Let S1, S2, S3, . . . be an infinite collection of non-empty sets of real numbers,and let S be their union:S =∞[i=1Si.(a) Suppose that S has a supremum. Explain briefly why each Si has asupremum. [3 marks](b) If each of the sets Si has a supremum, must S also have a supremum? Ifyes, give a proof. If no, give a counterexample and justify it

(a) If S has a supremum, it means there is a least upper bound for all the elements in the set S. Since each Si is a subset of S (because S is the union of all Si), the supremum of S would also serve as an upper bound for each Si. Therefore, each Si must also have a supremum, because the definition of supremum is the least upper bound of a set.

(b) If each of the sets Si has a supremum, it does not necessarily mean that S also has a supremum. For example, consider the sets Si = {1, 1/2, 1/3, ..., 1/i} for i = 1, 2, 3, .... Each Si has a supremum (which is 1), but the union of all Si, which is S = {1, 1/2, 1/3, ...}, does not have a supremum in the set of real numbers, because there is no least upper bound. The set S gets arbitrarily close to 0, but never reaches it, so there is no real number that serves as the least upper bound.

(a) If S has a supremum, it means there is a least upper bound for all the elements in the set S. Since each Si is a subset of S (because S is the union of all Si), the supremum of S would also serve as an upper bound for each Si. Therefore, each Si must also have a supremum.

(b) If each of the sets Si has a supremum, it does not necessarily mean that S also has a supremum. For example, consider the sets Si = {1/n} for n = i, i = 1, 2, 3, .... Each of these sets has a supremum (which is 1/n for Si), but the union of all these sets S = {1, 1/2, 1/3, ...} does not have a supremum in the set of real numbers, because there is no least upper bound. The number 1 is an upper bound, but not the least upper bound, because for any given upper bound, we can always find a smaller one (for example, 1 - ε for any positive ε). Therefore, even though each Si has a supremum, the set S does not.