Video obtainable 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 can write \$xin(Acup B)\$ if and also 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 displayed by the shaded area in the Venn diagram.

You are watching: Which venn diagram is not correct? Fig.1.4 - The shaded area shows the collection \$B cup A\$.

Similarly us can specify the union of three or an ext 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 set containing all aspects that space in at least one that the sets. We have the right to write this union much more 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 many sets \$A_1 cup A_2 cup A_3 cupcdots\$.

The intersection of 2 sets \$A\$ and also \$B\$, denoted by \$A cap B\$, consists of all aspects that are both in \$A\$ \$underline extrmand\$ \$B\$. For example, \$1,2\cap2,3=2\$. In figure 1.5, the intersection of set \$A\$ and also \$B\$ is shown by the shaded area using a Venn diagram. Fig.1.5 - The shaded area reflects the set \$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 room in every \$A_i\$"s. Number 1.6 mirrors the intersection of 3 sets. Fig.1.6 - The shaded area mirrors the set \$A cap B cap C\$.

The complement of a set \$A\$, denoted by \$A^c\$ or \$arA\$, is the collection of all elements that space in the universal set \$S\$ but are no in \$A\$. In number 1.7, \$arA\$ is presented by the shaded area using a Venn diagram. Fig.1.7 - The shaded area mirrors the set \$arA=A^c\$.

The distinction (subtraction) is identified as follows. The set \$A-B\$ consists of facets that room in \$A\$ yet not in \$B\$. For instance if \$A=1,2,3\$ and \$B=3,5\$, climate \$A-B=1,2\$. In number 1.8, \$A-B\$ is presented 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 execute not have any type of shared elements; i.e., your intersection is the empty set, \$A cap B=emptyset\$. An ext generally, numerous sets are dubbed disjoint if they space pairwise disjoint, i.e., no 2 of castle share a common elements. Number 1.9 shows three disjoint sets. Fig.1.9 - sets \$A, B,\$ and also \$C\$ space disjoint.

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

Here space some rules that space often useful when working through sets. We will see examples of their intake shortly.

Theorem : De Morgan"s law

For any type 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 type of sets \$A\$, \$B\$, and also \$C\$ we have \$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 collection is given by \$S=1,2,3,4,5,6\$, and \$A=1,2\$, \$B=2,4,5, C=1,5,6 \$ room 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\$. Inspect the distributive regulation by recognize \$A cap (B cup C)\$ and also \$(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 \$\$(A cup B)^c=1,2,4,5^c=3,6,\$\$ i beg your pardon is the same as\$\$A^c cap B^c=3,4,5,6 cap 1,3,6=3,6.\$\$We have \$\$A cap (B cup C)=1,2 cap 1,2,4,5,6=1,2,\$\$ i beg your pardon 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 collection containing ordered bag from \$A\$ and also \$B\$. That is, if \$C=A imes B\$, then each element of \$C\$ is the the kind \$(x,y)\$, where \$x in A\$ and \$y in B\$:\$\$A imes B = x in A extrm and also y in B .\$\$For example, if \$A=1,2,3\$ and also \$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 space ordered, so for example, \$(1,H) eq (H,1)\$. For this reason \$A imes B\$ is not the same as \$B imes A\$.

If you have actually two finite sets \$A\$ and \$B\$, wherein \$A\$ has actually \$M\$ elements and \$B\$ has \$N\$ elements, then \$A imes B\$ has \$M imes N\$ elements. This preeminence is referred to as the multiplication principle and also is an extremely useful in count the numbers of elements in sets. The variety of elements in a set is denoted by \$|A|\$, so here we create \$|A|=M, |B|=N\$, and \$|A imes B|=MN\$. In the over example, \$|A|=3, |B|=2\$, hence \$|A imes B|=3 imes 2 = 6\$. We 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 says that because that finite set \$A_1, A_2, cdots, A_n\$, if \$\$|A_1|=M_1, |A_2|=M_2, cdots, |A_n|=M_n,\$\$ then \$\$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 crucial example the sets derived using a Cartesian product is \$mathbbR^n\$, where \$n\$ is a herbal number. Because that \$n=2\$, us have
 \$mathbbR^2\$ \$= mathbbR imes mathbbR\$ \$= (x,y) \$.See more: Employees Are Likely To See An Adaptive Change As __________.
Thus, \$mathbbR^2\$ is the collection consisting of all points in the two-dimensional plane. Similarly, \$mathbbR^3=mathbbR imes mathbbR imes mathbbR\$ and also so on.