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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 $A cap (B cup C)=(A cap B) cup (Acap C)$; $A cup (B cap C)=(A cup B) cap (Acup C)$.


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 $(A cap B) cup (Acap C)$.


$A cup B=1,2,4,5$. $A cap B=2$. $overlineA=3,4,5,6$ ($overlineA$ 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 $.

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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.