Ashwini Asked a Question
August 11, 2020 5:35 pmpts 30 pts
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 Best Answer
    see attached
    • cropped-142479212.jpg
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    Deepak singh
    at any step , you have doubt then ask
  • Anonymous User thankyou
    since F is closed bounded set, therefore it is compact. Now see the attachment.
    • cropped8140399550214634764.jpg
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    Anonymous User
    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.
  • Vishal goswami
    what is your problem? plese explain clearly.
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    Ashwini
    I can't understand the solution...pls explain in easy way
  • Deepak singh
    F set is closed means set of limit points of F is contained in F..
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  • Vishal goswami
    see attachment
    • cropped446015211737346908.jpg
    Likes(0) Reply(0)
  • Vishal goswami
    if you have any doubt then please ask.
    Likes(0) Reply(0)
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