Function and subsets - inverses

[threeder]

Homework Statement

Let $f:X \rightarrow Y$ and $B_1, B_2 \in P(Y)$ where P(Y) is the power set.

Prove that $f^{-1}(B_1\cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$

Homework Equations

The book gives this definition:
Suppose $f:X \rightarrow Y$ is a function.

The function $f^{-1}:P(Y) \rightarrow P(X)$ is defined by $$f^{-1}(B)=\begin{cases} x\in X | f(x)\in B \end{cases}\}$$ for $B\in P(Y)$

The Attempt at a Solution

All I can do is just basically rewrite the definition:

Say $y_0 \in B_1\cap B_2$. Then $$f^{-1}(\{y_0\})=\{x \in X ~|~ y_0=f(x)\}$$

Then I make magic leap concluding that since $y_0 \in B_1\cap B_2$, $y\in B_1$ and $y\in B_2$. Hence, these sets will be equal and so $f^{-1}(B_1\cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$ . But I realize that a lot of grammar is missing. Can somebody help me out?

 

[HallsofIvy]

The "magic leap" can be given in more detail. You are basically trying to prove "$A= B$". To prove that for sets, first prove "$A\subset B$ and the prove $B\subset A$. And to prove $A\subset B$, start with "if $x\in A$" and use the properties of A and B to conclude "then $x\in B$".

Here, you want to prove that $f^{-1}(B_1\cap B_2)= f^{-1}(B_1)\cap f^{-1}(B_2)$. So start "if $x\in f^{-1}(B_1\cap B_2)$". Then, by definition of $f^{-1}$, there exist $y\in B_1\cap B_2$ such that f(x)= y. Since y is in $B_1\cap B_2$, y is in $B_1$ and y is in $B_2$.
1) If y is in $B_1$ and f(x)= y, then $x\in f^{-1}(B_1)$.
2) If y is in $B_2$ and f(x)= y, then $x\in f^{-1}(B_2)$.

Since both are true, $x\in f^{-1}(B_1)\cap f^{-1}(B_2)$.
That is, $f^{-1}(B_1\cap B_2)\subset f^{-1}(B_1)\cap f^{-1}(B_2)$.

Now, to prove the other way, that [itex]f^{-1}(B_1)\cap f^{-1}(B_2)\subset f^{-1}(B_1\cap B_2)/itex], start:<br /> If $x\in f^{-1}(B_1)\cap f^{-1}(B_2)$ and I will let you try the rest.[/itex]

 

[threeder]

Lets try it:

Say $x\in f^{-1}(B_1)\cap f^{-1}(B_2)$. Then there exists $y\in B_1$ and it also $y\in B_2$ such that $y=f(x)$. Because of $x$ belonging to both sets, and such $y$ existence, it also belongs to $y\in B_1\cap B_2$ such that $y=f(x)$. Hence $f^{-1}(B_1)\cap f^{-1}(B_2)$ is a subset of $f^{-1}(B_1\cap B_2)$

right?