If \[(A-B)\cup (B-A)=A\] for subsets \[A\] and $B$ of the universal set $\bigcup $, then which one of the following is correct?
A. $B$ is proper non-empty subset of \[A\]
B. \[A\] and $B$ are non-empty disjoint sets
C. $B=\varnothing $
D. None of these

Hint: In this question, we are to find the supporting condition for the given statement \[(A-B)\cup (B-A)=A\]. Since \[A\] and $B$ are non-empty sets and subsets of a universal set $\bigcup $, we can use the set theory definitions, we get the required answer.

Formula Used: Set: A collection of objects in which it is possible to decide whether a given object belongs to the collection or not is said to be a set. Those objects are nothing but the elements in the set.
Sets are represented in two ways: The roaster method and the Set builder form
Roaster method:
The set of vowels – \[\left\{ a,\text{ }e,\text{ }i,\text{ }o,\text{ }u \right\}\]
The set of natural numbers – \[\left\{ 1,\text{ }2,\text{ }3,\text{ }\ldots . \right\}\]
Set builder form:
The set of vowels – \[\left\{ x:x\text{ }is\text{ }a\text{ }vowel\text{ }in\text{ }English\text{ }alphabet \right\}\]
The set of natural numbers – \[\left\{ x:x\text{ }is\text{ }a\text{ }natural\text{ }number \right\}\]
Some of the important mathematical sets:
$N$ - the set of natural numbers - $N=\{1,2,3...\}$
$Z$- the set of integers - $Z=\{0,\pm 1,\pm 2,\pm 3,...\}$
If two sets $A, B$ where the set $A$ is said to be the subset of $B$ i.e., $A\subseteq B$ then every element of $A$ is in the set $B$ and the set $A$ is said to be the proper subset of $B$ i.e., \[A\subset B\] then $A\subseteq B$ and $A\ne B$.

Complete step by step solution: It is given that, \[A\] and $B$ are non-empty sets and subsets of universal set $\bigcup $.
I.e., $A\subseteq \bigcup ;B\subseteq \bigcup $
Given that \[(A-B)\cup (B-A)=A\].
If $B$ is proper non-empty subset of \[A\] i.e., \[B\subset A;B\ne A\], then
\[A-B\subset A\]
That means $A-B\ne A$
And
\[B-A=\varnothing \]
\[\therefore (A-B)\cup (B-A)\ne A\]
Thus, statement (1) is not correct.
If \[A\] and $B$ are non-empty disjoint sets i.e., \[A\cap B=\varnothing \], then
\[\begin{align}
  & A-B=\varnothing \\
 & B-A=\varnothing \\
 & \therefore (A-B)\cup (B-A)\ne A \\
\end{align}\]
Thus, statement (2) is also incorrect.
If \[B=\varnothing \], then
\[\begin{align}
  & A-B=A \\
 & B-A=\varnothing \\
 & \Rightarrow (A-B)\cup (B-A)=A\cup \varnothing =A \\
 & \therefore (A-B)\cup (B-A)=A \\
\end{align}\]

Option ‘C’ is correct

Note: Here we need to check all the options for the given sets. So, by eliminating the method we can choose the correct one.