last edited: Jan 1, 2008
2. Jan 1, 2008

HallsofIvy

Staff Emeritus
The statement, as given, is not true. Since f is only given as a "map", it may not have an inverse function and so f-1(y) may not exist. Your question, however, is about "inverse images", which exist whether or not f has an inverse. What is true is that f-1({y}) is a subset of f-1(B). {y} is, of course, the set containing the single point y. That statement follows directly from the definition of "inverse image". What is the definition of f-1(B)?

Last edited: Jan 1, 2008
3. Jan 1, 2008

jacobrhcp

f^-1 (B) := {x in X| f(x) in B}

The book told us that strictly speaking one should write f({y}), but in order to keep simplicity, they just use the notation f(y). What I'm trying to prove is what you have written down. I am sorry I didn't bring that part over correct.

but, back to the problem. this would make

f^-1 (y) = {x in X| f(x) in {y}}
f^-1 (B) = {x in X| f(x) in B}

because y is in B, this implies that f^-1 (y) is in f^-1 (B), is that right?

Anyway, thanks for your help... I'm not even sure if this is 'calc&beyond'... I'm studying physics and took mathematics as a major too since the start of this year, and this class is given at the same time as the first calculus and linear algebra classes.

4. Jan 1, 2008

HallsofIvy

Staff Emeritus
Hmmph! Well, go by your text book- but the part about "is a subset of" rather than "is in" is important.

Again, "is a subset of" not "is in" which means "is a member of". $f^{-1}(y)= f^{-1}(\{y\})$ is a subset of X, not a member of X! If $x\in f^{-1}(\{y\})$, then, by definition, $f(x)\in \{y\}$ which is a subset of B. Therefore, if $x\in f^{-1}({y})$, $f(x)\in B$.

5. Jan 1, 2008

jacobrhcp

haha, I love the fact you could respond so quickly.

The textbook used 'is a subset of' instead of 'is in' on many places (I think in all the places it is necessary), I just went over it too fast so I didn't copy it right because I hasn't realized it made such a difference. I get what you're explaining now, though.

Thanks, and SOLVED!