Question

Question: Let f be a function from the set A to the set B. Let S and T be subsets of A. show that

$$\begin{gathered} \mathit{a}\mathbf{\left)}\mathit{f}\mathbf{\right(}\mathit{S}\mathbf{\cup }\mathit{T}\mathbf{)}\mathbf{=}\mathit{f}\mathbf{(}\mathit{S}\mathbf{)}\mathbf{\cup }\mathit{f}\mathbf{(}\mathit{T}\mathbf{\left)} \\ \mathit{b}\mathbf{\right)}\mathit{f}\mathbf{(}\mathit{S}\mathbf{\cap }\mathit{T}\mathbf{)}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{S}\mathbf{\right)}\mathbf{\cap }\mathit{f}\mathbf{\left(}\mathit{T}\mathbf{\right)} \end{gathered}$$

Short answer

Answer:

The function is

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

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)

$$\begin{gathered} f:A\to B \\ S\subseteq A\text{and T}\subseteq \text{A} \end{gathered}$$

To proof:$f(S\cup T)=f\left(S\right)\cup f\left(T\right)$

PROOF

FIRST PART

Let $x\in f(S\cup T)$ . Then there exists an element $y\in f(S\cup T)$such that $f\left(y\right)=x$.

By the definition of the union

$y\in S\vee y\in T$

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

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

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

$f\left(y\right)\in f\left(S\right)\vee f\left(y\right)\in f\left(T\right)$

Since$f\left(y\right)=x$

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

By the definition of union:

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

By the definition of subset:

$f(S\cup T)=f\left(S\right)\cup f\left(T\right)$

Step 3:

SECOND PART

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

By the definition of the union

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

Then there exists an element $y\in S\text{or}y\in T$ such that$f\left(y\right)=x$

$y\in S\vee y\in T$

By the definition of union:

$y\in S\cup T$

$f(S\cup T)$contains all elements that are the image of an element in$S\cup T$

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

Since$f\left(y\right)=x$

By the definition of subset:

$f\left(S\right)\cup f\left(T\right)\subseteq f(S\cup T)$

Step 4:

Given: (a)

$f:A\to B$

$S\subseteq A\text{and T}\subseteq \text{A}$

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

PROOF

FIRST PART

Let $x\in f(S\cap T)$, then there exists an element $y\in S\cap T$such that$f\left(y\right)=x$.

By the definition of the union

$y\in S\wedge y\in T$

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

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

$f\left(y\right)\in f\left(S\right)\wedge f\left(y\right)\in f\left(T\right)$

Since$f\left(y\right)=x$

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

By the definition of union:

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

By the definition of subset:

$f(S\cap T)\subseteq f\left(S\right)\cap f\left(T\right)$

Hence, the solution is

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