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Every open set of real numbers is the union of
- a)countable collection of disjoint open intervals
- b)uncountable collection of disjoint open intervals
- c)countable collection of disjoint closed intervals.
- d)uncountable collection of disjoint closed intervals
Correct answer is option 'A'. Can you explain this answer?
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Every open set of real numbers is the union ofa)countable collection o...
Explanation:
In order to understand why the correct answer is option 'A', let's first define what an open set is and then discuss each option.
Open Set:
An open set in the real numbers is a set that does not contain its boundary points. In other words, for any point in the set, there exists an open interval around that point that is contained entirely within the set.
Option A:
A countable collection of disjoint open intervals can cover every point in an open set. This is because for any point in the open set, there exists an open interval around that point that is contained entirely within the set. By taking the union of all these open intervals, we can cover every point in the open set.
Option B:
An uncountable collection of disjoint open intervals is not necessary to cover every point in an open set. Since the open set contains only real numbers, which are countable, covering every point with an uncountable collection of intervals is not required.
Option C:
A countable collection of disjoint closed intervals cannot cover every point in an open set. This is because a closed interval includes its boundary points, while an open set does not contain its boundary points. So, there will always be points in the open set that are not covered by the closed intervals.
Option D:
An uncountable collection of disjoint closed intervals is not necessary to cover every point in an open set. Similar to option B, the open set contains only countable real numbers, so an uncountable collection of intervals is not required.
Therefore, the correct answer is option 'A' as a countable collection of disjoint open intervals is sufficient to cover every point in an open set.
In order to understand why the correct answer is option 'A', let's first define what an open set is and then discuss each option.
Open Set:
An open set in the real numbers is a set that does not contain its boundary points. In other words, for any point in the set, there exists an open interval around that point that is contained entirely within the set.
Option A:
A countable collection of disjoint open intervals can cover every point in an open set. This is because for any point in the open set, there exists an open interval around that point that is contained entirely within the set. By taking the union of all these open intervals, we can cover every point in the open set.
Option B:
An uncountable collection of disjoint open intervals is not necessary to cover every point in an open set. Since the open set contains only real numbers, which are countable, covering every point with an uncountable collection of intervals is not required.
Option C:
A countable collection of disjoint closed intervals cannot cover every point in an open set. This is because a closed interval includes its boundary points, while an open set does not contain its boundary points. So, there will always be points in the open set that are not covered by the closed intervals.
Option D:
An uncountable collection of disjoint closed intervals is not necessary to cover every point in an open set. Similar to option B, the open set contains only countable real numbers, so an uncountable collection of intervals is not required.
Therefore, the correct answer is option 'A' as a countable collection of disjoint open intervals is sufficient to cover every point in an open set.
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Every open set of real numbers is the union ofa)countable collection o...
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Every open set of real numbers is the union ofa)countable collection of disjoint open intervalsb)uncountable collection of disjoint open intervalsc)countable collection of disjoint closed intervals.d)uncountable collection of disjoint closed intervalsCorrect answer is option 'A'. Can you explain this answer?
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