Some Results of Subsets

Solution:

The statement A  Í B Þ that every element of A is also in the set B, which includes the possibility that A = B.

The statement A Ì B (A is a proper subset of B) Þ that A is a subset of B and A  ¹  B; hence there exists at least one element in B which is not in A.

Solution:

Solution:

If every element of A belongs to a set B, and every element of B belongs to C, then clearly every element of A belongs to C. In other words A Í B and B Í C then A Í C.

Proof:

We must prove two statements.

• If x  Î A and x   Ï B then, by the statement "two sets A and B are different if there exists an element which belongs to one set but not to the other" and by hypothesis, A = B is contradicted. Thus A  Í B. Similarly B Í A.
• If A  ≠ B, then there is an element in set (say A) that is not in the other. Thus A is not contained in B, which contradicts the hypothesis that each set is contained in the other. Since the assumption that A ≠ B leads to a contradiction, it follows that A = B.

Or

• If A Í B and B Í A then A and B have the same elements i.e. A = B. Conversely if A = B then A Í B and B Í A, since every set is a subset of itself.

Prove that Null Set is the subset of every set

Proof:

Let A be any set.

In order to prove that   f Í A we must show that there is no element of f which is not present in A. And since f contains no element at all, no such element can be found out. Hence f  Ì A.

Proof:

Every set A is a subset of universal set U since, by definition all elements of A belong to U. Also the null set f Í A.

Proof:

The null set f is a subset of every set, in particular f Í A. By hypothesis, A Í f. The two conditions imply A = f.

Total number of subsets in a set is 2ⁿ

The total number of all possible subsets of a given set containing n elements is 2ⁿ.

Proof:

r = (0, 1, 2, 3, . . . n)

Hence the total number of subsets is