Let a,b exist in the reals with a<b.
(i) for the open interval E=(a,b), prove that infE=a and supE=b.
(ii) for the closed interval F=[a,b], prove that the infF=a and infF=b.
Looks to me like you just have to write the definition of inf and sup.
a is not in E, but is greater than any other number less than any element of E.
Can you say:
Since x in E then E is not empty and is a subset of the reals. Since there exist and x in E where a<x then a is a lower bound of E. then let lambda be a different lower bound in E then let a precede lambda. this can't occur because we stated that a is a lower bound of E thus lambda precedes a. and thus infE=a. prove supE=b similarly.
The for the closed set, since a and b are in E if there exist and x in E its less than or equal to b and greater than or equal to a. then assume like the one above that there is another element in E where its either preceded for is preceded by a or b?
To prove the properties of intervals, we need to understand the concepts of infimum and supremum.
The infimum of a set E, denoted as infE, is the greatest lower bound of the set. In simpler terms, it is the largest number that is less than or equal to all elements in the set. Similarly, the supremum of a set E, denoted as supE, is the least upper bound of the set. It is the smallest number that is greater than or equal to all elements in the set.
(i) For the open interval E=(a,b):
To prove that infE=a and supE=b, we need to show two things:
1. For any number x<a, x is not a lower bound of E.
2. For any number y>b, y is not an upper bound of E.
Let's go step by step:
1. For any number x<a, x is not a lower bound of E:
- Since x<a, we can pick a number between x and a, let's call it x'.
- x < x' < a, where x' belongs to the set E.
- This means that any number x<a cannot be a lower bound of E.
2. For any number y>b, y is not an upper bound of E:
- Since y>b, we can pick a number between b and y, let's call it y'.
- b < y' < y, where y' belongs to the set E.
- This means that any number y>b cannot be an upper bound of E.
Therefore, we can conclude that infE = a and supE = b for the open interval E = (a, b).
(ii) For the closed interval F=[a,b]:
To prove that infF=a and supF=b, we need to show two things:
1. a is a lower bound of F, and no number less than a can be a lower bound.
2. b is an upper bound of F, and no number greater than b can be an upper bound.
Let's go step by step:
1. a is a lower bound of F, and no number less than a can be a lower bound:
- Since F is a closed interval, any number x in F satisfies a<=x<=b.
- This means a is a lower bound of F, as it is less than or equal to all elements in F.
- Also, any number less than a would be less than the lower bound of F.
2. b is an upper bound of F, and no number greater than b can be an upper bound:
- Since F is a closed interval, any number x in F satisfies a<=x<=b.
- This means b is an upper bound of F, as it is greater than or equal to all elements in F.
- Additionally, any number greater than b would be greater than the upper bound of F.
Therefore, we can conclude that infF = a and supF = b for the closed interval F = [a, b].