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We can extend the concept of subtraction, used in algebra, to the sets. If a set “B” is subtracted from set “A”, the resulting difference set consists of elements, which are exclusive to set “A”. We represent the symbol of difference of sets as “A-B” and pronounce the same as “A minus B”.
In the set builder form, the difference set is :
$$A-B=\{x:\phantom{\rule{1em}{0ex}}x\in A\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\notin B\}$$
and
$$B-A=\{x:\phantom{\rule{1em}{0ex}}x\in B\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\notin A\}$$
On Venn’s diagram, the difference "A-B" is the region of “A”, which excludes the common region with set “B”.
Let us examine the defining set of intersection :
$$A-B=\{x:\phantom{\rule{1em}{0ex}}x\in A\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\notin B\}$$
We consider an arbitrary element, say “x”, of the difference set. Then, we interpret the conditional meaning as :
$$If\phantom{\rule{1em}{0ex}}x\in A-B\Rightarrow x\in A\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\notin B.$$
The conditional statement is true in opposite direction as well. Hence,
$$If\phantom{\rule{1em}{0ex}}x\in A\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\notin B\Rightarrow x\in A-B.$$
We can summarize two statements with two ways arrow as :
$$If\phantom{\rule{1em}{0ex}}x\in A-B\iff x\in A\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\notin B.$$
From Venn’s diagram, we observe that if we derive union of ( $A\cap B$ ) to either of the difference sets, then we get the complete individual set.
$$A=\left(A-B\right)\cup \left(A\cap B\right)$$
and
$$B=\left(B-A\right)\cup \left(A\cap B\right)$$
The positions of sets about minus operator affect the result. It is clear from the figure above, where “A-B” and “B-A” represent different regions on Venn’s diagram. As such, the difference of sets is not commutative. Let us consider the example used earlier, where :
$$A=\left\{\mathrm{1,2,3,4,5,6}\right\}$$
$$A=\left\{\mathrm{4,5,6,7,8}\right\}$$
Then,
$$\Rightarrow A-B=\left\{\mathrm{1,2,3}\right\}$$
and
$$\Rightarrow B-A=\left\{\mathrm{7,8}\right\}$$
Clearly,
$$\Rightarrow A-B\ne B-A$$
From the Venn’s diagram, we can see that union of two sets is equal to three distinct regions. Alternatively, we can say that the region represented by the union of two sets is equal to the sum of the regions representing three “disjoint” sets (i) difference set A-B (ii) intersection set " $A\cap B$ " and (iii) difference set B-A.
We use the term “symmetric set” for combining two differences as marked on Venn’s diagram. It is denoted as “ $A\text{\Delta}B$ ”.
$$A\text{\Delta}B=\left(A-B\right)\cup \left(B-A\right)$$
The complement is a special case of the difference operation. The set in question is subtracted from universal set, “U”. Thus, one of the sets in difference operation is fixed. We define complement of a set as its difference with universal set, "U". The complement of a set is denoted by the same symbol as that of set, but with an apostrophe. Hence, complement of set A is set A’.
We write the complement set in terms of set builder form as :
$$A\prime =\{x:\phantom{\rule{1em}{0ex}}x\in U\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\notin A\}$$
Note that elements of A’ does not belong to set “A”. On Venn’s diagram, the complement of “A” is the remaining region of the universal set.
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