The sets $ A - B, $ $ B - A $ and $ A \cap B $ are mutually disjoint sets. Use example to observe if this is true.

Hint: Choose any two sets A and B. Then find $ A - B, $ $ B - A $ and $ A \cap B $ using the properties of sets. Once, to get the results, compare them to check if they are mutually disjoint. Any two sets are called mutually disjoint if they have no element common in them.

Complete step-by-step answer:
Let A is the collection of the first 10 natural numbers.
 $ \Rightarrow A = \{ 1,2,3,4,5,6,7,8,9,10\} $
Let B is the collection of integers whose square is less than 16.
 $ \Rightarrow B = \{ - 3, - 2, - 1,0,1,2,3\} $
We know that,
 $ A - B $ is the collection of elements that are in A but not in B.
 $ \Rightarrow A - B = \{ 4,5,6,7,8,9,10\} $ . . . (1)
 $ B - A $ is the collection of elements that are in B but not in A.
 $ \Rightarrow B - A = \{ - 3, - 2, - 1,0\} $ . . . (2)
And $ A \cap B $ is the collection of elements that is common in both, A and B.
 $ \Rightarrow A \cap B = \{ 1,2,3\} $ . . . (3)
We can clearly observe that the elements that are in equation (1) are not present in equations (2) and (3).
The elements that are present in equation (2) are not present in equations (1) and (3).
And the elements that are present in equation (3) are not present in equation (1) and (2).
Therefore, we can conclude that, $ A - B, $ $ B - A $ and $ A \cap B $ are mutually disjoint.
Hence the statement given in the question is true.

Note: Sometimes, it is possible that for some question, the example we use might satisfy the given conditions in the question. But the statement might not be true for all the possible examples. In this question, the definition of $ A - B, $ $ B - A $ and $ A \cap B $ itself clears that they are going to be mutually disjoint. So, it didn’t matter much for this question. But you cannot always use this method of taking one example to check if any mathematical condition is true or false. In set theory, if you are confused about whether to take any example or not, then it is better to check that condition using the Venn diagram.