For Every Set B in the Family a, B Is a Subset of Ua
Suppose $I$ is a set, called the index 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 past $\{A_i\}_{i\in I}$.
Example 1.6.1 Suppose $I$ is the days of the year, and for each $i\in I$, $A_i$ is the set of people whose birthday is $i$, so, for example, $\hbox{Beethoven}\in A_{({\scriptsize\hbox{Dec 16}})}$. $\foursquare$
Case one.6.2 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 $i|x$ ways that $x$ is a multiple of $i$; it is read "$i$ divides $10$''). $\foursquare$
Given an indexed family unit $\{A_i:i\in I\}$ we can ascertain the intersection and union of the sets $A_i$ using the universal and existential quantifiers: $$ \eqalign{\bigcap_{i\in I} A_i & = \{ ten: \forall i\in I\,(x\in A_i)\}\cr \bigcup_{i\in I} A_i & = \{ x: \exists i\in I\,(x\in A_i)\}.\cr} $$
Example i.6.3 If $\{A_i:i\in I\}$ is the indexed family unit of example 1.6.i, then $\bigcap_{i\in I} A_i$ is the empty gear up and $\bigcup_{i\in I} A_i$ is the set of all people. If $\{A_i:i\in I\}$ is the indexed family unit of example one.half-dozen.2, then $\bigcap_{i\in I} A_i$ is $\{0\}$ and $\bigcup_{i\in I} A_i$ is the set up of all integers. $\square$
Since the intersection and union of an indexed family are essentially "translations'' of the universal and existential quantifiers, information technology should not be besides surprising that there are De Morgan'southward laws that apply to these unions and intersections.
Theorem 1.6.4 If $\{A_i:i\in I\}$ is an indexed family unit of sets and 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 practise (a): $x\in (\bigcap_{i\in I} A_i)^c$ iff $\lnot (x\in \bigcap_{i\in I} A_i)$ iff $\lnot \forall i\,{\in}\, I\,(x\in A_i)$ iff $\exists i\,{\in}\, I\,(x\notin A_i)$ iff $\exists i\,{\in}\, I\,(ten\in A_i^c)$ iff $x\in \bigcup_{i\in I} A_i^c$. $\qed$
Yous may be puzzled by the inclusion of this theorem: is it not but part of theorem 1.five.6? No: theorem
Theorem one.six.five If $\{A_i:i\in I\}$ is an indexed family of sets and $B$ is any prepare, 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 then $B\subseteq \bigcap_{i\in I} A_i$,
d) if $A_i\subseteq B$, for all $i\in I$, and then $\bigcup_{i\in I} A_i\subseteq B$.
Proof. Part (a) is a example of specialization, that is, if $x\in \bigcap_{i\in I} A_i$, then $x\in A_i$ for all $i\in I$, in detail, when $i=j$. Role (d) also is like shooting fish in a barrel—if $x\in \bigcup_{i\in I} A_i$, then for some $i\in I$, $x\in A_i\subseteq B$, and 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 singled-out elements of $I$. The indexed family of example one.six.1 is pair-wise disjoint, but the ane in example 1.6.two is not. If $South$ is a set then an indexed family $\{A_i:i\in I\}$ of subsets of $S$ is a partition of $S$ if information technology is pair-wise disjoint and $S= \bigcup_{i\in I} A_i$. Partitions appear frequently in mathematics.
Example 1.6.vi Let $I=\{eastward,o\}$, $A_e$ be the set of fifty-fifty integers and $A_o$ be the fix of odd integers. And then $\{A_i:i\in I\}$ is a partition of $Southward=\Z$. $\square$
Case 1.6.seven Let $I=\R$, $Due south=\R^2$, and for each $i\in I$, let $A_i=\{(x,i):10\in \R\}$. Each $A_i$ is a horizontal line and the indexed family partitions the plane. $\square$
Sometimes we want to discuss a collection of sets (that is, a set of sets) even though at that place is no natural index present. In this case nosotros can use the collection itself as the index.
Example ane.six.eight If ${\cal South}=\{\{1,3,iv\}, \{2,three,4,6\}, \{3,4,5,7\}\}$, then $\bigcap_{A\in {\cal S}}A =\{3,iv\}$ and $\bigcup_{A\in {\cal South}}A =\{1,ii,3,4,5,6,7\}$. $\foursquare$
An especially useful drove of sets is the ability set of a set: If $10$ is any set, the power set of $X$ is ${\cal P}(X)=\{A:A\subseteq Ten\}$.
Example 1.6.nine If $X=\{ane,2\}$, and so ${\cal P}(X)=\{\emptyset,\{1\},\{2\},\{ane,2\}\}$. $\square$
Instance 1.6.ten ${\cal P}(\emptyset)=\{\emptyset\}$, that is, the ability set of the empty fix is non-empty. $\square$
Exercises i.half-dozen
Ex ane.6.1 Let $I=\{1,ii,iii\}$, $A_1=\{ane,3,4,six,7\}$ $A_2=\{one,four,5,7,8,nine\}$, $A_3= \{two,4,seven,10\}$. Find $\bigcap_{i\in I} A_i$ and $\bigcup_{i\in I} A_i$.
Ex 1.six.2 Suppose $I=[0,1]\subseteq \R$ and for each $i\in I$, allow $A_i=(i-1,i+1)\subseteq \R$. Notice $\bigcap_{i\in I} A_i$ and $\bigcup_{i\in I} A_i$.
Ex 1.6.iii Bear witness parts (b) and (c) of theorem 1.half dozen.5.
Ex 1.6.4 Suppose $U$ is the universe of soapbox and the index set $I=\emptyset$. What should we mean by $\bigcap_{i\in I} A_i$ and $\bigcup_{i\in I} A_i$? Show that Theorem 1.6.4 still holds using your definitions.
Ex one.6.5 Suppose $\{A_i\}_{i\in I}$ is a sectionalisation of a set $S$. If $T\subseteq S$, bear witness that $\{A_i\cap T\}_{i\in I}$ is a division of $T$.
Ex 1.vi.6 A collection of sets, $\cal S$, is totally ordered if for every $A,B\in {\cal S}$, either $A\subseteq B$ or $B\subseteq A$. If $\cal S$ is totally ordered, show that ${\cal S}^c=\{A^c: A\in {\cal Southward}\}$ is totally ordered.
Ex 1.six.7 Suppose $\cal S$ is a collection of sets and $B$ is another set. Prove that if $B$ is disjoint from every $A\in {\cal S}$ and then $B$ is disjoint from $\bigcup_{A\in {\cal S}} A$.
Source: https://www.whitman.edu/mathematics/higher_math_online/section01.06.html
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