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The union of two sets is a set containing all facets that room in \$A\$ or in \$B\$ (possibly both). Because that example, \$1,2\cup2,3=1,2,3\$. Thus, we have the right to write \$xin(Acup B)\$ if and only if \$(xin A)\$ or \$(xin B)\$. Note that \$A cup B=B cup A\$. In number 1.4, the union of set \$A\$ and \$B\$ is shown by the shaded area in the Venn diagram.

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Fig.1.4 - The shaded area shows the collection \$B cup A\$.

Similarly we can define the union of 3 or much more sets. In particular, if \$A_1, A_2, A_3,cdots, A_n\$ space \$n\$ sets, their union \$A_1 cup A_2 cup A_3 cdots cup A_n\$ is a collection containing all aspects that room in at the very least one the the sets. We can write this union an ext compactly through \$\$igcup_i=1^n A_i.\$\$For example, if \$A_1=a,b,c, A_2=c,h, A_3=a,d\$, then \$igcup_i A_i=A_1 cup A_2 cup A_3=a,b,c,h,d\$. We can similarly define the union of infinitely countless sets \$A_1 cup A_2 cup A_3 cupcdots\$.

The intersection of 2 sets \$A\$ and \$B\$, denoted by \$A cap B\$, consists of all elements that space both in \$A\$ \$underline extrmand\$ \$B\$. Because that example, \$1,2\cap2,3=2\$. In number 1.5, the intersection of to adjust \$A\$ and \$B\$ is presented by the shaded area utilizing a Venn diagram.

Fig.1.5 - The shaded area mirrors the collection \$B cap A\$.

More generally, for sets \$A_1,A_2,A_3,cdots\$, their intersection \$igcap_i A_i\$ is defined as the set consisting the the facets that space in every \$A_i\$"s. Number 1.6 shows the intersection of 3 sets.

Fig.1.6 - The shaded area shows the set \$A cap B cap C\$.

The complement that a set \$A\$, denoted through \$A^c\$ or \$arA\$, is the set of all aspects that are in the universal collection \$S\$ but are no in \$A\$. In number 1.7, \$arA\$ is shown by the shaded area utilizing a Venn diagram.

Fig.1.7 - The shaded area mirrors the collection \$arA=A^c\$.

The difference (subtraction) is characterized as follows. The collection \$A-B\$ consists of elements that room in \$A\$ however not in \$B\$. For example if \$A=1,2,3\$ and also \$B=3,5\$, climate \$A-B=1,2\$. In number 1.8, \$A-B\$ is shown by the shaded area utilizing a Venn diagram. Note that \$A-B=A cap B^c\$.

Fig.1.8 - The shaded area shows the set \$A-B\$.

Two set \$A\$ and also \$B\$ room mutually exclusive or disjoint if they do not have any shared elements; i.e., your intersection is the north set, \$A cap B=emptyset\$. More generally, several sets are dubbed disjoint if they space pairwise disjoint, i.e., no two of castle share a common elements. Figure 1.9 mirrors three disjoint sets.

Fig.1.9 - set \$A, B,\$ and \$C\$ space disjoint.

If the earth"s surface is our sample space, we can want to partition it to the various continents. Similarly, a country can be partitioned to various provinces. In general, a arsenal of nonempty sets \$A_1, A_2,cdots\$ is a partition that a set \$A\$ if they are disjoint and also their union is \$A\$. In number 1.10, the to adjust \$A_1, A_2, A_3\$ and also \$A_4\$ form a partition that the universal set \$S\$.

Fig.1.10 - The repertoire of sets \$A_1, A_2, A_3\$ and also \$A_4\$ is a partition of \$S\$.

Here space some rules that space often useful when working with sets. We will certainly see instances of their intake shortly.

Theorem : De Morgan"s law

For any kind of sets \$A_1\$, \$A_2\$, \$cdots\$, \$A_n\$, we have actually \$(A_1 cup A_2 cup A_3 cup cdots A_n)^c=A_1^c cap A_2^c cap A_3^ccdots cap A_n^c\$; \$(A_1 cap A_2 cap A_3 cap cdots A_n)^c=A_1^c cup A_2^c cup A_3^ccdots cup A_n^c\$.

