Question

How can the union and intersection of \(n\)sets that all are subsets of the universal set\(U\) be found using bit strings?

The successor of the set \(A\)is the set \(A \cup \{ A\} \)

Short answer

Let there be \(n\)sets.

We require that the universal set\(U\)is finite.

If the universal set\(U\)contains\(m\)elements, then the bit string corresponding with every set will contain\(m\)bits.

If the \(ith\)element of the universal set\(U\)is in the set, then\(ith\)bit of the string is a zero.

If the \(ith\)element of the universal set \(U\) is not in the set, then the \(ith\)bit of the string is a \(0\).

Step-by-step solution

Step 1

Union \(A \cup B\): all elements that are either in \(A\) OR in \(B\)

If either string contains a\(1\)on the\(ith\)bit, then the union contains a\(1\)on the\(ith\)bit as well.

If both strings contain a \(0\)on the \(ith\)bit, then the union contains a \(0\) on the \(ith\)bit.

Step 2

Intersection \(A \cap B\): all elements that are both in \(A\) AND in \(B\).

If both strings contain a\(1\)on the\(ith\)bit, then the intersection contains a\(1\)on the\(ith\)bit as well.

If both strings contain a \(0\)on the \(ith\)bit, then the intersection contains a \(0\) on the \(ith\)bit

Step 3

Hence, we conclude that:

Union: If either string contains a \(1\)on the \(ith\)bit, then the union contains a \(1\)on the \(ith\)bit as well. If both strings contain a \(0\)on the \(ith\)bit, then the union contains a \(0\)on the \(ith\) bit.

Intersection: If both strings contain a \(1\)on the \(ith\)bit, then the intersection contains a \(1\)on the \(ith\) bit as well. If either string contains a \(0\) on the \(ith\) bit, then the intersection contains a \(0\)on the \(ith\) bit.