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Prerna Chaudhary posted an Question
September 11, 2021 • 23:23 pm 30 points
  • IIT JAM
  • Mathematics (MA)

Let s be an infinite subset of r such that s nq = o which of the foliowing is true (a) smust have a limit point which belongs to q (b) smust have a limit point

Let S be an infinite subset of R such that S nQ = o which of the foliowing is true (a) Smust have a limit point which belongs to Q (b) Smust have a limit point which belongs to R/Q. (c) Scannot be a closed set in R. (d) R/Smust have a limit point which belongs to S.

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    Prerna chaudhary 1

    S is subset of R and S intersection rational is empty that means S does not contain any rational number , natural number,and integers form also so S will be the set of irrational numbers then we can say that rational intersection irrational is empty

  • comment-profile-img>
    Prerna chaudhary 1

    how can you say that intersection of same set is null

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    Arpan pal

    If any query my answer then you can told me freely again.

  • comment-profile-img>
    Arpan pal

    Option (d) is correct. Ans: S is the set of irrational numbers because S intersection Q is a empty set .If the set S is also a the set of irrational numbers and their intersection (S intersection Q) is an empty set. The set of rational numbers ( Q ) has limit points but it is not an closed set.So R/S must have limit points which belongs to S.

  • Anonymous User Best Answer

    S is a set of irrational number because S intersection Q is null set.every real number is limit point of S bcz between any two real number infinite irrational numbers.limit point of S not containd in S so S is not closed

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    But given , Option d is correct

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    if S is set of all irrational number R/S is set of rational number .every real number is limit point of so each real number limit point of R/S. root 2 is limit point of R/S which is in s

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    given Q intersection S equal to null set so s= compliment of Q . since Q is not open bcz nbd of 2 not containd in Q . there for compliment of Q is not closed hence S is not closed

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    sorry S is not closed if S= {root2/n} then limit point of S is 0 which is not in S

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    S={ root2+root 3 /n} then limit point of s is root 2 which is not in S

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