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Expert-verified Found in: Page 115 ### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095 # Question: a. Give an example to show that the inclusion in part (b) in exercise 40 may be proper. b. Show that if f is one-to-one, the inclusion in part(b) in exercise 40 is an equality.

The function is

$a\right)f\left(S\right)\cup f\left(T\right)=f\left(S\cup T\right)\phantom{\rule{0ex}{0ex}}b\right)f\left(S\cap T\right)\subseteq f\left(S\right)\cap f\left(T\right)$

See the 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)

$f:A\to B\phantom{\rule{0ex}{0ex}}S\subseteq A\text{and T}\subseteq \text{A}\phantom{\rule{0ex}{0ex}}f\left(S\cap T\right)=f\left(S\right)\cap f\left(T\right)$

For example:

Let$A=R,B=R,S=\left\{0,1\right\}\text{and T = { 1,2}}$

$\text{f(x) =}\left\{1\text{if x is even}\phantom{\rule{0ex}{0ex}}0\text{if x is odd}\right\$

Let us determine the image of every element is S and T.

$f\left(0\right)=1\phantom{\rule{0ex}{0ex}}f\left(1\right)=0\phantom{\rule{0ex}{0ex}}f\left(2\right)=1$

$f\left(S\right)$contains all elements that are

$f\left(S\right)\cap f\left(T\right)=\left\{0,1\right\}$

$S\cap T$contains all elements in both sets.

$f\left(S\cap T\right)=\left\{0\right\}$

We then note in this case $f\left(S\cap T\right)\subseteq f\left(S\right)\cap f\left(T\right)$, but not$f\left(S\cap T\right)=f\left(S\right)\cap f\left(T\right)$

## Step 3:

Given: (a)

$f:A\to B\phantom{\rule{0ex}{0ex}}S\subseteq A\text{and T}\subseteq \text{A}$

is one-to-one

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

PROOF

FIRST PART

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

role="math" localid="1668423347151" $x\in f\left(S\right)\cap f\left(T\right)$

By the definition of subset:

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

## Step 4:

FIRST PART

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

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

Hence, the solution is

Since $f\left(S\right)\cap f\left(T\right)\subseteq f\left(S\cap T\right)$ and $f\left(S\cap T\right)\subseteq f\left(S\right)\cap f\left(T\right)$the two sets have to be equal:

$f\left(S\right)\cap f\left(T\right)=f\left(S\cap T\right)$ ### Want to see more solutions like these? 