15.9. Proving Set Identities

So, the question here is how do we prove a set identity, if an identity is given to us, how do I prove that the two sets A and B which are given in the left hand side and in the right hand side they are the same. So, for that we have to understand here that two sets A and B are equal if they are respectively subsets of each other. Because the definition of A equal to B was for all x: x implies x belonging to A bi-implication x belonging to B.

To establish that two sets A and B are equal, you must demonstrate that each set is a subset of the other. This means that every element in A should also be found in B and vice versa. When we denote A equal to B, mathematically, this is represented as: for all elements x, if x is in A, then x also must be in B, and if x is in B, then x must also be in A. Essentially, you must show two conditions: 1) Every element in A is in B (A ⊆ B) and 2) Every element in B is in A (B ⊆ A).

Think of two containers filled with different fruits. For the containers to contain the same type of fruit, every fruit in the first container must be present in the second container and every fruit in the second container must be in the first. If you have an apple in the first container, and there isn’t an apple in the second, then both containers cannot be considered the same.

If I split this bi-implication, this means if x belongs to A it should belongs to B as well, and bi-implication can be split into conjunction of two implications. Now this condition means of course everything is with respect to for all x, this condition means A is a subset of B and this condition means B is a subset of A. That means to show that two sets A and B are equal, I have to show that A is a subset of B and B is a subset of A.

When we analyze the equality of two sets by splitting the bi-implication, we translate it into two separate statements, which are easier to handle individually. The first part states that any arbitrary element x that belongs to A must also belong to B. This ensures that A is a subset of B. The second part does the opposite, stating that if x is in B, then x must also be in A, which ensures that B is a subset of A. Therefore, both these conditions must be satisfied to conclude that A and B are indeed equal.

Consider two friends who are both members of the same book club. If each friend consistently reads the same list of books, you could say their reading lists are identical. To show this, you would show that every book on Friend A's list is also on Friend B's list and vice versa.

So, let me demonstrate what I said with respect to this example. I want to prove the De Morgan's law. The De Morgan’s law is there are two variants of De Morgan’s law. I am proving one of them. It says that if you take the complement of intersection of A and B that is same as the union of A and B.

De Morgan's Laws provide a foundation for working with the complements and intersections of sets. The law states that the complement of the intersection of two sets is equal to the union of their complements. Mathematically, this can be written as: (A ∩ B)' = A' ∪ B'. This means to find elements not in both A and B, you can find all the elements not in A and all the elements not in B, then unify those two sets. Understanding this relationship allows simplification and manipulation of complex set expressions.

Imagine a gym that has two types of memberships: one that includes weightlifting and another that includes swimming. If you want to identify individuals who do not participate in both activities, instead of checking those who do neither, you can look at all the people who don’t have weightlifting memberships and those who don’t have swimming memberships. Collectively, these individuals represent those who are not part of both groups, demonstrating De Morgan's Law in action.

--