Mathematics Exam > Mathematics Questions > Let S be a nonempty subset of R. If S is a fi...
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Let S be a nonempty subset of R. If S is a finite union of disjoint bounded intervals, then which one of the following is true?
- a)If S is not compact, then sup S ∉ S and inf S ∉ S
- b)Even if sup S ∈ S and inf S ∈ S, S need not be compact
- c)If sup S ∈ S and inf S ∈ S, then S is compact
- d)Even if S is compact, it is not necessary that sup S ∈ S and inf S ∈ S
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Let S be a nonempty subset of R. If S is a finite union of disjoint bo...
Example [1,2)∪(3,4] is fine and proves that (A) and (C) don't hold in general. It also shows that (B) is true.
If S is compact, then supS∈S because:
supS is the limit of a sequence of elements of S;
since S is compact, the limit of any convergent sequence of elements of S also belongs to S.
The same argument applies to infS.
So option B is correct
Most Upvoted Answer
Let S be a nonempty subset of R. If S is a finite union of disjoint bo...
Explanation:
To understand the answer, let's break down the options and analyze each one:
a) If S is not compact, then sup S ≤ S and inf S ≥ S
This statement is not necessarily true. If S is not compact, it means that S is either not closed or not bounded. In this case, the supremum and infimum of S may or may not be in S. Therefore, we cannot conclude that sup S ≤ S and inf S ≥ S.
c) If sup S ≤ S and inf S ≥ S, then S is compact
This statement is also not necessarily true. The compactness of a set is not solely determined by the supremum and infimum. A set can be compact even if its supremum and infimum are not in the set. Therefore, we cannot conclude that S is compact based on sup S ≤ S and inf S ≥ S.
d) Even if S is compact, it is not necessary that sup S ≤ S and inf S ≥ S
This statement is true. Compactness does not imply that the supremum and infimum of a set are in the set. A compact set is a set that is closed and bounded, but it does not necessarily contain its supremum and infimum. Therefore, the statement is true.
The correct answer is option 'B': Even if sup S ≤ S and inf S ≥ S, S need not be compact.
This answer is true because the supremum and infimum of a set alone do not determine its compactness. A set can have sup S ≤ S and inf S ≥ S, but it may not be compact if it is not closed or not bounded.
To summarize:
- Option a) is not necessarily true.
- Option c) is not necessarily true.
- Option d) is true.
- Option b) is the correct answer.
To understand the answer, let's break down the options and analyze each one:
a) If S is not compact, then sup S ≤ S and inf S ≥ S
This statement is not necessarily true. If S is not compact, it means that S is either not closed or not bounded. In this case, the supremum and infimum of S may or may not be in S. Therefore, we cannot conclude that sup S ≤ S and inf S ≥ S.
c) If sup S ≤ S and inf S ≥ S, then S is compact
This statement is also not necessarily true. The compactness of a set is not solely determined by the supremum and infimum. A set can be compact even if its supremum and infimum are not in the set. Therefore, we cannot conclude that S is compact based on sup S ≤ S and inf S ≥ S.
d) Even if S is compact, it is not necessary that sup S ≤ S and inf S ≥ S
This statement is true. Compactness does not imply that the supremum and infimum of a set are in the set. A compact set is a set that is closed and bounded, but it does not necessarily contain its supremum and infimum. Therefore, the statement is true.
The correct answer is option 'B': Even if sup S ≤ S and inf S ≥ S, S need not be compact.
This answer is true because the supremum and infimum of a set alone do not determine its compactness. A set can have sup S ≤ S and inf S ≥ S, but it may not be compact if it is not closed or not bounded.
To summarize:
- Option a) is not necessarily true.
- Option c) is not necessarily true.
- Option d) is true.
- Option b) is the correct answer.
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Let S be a nonempty subset of R. If S is a finite union of disjoint bounded intervals, then which one of the following is true?a)If S is not compact, then sup S ∉ S and inf S ∉ Sb)Even if sup S ∈ S and inf S ∈ S, S need not be compactc)If sup S ∈ S and inf S ∈S, then S is compactd)Even if S is compact, it is not necessary that sup S ∈ S and inf S ∈ SCorrect answer is option 'B'. Can you explain this answer?
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Let S be a nonempty subset of R. If S is a finite union of disjoint bounded intervals, then which one of the following is true?a)If S is not compact, then sup S ∉ S and inf S ∉ Sb)Even if sup S ∈ S and inf S ∈ S, S need not be compactc)If sup S ∈ S and inf S ∈S, then S is compactd)Even if S is compact, it is not necessary that sup S ∈ S and inf S ∈ SCorrect answer is option 'B'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let S be a nonempty subset of R. If S is a finite union of disjoint bounded intervals, then which one of the following is true?a)If S is not compact, then sup S ∉ S and inf S ∉ Sb)Even if sup S ∈ S and inf S ∈ S, S need not be compactc)If sup S ∈ S and inf S ∈S, then S is compactd)Even if S is compact, it is not necessary that sup S ∈ S and inf S ∈ SCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let S be a nonempty subset of R. If S is a finite union of disjoint bounded intervals, then which one of the following is true?a)If S is not compact, then sup S ∉ S and inf S ∉ Sb)Even if sup S ∈ S and inf S ∈ S, S need not be compactc)If sup S ∈ S and inf S ∈S, then S is compactd)Even if S is compact, it is not necessary that sup S ∈ S and inf S ∈ SCorrect answer is option 'B'. Can you explain this answer?.
Let S be a nonempty subset of R. If S is a finite union of disjoint bounded intervals, then which one of the following is true?a)If S is not compact, then sup S ∉ S and inf S ∉ Sb)Even if sup S ∈ S and inf S ∈ S, S need not be compactc)If sup S ∈ S and inf S ∈S, then S is compactd)Even if S is compact, it is not necessary that sup S ∈ S and inf S ∈ SCorrect answer is option 'B'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let S be a nonempty subset of R. If S is a finite union of disjoint bounded intervals, then which one of the following is true?a)If S is not compact, then sup S ∉ S and inf S ∉ Sb)Even if sup S ∈ S and inf S ∈ S, S need not be compactc)If sup S ∈ S and inf S ∈S, then S is compactd)Even if S is compact, it is not necessary that sup S ∈ S and inf S ∈ SCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let S be a nonempty subset of R. If S is a finite union of disjoint bounded intervals, then which one of the following is true?a)If S is not compact, then sup S ∉ S and inf S ∉ Sb)Even if sup S ∈ S and inf S ∈ S, S need not be compactc)If sup S ∈ S and inf S ∈S, then S is compactd)Even if S is compact, it is not necessary that sup S ∈ S and inf S ∈ SCorrect answer is option 'B'. Can you explain this answer?.
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