Theorem : Distributive law

For any kind of sets \$A\$, \$B\$, and also \$C\$ we have actually \$A cap (B cup C)=(A cap B) cup (Acap C)\$; \$A cup (B cap C)=(A cup B) cap (Acup C)\$.

Example

If the universal set is given by \$S=1,2,3,4,5,6\$, and \$A=1,2\$, \$B=2,4,5, C=1,5,6 \$ space three sets, discover the following sets: \$A cup B\$ \$A cap B\$ \$overlineA\$ \$overlineB\$ examine De Morgan"s law by recognize \$(A cup B)^c\$ and \$A^c cap B^c\$. Check the distributive law by finding \$A cap (B cup C)\$ and \$(A cap B) cup (Acap C)\$.

Solution

\$A cup B=1,2,4,5\$. \$A cap B=2\$. \$overlineA=3,4,5,6\$ (\$overlineA\$ is composed of elements that are in \$S\$ however not in \$A\$). \$overlineB=1,3,6\$.We have actually \$\$(A cup B)^c=1,2,4,5^c=3,6,\$\$ i m sorry is the very same as\$\$A^c cap B^c=3,4,5,6 cap 1,3,6=3,6.\$\$We have actually \$\$A cap (B cup C)=1,2 cap 1,2,4,5,6=1,2,\$\$ which is the exact same as\$\$(A cap B) cup (Acap C)=2 cup 1=1,2.\$\$

A Cartesian product of 2 sets \$A\$ and also \$B\$, written as \$A imes B\$, is the set containing ordered bag from \$A\$ and \$B\$. The is, if \$C=A imes B\$, then each element of \$C\$ is that the kind \$(x,y)\$, whereby \$x in A\$ and \$y in B\$:\$\$A imes B = (x,y) .\$\$For example, if \$A=1,2,3\$ and \$B=H,T\$, then\$\$A imes B=(1,H),(1,T),(2,H),(2,T),(3,H),(3,T).\$\$Note that right here the pairs are ordered, so because that example, \$(1,H) eq (H,1)\$. Hence \$A imes B\$ is not the very same as \$B imes A\$.

If you have two limited sets \$A\$ and \$B\$, where \$A\$ has actually \$M\$ elements and \$B\$ has actually \$N\$ elements, then \$A imes B\$ has \$M imes N\$ elements. This rule is referred to as the multiplication principle and is an extremely useful in counting the numbers of facets in sets. The number of elements in a collection is denoted by \$|A|\$, so below we write \$|A|=M, |B|=N\$, and \$|A imes B|=MN\$. In the over example, \$|A|=3, |B|=2\$, thus \$|A imes B|=3 imes 2 = 6\$. Us can likewise define the Cartesian product that \$n\$ to adjust \$A_1, A_2, cdots, A_n\$ as\$\$A_1 imes A_2 imes A_3 imes cdots imes A_n = (x_1, x_2, cdots, x_n) .\$\$The multiplication principle states that for finite set \$A_1, A_2, cdots, A_n\$, if \$\$|A_1|=M_1, |A_2|=M_2, cdots, |A_n|=M_n,\$\$ climate \$\$mid A_1 imes A_2 imes A_3 imes cdots imes A_n mid=M_1 imes M_2 imes M_3 imes cdots imes M_n.\$\$

An vital example of sets acquired using a Cartesian product is \$mathbbR^n\$, wherein \$n\$ is a herbal number. Because that \$n=2\$, we have
 \$mathbbR^2\$ \$= mathbbR imes mathbbR\$ \$= x in mathbbR, y in mathbbR \$.See more: Dru Hill Never Make A Promise Lyrics, Never Make A Promise Lyrics (Video)
Thus, \$mathbbR^2\$ is the set consisting of every points in the two-dimensional plane. Similarly, \$mathbbR^3=mathbbR imes mathbbR imes mathbbR\$ and so on.