Question

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

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

Short answer

The function is

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

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)

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

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(T) 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(T) contains all elements that are the image of all an element in S.

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

Since

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

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 role="math" localid="1668673682522" $y\in Sory\in T$ 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

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

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)

$$\begin{gathered} f:A\to B \\ S\subseteq AandT\subseteq A \end{gathered}$$

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

PROOF

FIRST PART

Let $x\in f(S\cap T)$, 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(S) contains all elements that are the image of all an element in S.

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

data-custom-editor="chemistry" $\mathrm{f}\left(\mathrm{y}\right)\in \mathrm{f}\left(\mathrm{S}\right)\wedge \mathrm{f}\left(\mathrm{y}\right)\in \mathrm{f}\left(\mathrm{T}\right)$

Sincedata-custom-editor="chemistry" $\mathrm{f}\left(\mathrm{y}\right)=\mathrm{x}$

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

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

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