Find the Union and Intersection of Each of the Following Families

Suppose $I$ is a fix, called the alphabetize set, and with each $i\in I$ we associate a set $A_i$. We call $\{A_i:i\in I\}$ an indexed family unit of sets. Sometimes this is denoted by $\{A_i\}_{i\in I}$.

Case 1.6.1 Suppose $I$ is the days of the twelvemonth, and for each $i\in I$, $A_i$ is the fix of people whose altogether is $i$, and then, for example, $\hbox{Beethoven}\in A_{({\scriptsize\hbox{December 16}})}$. $\square$

Case 1.vi.two Suppose $I$ is the integers and for each $i\in I$, $A_i$ is the set of multiples of $i$, that is, $A_i=\{10\in \Z: i\vert x\}$ (recall that $i|x$ means that $x$ is a multiple of $i$; information technology is read "$i$ divides $10$''). $\square$

Given an indexed family $\{A_i:i\in I\}$ we tin define the intersection and union of the sets $A_i$ using the universal and existential quantifiers: $$ \eqalign{\bigcap_{i\in I} A_i & = \{ ten: \forall i\in I\,(x\in A_i)\}\cr \bigcup_{i\in I} A_i & = \{ x: \exists i\in I\,(10\in A_i)\}.\cr} $$

Example 1.half-dozen.three If $\{A_i:i\in I\}$ is the indexed family of case 1.6.ane, then $\bigcap_{i\in I} A_i$ is the empty set and $\bigcup_{i\in I} A_i$ is the set of all people. If $\{A_i:i\in I\}$ is the indexed family of example 1.6.2, then $\bigcap_{i\in I} A_i$ is $\{0\}$ and $\bigcup_{i\in I} A_i$ is the set of all integers. $\square$

Since the intersection and union of an indexed family unit are essentially "translations'' of the universal and existential quantifiers, information technology should not be likewise surprising that there are De Morgan's laws that employ to these unions and intersections.

Theorem 1.6.iv If $\{A_i:i\in I\}$ is an indexed family of sets then

a) $(\bigcap_{i\in I} A_i)^c = \bigcup_{i\in I} A_i^c$,

b) $(\bigcup_{i\in I} A_i)^c = \bigcap_{i\in I} A_i^c$.

Proof. We'll practice (a): $x\in (\bigcap_{i\in I} A_i)^c$ iff $\lnot (x\in \bigcap_{i\in I} A_i)$ iff $\lnot \forall i\,{\in}\, I\,(10\in A_i)$ iff $\exists i\,{\in}\, I\,(10\notin A_i)$ iff $\exists i\,{\in}\, I\,(x\in A_i^c)$ iff $x\in \bigcup_{i\in I} A_i^c$. $\qed$

You may be puzzled past the inclusion of this theorem: is it not simply role of theorem i.five.6? No: theorem

Theorem 1.6.v If $\{A_i:i\in I\}$ is an indexed family of sets and $B$ is whatever gear up, then

a) $\bigcap_{i\in I} A_i\subseteq A_j$, for each $j\in I$,

b) $A_j\subseteq \bigcup_{i\in I} A_i$, for each $j\in I$.

c) if $B\subseteq A_i$, for all $i\in I$, and so $B\subseteq \bigcap_{i\in I} A_i$,

d) if $A_i\subseteq B$, for all $i\in I$, so $\bigcup_{i\in I} A_i\subseteq B$.

