In each case below, either give an example of the desired object, or explainwhy no such object exists. Fully explain your answers.(a) A non-empty set of irrational numbers with rational supremum. [5 marks](b) A pair of non-empty sets X ⊆ Y , where Y has an infimum and X doesnot.

(a) A non-empty set of irrational numbers with a rational supremum does exist. For example, consider the set S = {x in R: x^2 < 2}. This set contains only irrational numbers. The supremum of this set is sqrt(2), which is an irrational number. However, if we consider the set S' = {x in R: x^2 < 2 and x is irrational}, the supremum of this set is indeed a rational number, which is 1.41421356 (approximation of sqrt(2)).

(b) A pair of non-empty sets X ⊆ Y, where Y has an infimum and X does not, can exist. For example, let Y = [0, 1] and X = (0, 1). Here, Y is the closed interval from 0 to 1, and X is the open interval from 0 to 1. The infimum of Y is 0, but X does not have an infimum in the set of real numbers because there is no largest number that is less than every number in X.

(a) A non-empty set of irrational numbers with a rational supremum does exist. For example, consider the set S = {x in R: x^2 < 2}. This set is non-empty and consists of irrational numbers. The supremum of this set is sqrt(2), which is an irrational number. However, if we consider the set in the rational number system (Q), the supremum is a rational number, which is 1.414 (approximation of sqrt(2)).

(b) A pair of non-empty sets X ⊆ Y, where Y has an infimum and X does not, can exist. For example, let Y = [0, 1] be a closed interval in R, and X = (0, 1) be an open interval in R. Here, Y has an infimum, which is 0, but X does not have an infimum in the set X itself because for any element you choose in X, you can always find a smaller one in X. However, in the context of the real numbers, X does have an infimum, which is 0. But 0 is not an element of X, so X does not have an infimum in the set X itself.