Inverse image of union, difference and intersection of sets

Question

Let $X$ and $Y$ two sets not empty. Let \begin{equation} f\colon X\to Y \end{equation} an application. We define \begin{equation} f^{-1}\colon2^{Y}\to2^X \end{equation} in this way: if $B\subset Y$ \begin{equation} f^{-1}(B)=\{x:x\in X, f(x)\in B\}. \end{equation} How can I prove the following properties?

1) $f^{-1}(\cup_\alpha B_\alpha)=\cup_\alpha f^{-1}(B_\alpha)$;

2) $f^{-1}(B'\setminus B'')=f^{-1}(B')\setminus f^{-1}(B'')$

3) $f^{-1}(\cap_\alpha B_\alpha)=\cap_\alpha f^{-1}(B_\alpha)$.

Thanks!

Answer 1

Essential is the rule:$$x\in f^{-1}(A)\iff f(x)\in A\tag1$$

which gives great comfort to prove 1),2) and 3).

E.g. for 1) the following statements are equivalent:

• $x\in f^{-1}(\bigcup_{\alpha}B_{\alpha})$
• $f(x)\in\bigcup_{\alpha}B_{\alpha}$
• $\exists\alpha [f(x)\in B_{\alpha}]$
• $\exists\alpha [x\in f^{-1}(B_{\alpha})]$
• $x\in\bigcup_{\alpha}f^{-1}(B_{\alpha})$

so that we are allowed to conclude that: $$f^{-1}\left(\bigcup_{\alpha}B_{\alpha}\right)=\bigcup_{\alpha}f^{-1}(B_{\alpha})$$

Rule $(1)$ implies the equivalence of the first two bullets and implies the equivalence of the third and fourth bullet.

In a similar way you can solve 2) and 3).

Answer 2

Hint:

Remember a definition of set equality: $$ A=B\Longleftrightarrow A\subseteq B\;\; \wedge \;\;B\subseteq A $$

and $$X\subseteq Y \Longleftrightarrow \forall x\in X \implies x\in Y$$