Proof. Part (a) is a case of specialization, that is, if $x\in \bigcap_{i\in I} A_i$, then $ten\in A_i$ for all $i\in I$, in particular, when $i=j$. Part (d) also is easy—if $x\in \bigcup_{i\in I} A_i$, then for some $i\in I$, $ten\in A_i\subseteq B$, then $x\in B$. Parts (b) and (c) are left as exercises. $\qed$

An indexed family $\{A_i:i\in I\}$ is pair-wise disjoint if $A_i\cap A_j=\emptyset$ whenever $i$ and $j$ are distinct elements of $I$. The indexed family of case ane.6.i is pair-wise disjoint, but the 1 in example 1.6.ii is not. If $S$ is a set and then an indexed family unit $\{A_i:i\in I\}$ of subsets of $Southward$ is a partition of $S$ if information technology is pair-wise disjoint and $Southward= \bigcup_{i\in I} A_i$. Partitions appear frequently in mathematics.

Example one.half dozen.6 Let $I=\{e,o\}$, $A_e$ be the prepare of even integers and $A_o$ be the ready of odd integers. Then $\{A_i:i\in I\}$ is a partition of $S=\Z$. $\square$

Example 1.6.7 Permit $I=\R$, $S=\R^two$, and for each $i\in I$, let $A_i=\{(x,i):x\in \R\}$. Each $A_i$ is a horizontal line and the indexed family partitions the airplane. $\square$

Sometimes we want to talk over a collection of sets (that is, a set of sets) even though there is no natural alphabetize nowadays. In this case nosotros can use the collection itself as the index.

Instance 1.6.viii If ${\cal S}=\{\{1,3,4\}, \{ii,3,4,6\}, \{3,4,5,7\}\}$, so $\bigcap_{A\in {\cal Southward}}A =\{3,4\}$ and $\bigcup_{A\in {\cal South}}A =\{ane,2,3,4,5,6,seven\}$. $\foursquare$

An especially useful collection of sets is the power gear up of a set: If $X$ is whatever set up, the power set of $Ten$ is ${\cal P}(X)=\{A:A\subseteq X\}$.

Example 1.6.ix If $X=\{1,2\}$, so ${\cal P}(10)=\{\emptyset,\{ane\},\{two\},\{one,two\}\}$. $\square$

Example 1.6.ten ${\cal P}(\emptyset)=\{\emptyset\}$, that is, the power ready of the empty set is non-empty. $\square$

Exercises ane.vi

Ex 1.6.one Let $I=\{1,2,iii\}$, $A_1=\{1,3,4,6,seven\}$ $A_2=\{1,4,5,seven,8,nine\}$, $A_3= \{2,four,7,10\}$. Find $\bigcap_{i\in I} A_i$ and $\bigcup_{i\in I} A_i$.

Ex ane.half dozen.ii Suppose $I=[0,1]\subseteq \R$ and for each $i\in I$, permit $A_i=(i-1,i+1)\subseteq \R$. Find $\bigcap_{i\in I} A_i$ and $\bigcup_{i\in I} A_i$.

Ex 1.6.3 Prove parts (b) and (c) of theorem ane.6.five.

Ex 1.six.4 Suppose $U$ is the universe of discourse and the index set up $I=\emptyset$. What should we mean by $\bigcap_{i\in I} A_i$ and $\bigcup_{i\in I} A_i$? Show that Theorem i.6.4 still holds using your definitions.

Ex 1.6.5 Suppose $\{A_i\}_{i\in I}$ is a partition of a set $South$. If $T\subseteq South$, show that $\{A_i\cap T\}_{i\in I}$ is a partition of $T$.

Ex 1.6.half-dozen A drove of sets, $\cal S$, is totally ordered if for every $A,B\in {\cal Due south}$, either $A\subseteq B$ or $B\subseteq A$. If $\cal S$ is totally ordered, show that ${\cal South}^c=\{A^c: A\in {\cal South}\}$ is totally ordered.

Ex ane.vi.7 Suppose $\cal South$ is a collection of sets and $B$ is some other prepare. Show that if $B$ is disjoint from every $A\in {\cal S}$ then $B$ is disjoint from $\bigcup_{A\in {\cal S}} A$.

ortizdonsfult1950.blogspot.com

Source: https://www.whitman.edu/mathematics/higher_math_online/section01.06.html

Share this post