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As a set includes only distinct elements, the common elements are represented only once in the union set. Thus, union set consists of elements of both sets without repeating an element. Now, the set is represented on Venn diagram as shown here.
For illustration of working with union, let us consider two sets of positive integers as given here,
$$A=\left\{\mathrm{1,2,3,4,5,6}\right\}$$
$$B=\left\{\mathrm{4,5,6,7,8}\right\}$$
The union of two sets is :
$$\Rightarrow A\cup B=\left\{\mathrm{1,2,3,4,5,6,4,5,6,7,8}\right\}$$
But repetition of elements in a set does not change it. Hence, we need not repeat elements in the resulting union.
$$\Rightarrow A\cup B=\left\{\mathrm{1,2,3,4,5,6,7,8}\right\}$$
Here, universal set is natural numbers. The representation of union of joint sets is shown in the figure. We can observe that very construction of union on Venn diagram ensures that elements are not repeated.
Let us examine the defining set of union :
$$A\cup B=\{x:x\in A\phantom{\rule{1em}{0ex}}or\phantom{\rule{1em}{0ex}}x\in B\}$$
We consider an arbitrary element, say “x”, of the union set. Then, we interpret the conditional meaning as :
$$If\phantom{\rule{1em}{0ex}}x\in A\cup B,\phantom{\rule{1em}{0ex}}then\phantom{\rule{1em}{0ex}}x\in A\phantom{\rule{1em}{0ex}}or\phantom{\rule{1em}{0ex}}x\in B.$$
Can we emphasize this conditional meaning in reverse order :
$$If\phantom{\rule{1em}{0ex}}x\in A\phantom{\rule{1em}{0ex}}or\phantom{\rule{1em}{0ex}}x\in B,\phantom{\rule{1em}{0ex}}then\phantom{\rule{1em}{0ex}}x\in A\cup B.$$
Yes, we can agree with the second conditional meaning as well. We, therefore, conclude that the statements work in both ways. We write two statements together as :
$$\phantom{\rule{1em}{0ex}}x\in A\cup B\iff x\in A\phantom{\rule{1em}{0ex}}or\phantom{\rule{1em}{0ex}}x\in B$$
We can reach yet another conclusion by observing representation of union set on Venn diagram. Now, if an arbitrary element “x” does not belong to union set, then it is clear that it does not belong to the region represented by the union set on the Venn’s diagram. Hence,
$$If\phantom{\rule{1em}{0ex}}x\notin \phantom{\rule{1em}{0ex}}\left(\text{does not belong}\right)\phantom{\rule{1em}{0ex}}A\cup B\Rightarrow x\notin A\phantom{\rule{1em}{0ex}}\mathrm{and}\phantom{\rule{1em}{0ex}}x\notin B.$$
The important thing to note here is the word “and” in place of “or” used before. Think about it. Here two conditions follow simultaneously. If an element does not belong to an union set, then it will not belong to either of individual sets simultaneously. Now, the next thing to consider is whether this conditional statement will be true other way round as well?
$$If\phantom{\rule{1em}{0ex}}x\notin A\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\notin B\Rightarrow x\notin A\cup B.$$
Yes, we can agree to this statement. We, therefore, conclude that the statements work in both ways. We write two statements together with the help of two ways arrow sign as :
$$x\notin A\cup B\iff x\notin A\phantom{\rule{1em}{0ex}}\mathrm{and}\phantom{\rule{1em}{0ex}}x\notin B.$$
Consider students in class X and class XI. Let us denote the respective sets as "T" for tenth and "E" for eleventh class. Clearly, union i.e. combination of two sets should include elements from each of the sets. Hence,
$$T\cup E=\text{students in class X and XI}$$
This is a straight forward union of two sets. The resulting set comprises of all elements present in both the sets. Since it is not possible that students studying in class X are also students of XI, we are sure that the numbers of elements in the union is sum of numbers of students in each class. As there is no commonality between two sets, it is a union of two “disjoint” sets. We conclude here that union of two disjoint sets has no common elements.
The set “B” consists of all elements of its subset “A”. In other words, the elements of a subset “A” also belongs to the set “B”. The operation of union is combining elements of two sets. The union with a subset, therefore, consists of elements from both “A” and “B”. However, all elements of “A” are also the elements of “B”. Therefore, we find that union set is same as the superset “B”. Symbolically,
$$If\phantom{\rule{1em}{0ex}}A\subset B,\phantom{\rule{1em}{0ex}}\mathrm{then}\phantom{\rule{1em}{0ex}}B\cup A=B.$$
We can check this deduction with the help of an example. Let us consider two sets as :
$$A=\left\{\mathrm{4,5,6}\right\}$$
$$B=\left\{\mathrm{1,2,3,4,5,6}\right\}$$
Here, we see that A $\subset $ B. Now,
$$B\cup A=\left\{\mathrm{1,2,3,4,5,6,4,5,6}\right\}=\left\{\mathrm{1,2,3,4,5,6}\right\}=B$$
If ${A}_{1},{A}_{2},{A}_{3},\dots \dots \dots ,{A}_{n}$ is a finite family of sets, then their unions, one after another, is denoted as :
$${A}_{1}\cup {A}_{2}\cup {A}_{3}\cup \dots \dots \dots \cup {A}_{n}$$
In this section we shall discuss some of the important characteristics/ deductions for the union operation.
The literal meaning of the word “idempotent” is “unchanged when multiplied by itself”. Following the clue, the union of a set with itself is the set itself. This is an equivalent statement conveying the meaning of “idempotent” in the context of union. Symbolically,
$$A\cup A=A$$
The union set consists of distinct elements and common elements taken once. Between two sets here, all elements are common. The union set consists of all elements of either set.
The algebraic operators like addition and multiplication have defined identities, which does not change the other operand of the operator. For example, if we add “0” to a number, it remains same. Hence, “0” is additive identity. Similarly, “1” is multiplicative identity.
In the case of union, we find that union of a set with empty set does not change the set. Hence, empty set is union identity.
$$A\cup \phi =A$$
As there is no element in empty set, union has same elements as that in “A”.
All sets are subsets of universal set for a given context. We have seen that union with subset results in the set itself. Clearly, union of universal set with its subset will result in the universal set itself.
$$U\cup A=U$$
In order to assess whether commutative property holds or not, we consider the example, used earlier. Let the sets be :
$$A=\left\{\mathrm{1,2,3,4,5,6}\right\}$$
$$B=\left\{\mathrm{4,5,6,7,8}\right\}$$
Then,
$$A\cup B=\left\{\mathrm{1,2,3,4,5,6,4,5,6,7,8}\right\}=\left\{\mathrm{1,2,3,4,5,6,7,8}\right\}$$
$$B\cup A=\left\{\mathrm{4,5,6,7,8,1,2,3,4,5,6}\right\}=\left\{\mathrm{1,2,3,4,5,6,7,8}\right\}$$
Thus, we see that order of operands with respect to the union operator is not differentiating. We can also appreciate this law on Venn diagram, which does not change by changing positions of sets across union operator.
The associative property also holds with respect to union operator. We know that associative property is about changing the place of parentheses as here :
$$\left(A\cup B\right)\cup C=A\cup \left(B\cup C\right)$$
The parentheses simply change the precedence of operation. On Venn diagram, union involving three sets appears same, irrespective of whether we apply union operation in a particular sequence.
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