
Patrick B. answered 09/08/19
Math and computer tutor/teacher
Barring any sneaky tricks with the empty set, the answer is no.
Given: A is a proper subset of B.
Suppose B is a proper subset of A.
Since A is a proper subset of B, the every element of A is also in B AND
B contains at least one element that is not in A.
By supposition, sine B is a proper subset of A, every element of B
is also an element of A, which contradicts the previous statement.