Subset vs Superset
In mathematics, the concept of set is fundamental. The modern study of set theory was formalized in the late 1800s. Set theory is a fundamental language of mathematics, and repository of the basic principles of modern mathematics. On the other hand, it’s a branch of mathematics in its own rights, which is classified as a branch of mathematical logic in modern mathematics.
A set is a well defined collection of objects. Welldefined means, that there exists a mechanism by which one is able to determine whether a given object belongs to a particular set or not. Objects that belong to a set are called elements or members of the set. Sets are usually denoted by capital letters and lower case letters are used to represent elements.
A set A is said to be a subset of a set B; if and only if, every element of set A is also an element of set B. Such a relation between sets is denoted by A ⊆ B. It can also be read as ‘A is contained in B’. The set A is said to be a proper subset if A ⊆ B and A ≠B, and denoted by A ⊂ B. If there is even one member in A that is not a member of B, then A cannot be a subset of B. Empty set is a subset of any set, and a set itself is a subset of same set.
If A is a subset of B, then A is contained in B. It implies that B contains A, or in other words, B is a superset of A. We write A ⊇ B to denote that B is a superset of A.
For an example, A = {1, 3} is a subset of B = {1, 2, 3}, since all the elements in A contained in B. B is a superset of A, because B contains A. Let A={1, 2, 3} and B={3, 4, 5}. Then A∩B={3} . Therefore, both A and B are supersets of A∩B. The set A∪B, is a superset of both A and B, because A∪B, contains all the elements in A and B.
If A is a superset of B and B is a superset of C, then A is a superset of C. Any set A is a superset of empty set and any set itself a superset of that set.
‘A is a subset of B’ is also read as ‘A is contained in B’, denoted by A ⊆ B. ‘B is a superset of A’ is also read as ‘B is contains in A’, denoted by A ⊇ B.

sureshkumar says
It is solved my problem