Question

For any two sets A and B, prove that:
(i) \[(A-B)\bigcup (B-A)=(A\bigcup B)-(A\bigcap B)\]
(ii) \[(A-B)=A\Leftrightarrow A\bigcap B=\phi \]

Answer 1

Hint: We will first proceed with the left hand side of the expression and then conversely with the right hand side of the expression. Hence we will prove both the parts similarly. \[(A\bigcup B)\] means union of sets A and B. \[(A\bigcap B)\] means intersection of sets A and B. \[(A-B)\] is the set of all elements in A that are not in set B.

Complete step-by-step answer:
Before proceeding with the question we should understand some basic concepts of sets.

\[(A\bigcup B)\] means union of sets A and B. That is it includes all the elements of both set A and set B.

\[(A\bigcap B)\] means intersection of sets A and B. That is it includes only the common elements of set A and set B.

\[(A-B)\] is the set of all elements in A that are not in set B.

A set which does not contain any element is called an empty set or void set or null set. It is denoted by { } or \[\phi \].

(i) The expression is \[(A-B)\bigcup (B-A)=(A\bigcup B)-(A\bigcap B)\].

Now let x be an arbitrary element of \[(A-B)\bigcup (B-A)\]. Then, x belongs to \[(A-B)\bigcup (B-A)\].

So this implies \[x\in A-B\] or \[x\in B-A\].

Now from the above definitions we can say that \[x\in A\] or \[x\notin B\] and \[x\in B\] or \[x\notin A\].

Now rearranging we get, \[x\in A\] or \[x\in B\] and \[x\notin B\] or \[x\notin A\].

Hence on recombining, we get \[x\in (A\bigcup B)-(A\bigcap B)\].

Thus, \[(A-B)\bigcup (B-A)\subseteq (A\bigcup B)-(A\bigcap B)............(1)\]
Now let y be an arbitrary element of \[(A\bigcup B)-(A\bigcap B)\]. Then, y belongs to \[(A\bigcup B)-(A\bigcap B)\].

So this implies \[y\in A\bigcup B\] or \[y\notin A\bigcap B\].

Now from the above definitions we can say that \[y\in A\] or \[y\in B\] and \[y\notin A\] or \[y\notin B\].

Now rearranging we get, \[y\in A\] or \[y\notin B\] and \[y\in B\] or \[y\notin A\].

Hence on recombining, we get \[y\in (A-B)\bigcup (B-A)\].

Thus, \[(A\bigcup B)-(A\bigcap B)\subseteq (A-B)\bigcup (B-A)............(2)\]

Now from equation (1) and equation (2) we have proved \[(A-B)\bigcup (B-A)=(A\bigcup B)-(A\bigcap B)\].

(ii) The expression is \[(A-B)=A\Leftrightarrow A\bigcap B=\phi \].

Suppose \[(A-B)=A\]. Now let \[x\in A\bigcap B\] which implies \[x\in A\] and \[x\in B\].

Since \[x\in A\] and \[A=A-B\]. This implies \[x\in A-B\] that means \[x\in A\] and \[x\notin B\]. This gives a contradiction to our supposition and hence \[A\bigcap B=\phi \].

Now checking conversely,

Let \[A-B\ne A\]. Now since \[A=(A-B)\bigcup (A\bigcap B)\].

We know that \[A\bigcap B=\phi \] and hence substituting this in equation (1) we get,
\[A=A-B\].

Hence we have proved \[(A-B)=A\Leftrightarrow A\bigcap B=\phi \].

Note: We have to remember the definition of the basic concepts related to sets to solve this question. Also we have to remember the symbol of the null set or else we might get confused. We in a hurry can make a mistake with the symbols used in sets.