Question

Let f be a function from A to B. Let S and T be subsets of B .

Show that

$$\begin{gathered} a\left)f^{-1}\right(S\cup T)=f^{-1}(S)\cup f^{-1}(T\left) \\ b\right)f^{-1}(S\cap T)=f^{-1}\left(S\right)\cap f^{-1}\left(T\right) \end{gathered}$$

Short answer

The function is

$$\begin{gathered} a\left)f^{-1}\right(S\cup T)=f^{-1}(S)\cup f^{-1}(T\left) \\ b\right)f^{-1}(S\cap T)=f^{-1}\left(S\right)\cap f^{-1}\left(T\right) \end{gathered}$$

Step-by-step solution

Step 1:

Union $A\cup B$: all elements that are either in A or in B.

Intersection $A\cap B$: all elements that are both in A and in B.

X is a subset of Y if every element of X is also an element of Y.

Notation:$X\subseteq Y$

Step 2:

Given: (a)

$$\begin{gathered} f:A\to B \\ S\subseteq BandT\subseteq B \\ a\left)f^{-1}\right(S\cup T)=f^{-1}(S)\cup f^{-1}(T) \end{gathered}$$

To proof:$f(S\cap T)\subseteq f\left(S\right)\cap f\left(T\right)$

PROOF

FIRST PART

Let $x\in f^{-1}(S\cup T)$. then there exists an element$y\in S\cup T$such that $f\left(y\right)=x$.

By the definition of the union

$$\begin{gathered} y\in S\vee y\in T \\ f\left(x\right)\in S\vee f\left(x\right)\in T \end{gathered}$$

$f^{-1}\left(S\right)$contains all elements that are the image of all an element in S.

$f^{-1}\left(T\right)$contains all elements that are the image of all an element in S.

$x\in f^{-1}\left(S\right)\wedge x\in f^{-1}\left(T\right)$

By the definition of union:

$x\in f^{-1}\left(S\right)\cup f^{-1}\left(T\right)$

By the definition of subset:

$f^{-1}(S\cup T)=f^{-1}\left(S\right)\cup f^{-1}\left(T\right)$

Step 3:

FIRST PART

Let $x\in f^{-1}\left(S\right)\cup f^{-1}\left(T\right)$.

By the definition of the union

$\begin{array}{c}y\in S\wedge y\in T\\ f\left(x\right)\in S\vee f\left(x\right)\in T\\ \end{array}$

By the definition of union

$f\left(x\right)\in S\cup T$

$f^{-1}(S\cup T)$contains all elements of A that have an element of$S\cup T$as image.

$x\in f^{-1}(S\cup T)$

By the definition of union:

$f^{-1}\left(S\right)\cup f^{-1}\left(T\right)\subseteq f^{-1}(S\cup T)$

The two sets have to be equal:

$f^{-1}(S\cup T)=f^{-1}\left(S\right)\cup f^{-1}\left(T\right)$

Step 4:

Given: (b)

$$\begin{gathered} f:A\to B \\ S\subseteq BandT\subseteq B \\ b\left)f^{-1}\right(S\cap T)=f^{-1}(S)\cap f^{-1}(T) \end{gathered}$$

To proof:$f^{-1}(S\cap T)=f^{-1}\left(S\right)\cap f^{-1}\left(T\right)$

PROOF

FIRST PART

Let $x\in f^{-1}(S\cap T)$ . then there exists an element $y\in S\cup T$such that $f\left(y\right)=x$.

By the definition of the union

$$\begin{gathered} y\in S\wedge y\in T \\ f\left(x\right)\in S\wedge f\left(x\right)\in T \end{gathered}$$

$f^{-1}\left(S\right)$contains all elements that are the image of all an element in S.

$f^{-1}\left(T\right)$contains all elements that are the image of all an element in S.

$x\in f^{-1}\left(S\right)\wedge x\in f^{-1}\left(T\right)$

By the definition of intersection:

$x\in f^{-1}\left(S\right)\wedge f^{-1}\left(T\right)$

By the definition of subset:

$f^{-1}(S\cap T)\subseteq f^{-1}\left(S\right)\cap f^{-1}\left(T\right)$

Step 5:

SECOND PART

Let $x\in f^{-1}\left(S\right)\cap f\left(T\right)$.

By the definition of the intersection

$\begin{array}{c}x\in f^{-1}\left(S\right)\wedge x\in f^{-1}\left(T\right)\\ f\left(x\right)\in S\wedge f\left(x\right)\in T\end{array}$

By the definition of intersection

$f\left(x\right)\in S\cap T$

contains all elements of A that have an element ofas image.

$x\in f^{-1}(S\cap T)$

By the definition of union:

$f^{-1}\left(S\right)\cap f^{-1}\left(T\right)\subseteq f^{-1}(S\cap T)$

The two sets have to be equal:

$f^{-1}(S\cap T)=f^{-1}\left(S\right)\cap f^{-1}\left(T\right)$

Hence, the solution is

data-custom-editor="chemistry" $$\begin{gathered} \text{a)}{f}^{-1}(S\cup T)=f^{-1}\left(S\right)\cup f^{-1}\left(T\right) \\ \text{b)}{f}^{-1}(S\cap T)=f^{-1}\left(S\right)\cap f^{-1}\left(T\right) \end{gathered}$$