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Ashwini posted an Question
August 11, 2020 • 23:05 pm 30 points
  • IIT JAM
  • Mathematics (MA)

Since s tol a noa. or tne ponis u, 1, z, and 3, is tol an oopen sel. ex. it f is a closed bounded set, then every infinite subset s of f has limit point in f. s

SInce S TOl a noa. Or tne ponis U, 1, Z, and 3, IS TOL an oOpen seL. Ex. It F is a closed bounded set, then every infinite subset S of F has limit point in F. Since S is a subset of the bounded set F, so S is a bounded and infinite set. By Bolzano Weierstrass Theorem, S has a limit point, say p. Since Sc F, so p is also a limit point of F. Since F is a closed set and p is a limit point of F, so p E F. Hence S has its limit point in F. Sol. 1E A and D ic dlacAd chow that A e D ie a A ic alcsas and

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    Deepak singh 1

    F set is closed means set of limit points of F is contained in F..

  • Shashi ranjan sinha best-answer

    since F is closed bounded set, therefore it is compact. Now see the attachment.

    cropped8140399550214634764.jpg
    eduncle-logo-app

    The proof given in your question is right. But if you are not able to understand it, either ask your doubt or see my proof based on method of contradiction.

  • Kiran goswami

    what is your problem? plese explain clearly.

    eduncle-logo-app

    I can't understand the solution...pls explain in easy way

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