Sets and functions are intimately related, as we will see throughout the chapter. Here we begin to explore some basic connections. Suppose $f\colon A\to B$ is a function. If $X\subseteq A$, define $$ f(X)=\{b\in B: \exists\, a\in X (b=f(a))\}\subseteq B, $$ called the image of $X$. If $Y\subseteq B$, define $$ f^{-1}(Y)=\{a\in A:f(a)\in Y\}\subseteq A, $$ called the preimage of $Y$. There is real opportunity for confusion here: the letter $f$ is being used in two different, though related, ways. We can apply $f$ to elements of $A$ to get elements of $B$, or we can apply it to subsets of $A$ to get subsets of $B$. Similarly, we can talk about the images or preimages of either elements or subsets. Context should always make it clear what is meant, but you should be aware of the problem.
Suppose $A=\{1,2,3,4,5,6\}$, $B=\{r,s,t,u,v,w\}$ and
$$ \begin{array}{} f(1)=r&f(3)=v&f(5)=r\\ f(2)=s&f(4)=t&f(6)=v\\ \end{array} $$
Then $$ \eqalign{f(\{1,3,5\}) &=\{r,v\},\cr f(\{4,5,6\})&=\{t,r,v\},\cr} $$ and $$ \eqalign{f^{-1}(\{r,t,u\})&=f^{-1}(\{r,t\}) = \{1,4,5\},\cr f^{-1}(\{u,w\})&=\emptyset.\cr} $$$\square$
Example 4.2.2 Suppose $f\colon \R\to \R$ is given by $f(x)=x^2$. Then $$ \eqalign {f([2,3])&=[4,9],\cr f((-2,1])&=[0,4),\cr f(\{1,2,3\})&=\{1,4,9\},\cr} $$ and $$ \eqalign{f^{-1}([0,1])&=[-1,1],\cr f^{-1}([-1,0])&=\{0\},\cr f^{-1}((-\infty,0))&=\emptyset.\cr} $$$\square$
By the range (or image) of a function $f\colon A\to B$, we mean $$ f(A)=\{b\in B:\exists a\in A\,(b=f(a))\}. $$ The range may be considerably smaller than the codomain.
Example 4.2.3 The range of the function in example 4.2.1 is $\{r,s,v,t\}$, which is a proper subset of the codomain. $\square$
Example 4.2.4 The range of $\sin\colon \R\to \R $ is $[-1,1]$. The range of $f\colon \R\to\R$ given by $f(x)=x^2$ is $[0,\infty)$. $\square$
The next two theorems show how induced set functions behave with respect to intersection and union.
Theorem 4.2.5 Suppose $f\colon A\to B$ is a function and $Y$ and $Z$ are subsets of $B$. Then
a) $f^{-1}(Y\cup Z)= f^{-1}(Y)\cup f^{-1}(Z)$,
b) $f^{-1}(Y\cap Z)= f^{-1}(Y)\cap f^{-1}(Z)$.
Proof. We prove part (b) and leave part (a) as an exercise. If $a\in A$, then $a\in f^{-1}(Y\cap Z)$ if and only if $f(a)$ is in $Y\cap Z$. This is true if and only if $f(a)\in Y$ and $f(a)\in Z$. This, in turn, is equivalent to $a\in f^{- 1}(Y) $ and $a\in f^{-1}(Z)$. Finally, this is true if and only if $a\in f^{-1}(Y)\cap f^{-1}(Z)$. $\qed$
Theorem 4.2.6 Suppose $f\colon A\to B$ is a function and $W$ and $X$ are subsets of $A$. Then
a) $f(W\cup X)=f(W)\cup f(X)$,
b) $f(W\cap X)\subseteq f(W)\cap f(X)$.
Proof. We'll do part (b). If $b\in B$ is in $f(W\cap X)$, then $b=f(a)$ for some $a\in W\cap X$. Since $a\in W\cap X$, $a$ is in both $W$ and $X$. Therefore, $b=f(a)$ is in both $f(W)$ and $f(X)$, that is, $b\in f(W)\cap f(X)$. $\qed$
It is perhaps surprising to compare these two theorems and observe that of the two induced set functions, it is $f^{-1}$ that is "better behaved'' with respect to the usual set operations.
Exercises 4.2
In the first two exercises, use the function $f\colon \{1,2,3,4,5,6,7\}\to \{a,b,c,d,e,f,g,h\}$ given by:
$$ \begin{array}{} f(1)=d&f(4)=a&f(6)=e\\ f(2)=e&f(5)=b&f(7)=f\\ f(3)=f&&\\ \end{array} $$
Ex 4.2.1 Find the following:
a) $f(\{2,4,6\})$
b) $f(\emptyset)$
c) $f^{-1}(\{d,e,h\})$
d) $f^{-1}(\{a, b, f, c\})$
Ex 4.2.2 Use the function $f$ given above.
a) If $Y=\{a,b,c,e,f\}$ and $Z=\{a,b,e,g,h\}$, verify the statements in Theorem 4.2.5.
b) If $W=\{1,2,3,4\}$ and $X=\{2,4,6,7\}$, verify the statements in Theorem 4.2.6.
Ex 4.2.3 Suppose $f\colon \R \to \R$ is given by $f(x)= |x-1|$. Find the following:
a) $f([-1,1])$
b) $f(\{-4, -2, 0, 1, 5\})$
c) $f^{-1}((0,2))$
d) $f^{-1}(\{-2, 0, 4, 5\})$
Ex 4.2.4 Suppose $f\colon \R\to \R$ is given by $f(x)=x^2$.
a) If $Y=(1,\infty)$ and $Z= (-\infty, 4)$, verify the statements in Theorem 4.2.5.
b) If $W=[-3, 2]$, and $X=(0, 4]$, verify the statements in Theorem 4.2.6.
Ex 4.2.5 Prove 4.2.5(a).
Ex 4.2.6 Prove 4.2.6(a).
In the next two exercises suppose $f\colon A\to B$ is a function and $\{X_i\}_{i\in I}$ is a family of subsets of $A$.
Ex 4.2.7 Prove $f(\bigcup_{i\in I}X_i)= \bigcup_{i\in I}f(X_i)$.
Ex 4.2.8 Prove $f(\bigcap_{i\in I}X_i)\subseteq \bigcap_{i\in I}f(X_i)$.
In the next two exercises suppose $\{Y_i\}_{i\in I}$ is an indexed family of subsets of $B$ and $f\colon A\to B$ is a function.
Ex 4.2.9 Prove $f^{-1}(\bigcup_{i\in I} Y_i)=\bigcup_{i\in I} f^{-1}(Y_i)$.
Ex 4.2.10 Prove $f^{-1}(\bigcap_{i\in I} Y_i)=\bigcap_{i\in I} f^{-1}(Y_i)$.