Prove:

Let a,b exist in the reals with a<b,
(i) For the open interval E=(a,b) prove that the inf E=a and sup E=b
(ii) for the closed interval F=[a,b], prove that inf F=a and sup F=b.

To prove the statements (i) and (ii), we will use the definitions of infimum and supremum.

(i) For the open interval E=(a, b):

1. Infimum (inf) of E:
- Definition: The infimum of a set E is the greatest lower bound of E, denoted as inf E.
- Step 1: We need to show that a is a lower bound of E.
- By the definition of the open interval E=(a, b), every element x in E satisfies a < x.
- Therefore, a is a lower bound of E.
- Step 2: We need to show that no greater number than a can be a lower bound of E.
- Assume there exists a lower bound L of E such that L > a.
- Since L is a lower bound of E, it should satisfy L ≤ x for every x in E.
- However, since E=(a, b), there exists an element y in E such that a < y < L (by the denseness property of real numbers).
- This contradicts the assumption that L is a lower bound of E.
- Conclusion: Since a is a lower bound of E and there is no greater lower bound, a is the infimum of E, inf E = a.

2. Supremum (sup) of E:
- Definition: The supremum of a set E is the least upper bound of E, denoted as sup E.
- Step 1: We need to show that b is an upper bound of E.
- By the definition of the open interval E=(a, b), every element x in E satisfies x < b.
- Therefore, b is an upper bound of E.
- Step 2: We need to show that no smaller number than b can be an upper bound of E.
- Assume there exists an upper bound U of E such that U < b.
- Since U is an upper bound of E, it should satisfy x ≤ U for every x in E.
- However, since E=(a, b), there exists an element z in E such that U < z < b (again, by the denseness property of real numbers).
- This contradicts the assumption that U is an upper bound of E.
- Conclusion: Since b is an upper bound of E and there is no smaller upper bound, b is the supremum of E, sup E = b.

(ii) For the closed interval F=[a, b]:

1. Infimum (inf) of F:
- Definition: The infimum of a set F is the greatest lower bound of F, denoted as inf F.
- Step 1: We need to show that a is a lower bound of F.
- By the definition of the closed interval F=[a, b], every element x in F satisfies a ≤ x.
- Therefore, a is a lower bound of F.
- Step 2: We need to show that no greater number than a can be a lower bound of F.
- Assume there exists a lower bound L of F such that L > a.
- Since L is a lower bound of F, it should satisfy L ≤ x for every x in F.
- However, since F=[a, b], there exists an element y in F such that a ≤ y < L (by the density property).
- This contradicts the assumption that L is a lower bound of F.
- Conclusion: Since a is a lower bound of F and there is no greater lower bound, a is the infimum of F, inf F = a.

2. Supremum (sup) of F:
- Definition: The supremum of a set F is the least upper bound of F, denoted as sup F.
- Step 1: We need to show that b is an upper bound of F.
- By the definition of the closed interval F=[a, b], every element x in F satisfies x ≤ b.
- Therefore, b is an upper bound of F.
- Step 2: We need to show that no smaller number than b can be an upper bound of F.
- Assume there exists an upper bound U of F such that U < b.
- Since U is an upper bound of F, it should satisfy x ≤ U for every x in F.
- However, since F=[a, b], there exists an element z in F such that U < z ≤ b (again, by the density property).
- This contradicts the assumption that U is an upper bound of F.
- Conclusion: Since b is an upper bound of F and there is no smaller upper bound, b is the supremum of F, sup F = b.

By proving the above statements, we have shown that (i) for the open interval E=(a,b), inf E = a and sup E = b, and (ii) for the closed interval F=[a,b], inf F = a and sup F = b.

To prove these statements, we need to understand what is meant by the infimum and supremum of a set.

(i) For the open interval E = (a, b):

First, let's prove that the infimum of E is a.

Infimum (inf) is defined as the greatest lower bound of a set. In other words, it is the largest number that is less than or equal to all the numbers in the set.

Since a < b, any number less than a cannot be a lower bound for E. Therefore, a is a lower bound for E.

Next, we need to show that a is the greatest lower bound. To do this, we need to prove two things:
1. a is a lower bound for E.
2. If c is any lower bound for E, then c ≤ a.

For the first condition, we have already established that a is a lower bound for E.

For the second condition, assume that c is a lower bound for E. Since c is less than or equal to every element in E, c ≤ a must also hold, since a is the smallest element in E.

Therefore, a is the infimum of E.

Next, let's prove that the supremum of E is b.

Supremum (sup) is defined as the least upper bound of a set. In other words, it is the smallest number that is greater than or equal to all the numbers in the set.

Again, since a < b, any number greater than b cannot be an upper bound for E. Therefore, b is an upper bound for E.

Next, we need to show that b is the least upper bound. To do this, we need to prove two things:
1. b is an upper bound for E.
2. If c is any upper bound for E, then b ≤ c.

For the first condition, we have already established that b is an upper bound for E.

For the second condition, assume that c is an upper bound for E. Since c is greater than or equal to every element in E, b must also be less than or equal to c, since b is the largest element in E.

Therefore, b is the supremum of E.

(ii) For the closed interval F = [a, b]:

The proofs for the closed interval are similar to the open interval, but with some minor changes.

To prove that the infimum of F is a, we need to show two things:
1. a is a lower bound for F.
2. If c is any lower bound for F, then c ≤ a.

For the first condition, since a is the smallest value in the closed interval, a is also a lower bound for F.

For the second condition, assume that c is a lower bound for F. Since c must be less than or equal to every element in F, c ≤ a must also hold, as a is the smallest element in F.

Therefore, a is the infimum of F.

To prove that the supremum of F is b, we need to show two things:
1. b is an upper bound for F.
2. If c is any upper bound for F, then b ≤ c.

For the first condition, since b is the largest value in the closed interval, b is also an upper bound for F.

For the second condition, assume that c is an upper bound for F. Since c must be greater than or equal to every element in F, b must also be less than or equal to c, as b is the largest element in F.

Therefore, b is the supremum of F.

By proving these statements, we have shown that for both open and closed intervals, the infimum of E/F is a and the supremum of E/F is b.