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[parent] the inverse image commutes with set operations (Proof)

Theorem. Let $ f$ be a mapping from $ X$ to $ Y$. If $ \{B_i\}_{i\in I}$ is a (possibly uncountable) collection of subsets in $ Y$, then the following relations hold for the inverse image:

(1)
$ \displaystyle f^{-1}\big(\bigcup_{i\in I} B_i\big) = \bigcup_{i\in I} f^{-1}\big(B_i\big) $
(2)
$ \displaystyle f^{-1}\big(\bigcap_{i\in I} B_i\big) = \bigcap_{i\in I} f^{-1}\big(B_i\big) $
If $ A$ and $ B$ are subsets in $ Y$, then we also have:
(3)
For the set complement,
$\displaystyle \big(f^{-1}(A)\big)^\complement=f^{-1}(A^\complement).$
(4)
For the set difference,
$\displaystyle f^{-1}(A\setminus B) = f^{-1}(A)\setminus f^{-1}(B).$
(5)
For the symmetric difference,
$\displaystyle f^{-1}(A \bigtriangleup B)=f^{-1}(A) \bigtriangleup f^{-1}(B).$

Proof. For part (1), we have

$\displaystyle f^{-1}\big(\bigcup_{i\in I} B_i\big)$ $\displaystyle =$ $\displaystyle \Big\{ x\in X\mid f(x) \in \bigcup_{i\in I} B_i\Big\}$  
  $\displaystyle =$ $\displaystyle \left\{ x\in X \mid f(x) \in B_i\ \mbox{for some}\ i\in I\right\}$  
  $\displaystyle =$ $\displaystyle \bigcup_{i\in I}\left\{ x\in X \mid f(x) \in B_i \right\}$  
  $\displaystyle =$ $\displaystyle \bigcup_{i\in I} f^{-1}\big(B_i\big).$  

Similarly, for part (2), we have
$\displaystyle f^{-1}\big(\bigcap_{i\in I} B_i\big)$ $\displaystyle =$ $\displaystyle \big\{ x\in X \mid f(x) \in \bigcap_{i\in I} B_i\big\}$  
  $\displaystyle =$ $\displaystyle \left\{ x\in X \mid f(x) \in B_i\ \mbox{for all}\ i\in I\right\}$  
  $\displaystyle =$ $\displaystyle \bigcap_{i\in I}\left\{ x\in X \mid f(x) \in B_i \right\}$  
  $\displaystyle =$ $\displaystyle \bigcap_{i\in I} f^{-1}\big(B_i\big).$  

For the set complement, suppose $ x\notin f^{-1}(A)$. This is equivalent to $ f(x)\notin A$, or $ f(x)\in A^\complement$, which is equivalent to $ x\in f^{-1}(A^\complement)$. Since the set difference $ A\setminus B$ can be written as $ A\cap B^c$, part (4) follows from parts (2) and (3). Similarly, since $ A\bigtriangleup B=(A\setminus B) \cup (B\setminus A)$, part (5) follows from parts (1) and (4). $ \Box$



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Cross-references: equivalent, proof, symmetric difference, set difference, complement, inverse image, relations, subsets, collection, uncountable, mapping
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This is version 8 of the inverse image commutes with set operations, born on 2003-04-26, modified 2003-07-30.
Object id is 4213, canonical name is InverseImageCommutesWithSetOperations.
Accessed 3042 times total.

Classification:
AMS MSC: 03E20 (Mathematical logic and foundations :: Set theory :: Other classical set theory )
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