Exercise 1.2.9

Question

Given a function $f: D \rightarrow \mathbf{R}$ and a subset $B \subseteq \mathbf{R}$, let $f^{-1}(B)$ be the set of all points from the domain $D$ that get mapped into $B ;$ that is, $f^{-1}(B)=\{x \in D: f(x) \in B\} .$ This set is called the \emph{preimage} of $B$.
  \begin{enumerate}[label=(\alph*)] 
  \item Let $f(x)=x^{2} .$ If $A$ is the closed interval $[0,4]$ and $B$ is the closed interval $[-1,1]$, find $f^{-1}(A)$ and $f^{-1}(B)$. Does $f^{-1}(A \cap B)=f^{-1}(A) \cap f^{-1}(B)$ in this case? Does $f^{-1}(A \cup B)=f^{-1}(A) \cup f^{-1}(B) ?$
  \item The good behavior of preimages demonstrated in (a) is completely general. Show that for an arbitrary function $g: \mathbf{R} \rightarrow \mathbf{R}$, it is always true that $g^{-1}(A \cap B)=g^{-1}(A) \cap g^{-1}(B)$ and $g^{-1}(A \cup B)=g^{-1}(A) \cup g^{-1}(B)$ for all sets $A, B \subseteq \mathbf{R}$
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item $f^{-1}(A) = [-2, 2]$, $f^{-1}(B) = [-1, 1]$, $f^{-1}(A\cap B) = [-1, 1] = f^{-1}(A) \cap f^{-1}(B)$, $f^{-1}(A \cup B) = [-2, 2] = f^{-1}(A) \cup f^{-1}(B)$
  \item First let $x \in g^{-1}(A \cap B)$ meaning $g(x) \in A \cap B$ implying $g(x) \in A$ and $g(x) \in B$ which is the same as $x \in g^{-1}(A)$ and $x \in g^{-1}(B)$ meaning $x \in g^{-1}(A) \cap g^{-1}(B)$. \par
    Second let $x \in g^{-1}(A) \cap g^{-1}(B)$, this is the same as $x \in g^{-1}(A)$ and $x \in g^{-1}(B)$ which is the same as $g(x) \in A$ and $g(x) \in B$ implying $g(x) \in A \cap B$ and thus $x \in g^{-1}(A \cap B)$. Thus $g^{-1}(A \cap B) = g^{-1}(A) \cap g^{-1}(B)$.

Seeing $g^{-1}(A \cup B) = g^{-1}(A) \cup g^{-1}(B)$ is obvious. see 1.2.7 (d).
   \end{enumerate}

Exercise 1.2.9

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Exercise 1.2.9

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