Exercise 1.2.7

Question

Given a function $f$ and a subset $A$ of its domain, let $f(A)$ represent the range of $f$ over the set $A$; that is, $f(A)=\{f(x): x \in A\}$.
  \begin{enumerate}[label=(\alph*)] 
  \item Let $f(x)=x^{2} .$ If $A=[0,2]$ (the closed interval $\left.\{x \in \mathbf{R}: 0 \leq x \leq 2\}ight)$ and $B=[1,4]$, find $f(A)$ and $f(B) .$ Does $f(A \cap B)=f(A) \cap f(B)$ in this case? Does $f(A \cup B)=f(A) \cup f(B) ?$
  \item Find two sets $A$ and $B$ for which $f(A \cap B) 
eq f(A) \cap f(B)$.
  \item  Show that, for an arbitrary function $g: \mathbf{R} ightarrow \mathbf{R}$, it is always true that $g(A \cap B) \subseteq g(A) \cap g(B)$ for all sets $A, B \subseteq \mathbf{R}$
  \item Form and prove a conjecture about the relationship between $g(A \cup B)$ and $g(A) \cup g(B)$ for an arbitrary function $g$
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item $f(A)=[0, 4]$, $f(B)=[1,16]$, $f(A\cap B) = [1,4] = f(A)\cap f(B)$ and $f(A\cup B) = [0, 16] = f(A) \cup f(B)$
  \item $A = \{-1\}$, $B = \{1\}$ thus $f(A \cap B) = \emptyset$ but $f(A) \cap f(B) = \{1\}$
  \item Suppose $y \in g(A \cap B)$, then $\exists x \in A \cap B$ such that $g(x) = y$. But if $x \in A \cap B$ then $x \in A$ and $x \in B$, meaning $y \in g(A)$ and $y \in g(B)$ implying $y \in g(A) \cap g(B)$ and thus $g(A \cap B) \subseteq g(A) \cap g(B)$. \par
    Notice why it is possible to have $x \in g(A) \cap g(B)$ but $x 
otin g(A \cap B)$, this happens when something in $A \setminus B$ and something in $B \setminus A$ map to the same thing. If $g$ is 1-1 this does not happen.
  \item I conjecture that $g(A \cup B) = g(A) \cup g(B)$. To prove this we show inclusion both ways,
    First suppose $y \in g(A\cup B)$. then either $y \in g(A)$ or $y \in g(B)$, implying $y \in g(A) \cup g(B)$. Now suppose $y \in g(A) \cup g(B)$ meaning either $y \in g(A)$ or $y \in g(B)$ which is the same as $y \in g(A \cup B)$ as above.
   \end{enumerate}

Exercise 1.2.7

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Exercise 1.2.7

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