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Q41E

Expert-verifiedFound in: Page 115

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.**

**Answer: **

The function is

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

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:${X}{\subseteq}{Y}$

Given: (a)

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

For example:

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

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

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

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

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

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

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

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

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:

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

By the definition of subset:

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

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

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

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

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