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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 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.

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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.

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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.

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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.

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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 $A-B=A cap B^c$.

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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.

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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 $S$.

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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) $.

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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.