Subset and set inclusion

The governing concept is subset inclusion. A set H is a subset of I, written H ⊆ I, when every element of H is also an element of I; H may have fewer elements than I. Here H contains 3 and 6, and both of these elements appear in I. Therefore H ⊆ I is true, making option A correct. The reverse statement I ⊆ H is false because I also contains 9 and 12, which are not elements of H. Option C is false because 9 is not in H. Option D is false because equal sets must contain exactly the same elements, while I has two additional elements. The order of elements is irrelevant in sets, but membership and completeness are decisive.

A subset can contain one or more elements only if every one of those elements belongs to the original set. The set J is {1, 3, 5, 7}. Option A contains 3 and 7, and both are members of J, so {3, 7} ⊆ J. Hence A is correct. Option B fails because 9 is not in J. Option C fails because 2 is not in J, and option D fails because 8 is not in J. It is enough for one element to be outside J to disqualify an option as a subset. The distinction is not about the number of elements: a subset may have fewer elements than J, but it cannot introduce any new element. Thus checking membership of every listed element gives a unique answer.