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Let S be an infinite set and S1, S2, S3, ..., Sn be sets such that S1 ∪S2 ∪S3∪ .......Sn = S then
- a)at least one of the sets Si is a finite set
- b)not more than one of the set Si can be inite
- c)at least one of the sets Si is an ininite set
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Let S be an infinite set and S1, S2, S3, ..., Snbe sets such that S1&c...
Let S = S1 ∪ S2 ∪ S3 ∪ .... Sn .
For S to be infinite set, atleast one of sets Si must be infinite,
if all Si were finite, then S will also be finite.
For S to be infinite set, atleast one of sets Si must be infinite,
if all Si were finite, then S will also be finite.
Most Upvoted Answer
Let S be an infinite set and S1, S2, S3, ..., Snbe sets such that S1&c...
To prove that the union of S1, S2, S3, ..., Sn is a subset of S, we can use mathematical induction.
Base Case: For n = 1, we have S1 ⊆ S by assumption.
Inductive Step: Assume that the union of S1, S2, S3, ..., Sn is a subset of S. We need to show that the union of S1, S2, S3, ..., Sn, Sn+1 is also a subset of S.
By definition, the union of S1, S2, S3, ..., Sn, Sn+1 is the set of all elements that are in at least one of these sets. Let x be an arbitrary element in the union of S1, S2, S3, ..., Sn, Sn+1. This means that x is in at least one of these sets.
Case 1: x is in Sn+1. Since Sn+1 ⊆ S by assumption, x is also in S.
Case 2: x is not in Sn+1. In this case, x must be in the union of S1, S2, S3, ..., Sn. By the induction hypothesis, x is in S.
In both cases, we see that x is in S. Since x was an arbitrary element in the union of S1, S2, S3, ..., Sn, Sn+1, we can conclude that the union of S1, S2, S3, ..., Sn, Sn+1 is a subset of S.
Therefore, by mathematical induction, the union of S1, S2, S3, ..., Sn is a subset of S for any positive integer n.
Base Case: For n = 1, we have S1 ⊆ S by assumption.
Inductive Step: Assume that the union of S1, S2, S3, ..., Sn is a subset of S. We need to show that the union of S1, S2, S3, ..., Sn, Sn+1 is also a subset of S.
By definition, the union of S1, S2, S3, ..., Sn, Sn+1 is the set of all elements that are in at least one of these sets. Let x be an arbitrary element in the union of S1, S2, S3, ..., Sn, Sn+1. This means that x is in at least one of these sets.
Case 1: x is in Sn+1. Since Sn+1 ⊆ S by assumption, x is also in S.
Case 2: x is not in Sn+1. In this case, x must be in the union of S1, S2, S3, ..., Sn. By the induction hypothesis, x is in S.
In both cases, we see that x is in S. Since x was an arbitrary element in the union of S1, S2, S3, ..., Sn, Sn+1, we can conclude that the union of S1, S2, S3, ..., Sn, Sn+1 is a subset of S.
Therefore, by mathematical induction, the union of S1, S2, S3, ..., Sn is a subset of S for any positive integer n.
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Let S be an infinite set and S1, S2, S3, ..., Snbe sets such that S1∪S2∪S3∪ .......Sn= S thena)at least one of the sets Siis a finite setb)not more than one of the set Sican be initec)at least one of the sets Siis an ininite setd)none of theseCorrect answer is option 'C'. Can you explain this answer?
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