To find upper bounds for a set K in the power set of N (P(N)), we need to find sets A that contain all the elements of K. Conversely, to find lower bounds for K, we need to find sets A that are contained within each element of K.
First, let's find three upper bounds for K in P(N):
1. The set A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 13, 20} is an upper bound for K. It contains all the elements from each subset in K.
2. The set A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 13} is another upper bound for K. It contains all the elements from each subset in K except for the element 20.
3. The set A = {1, 2, 3, 4, 5, 6, 7, 8, 9} is a third upper bound for K. It contains all the elements from each subset in K except for the elements 12, 13, and 20.
Now let's find three lower bounds for K in P(N):
1. The set A = {2, 5, 7} is a lower bound for K. It is a subset contained within each subset in K.
2. The set A = {2, 5} is another lower bound for K. It is a subset contained within each subset in K except for the subset {1, 2, 3, 5, 7, 9, 13, 20}.
3. The set A = {2, 7} is a third lower bound for K. It is a subset contained within each subset in K except for the subset {2, 5, 7, 8, 12}.
Now, let's analyze the existence of a least upper bound and a greatest lower bound for K:
- Since K is a collection of subsets, and the power set of N (P(N)) is partially ordered by set inclusion, a least upper bound (supremum) for K must exist if all upper bounds for K have a common superset. However, since the upper bounds we found for K do not have a common superset, K does not have a least upper bound.
- Similarly, a greatest lower bound (infimum) for K must exist if all lower bounds for K have a common subset. However, since the lower bounds we found for K do not have a common subset, K does not have a greatest lower bound.