For a subset $A$ of a metric space which of the following implies the other three ?











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For a subset $A$ of a metric space which of the following implies the other three ?



$a)$ $A$ is closed



$b)$ $A$ is bounded



$c)$Closure of $B$ is compact ,for every $B subseteq A$



$d)$ $A$ is compact



I thinks option d) will correct .



is it true??










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  • 1




    Yes, it's true. (d) is the strongest condition you can have among them, (d) implies all the other.
    – John_Wick
    Nov 23 at 13:50










  • @John_Wick thanks u
    – Messi fifa
    Nov 23 at 13:51










  • Obviously If $A$ is compact then it implies $A$ is closed and bounded then there is a subset B whose closure is compact in fact any closed closed subset of a compact set in a (metric) space X is compact.
    – John Nash
    Nov 23 at 13:54

















up vote
0
down vote

favorite












For a subset $A$ of a metric space which of the following implies the other three ?



$a)$ $A$ is closed



$b)$ $A$ is bounded



$c)$Closure of $B$ is compact ,for every $B subseteq A$



$d)$ $A$ is compact



I thinks option d) will correct .



is it true??










share|cite|improve this question


















  • 1




    Yes, it's true. (d) is the strongest condition you can have among them, (d) implies all the other.
    – John_Wick
    Nov 23 at 13:50










  • @John_Wick thanks u
    – Messi fifa
    Nov 23 at 13:51










  • Obviously If $A$ is compact then it implies $A$ is closed and bounded then there is a subset B whose closure is compact in fact any closed closed subset of a compact set in a (metric) space X is compact.
    – John Nash
    Nov 23 at 13:54















up vote
0
down vote

favorite









up vote
0
down vote

favorite











For a subset $A$ of a metric space which of the following implies the other three ?



$a)$ $A$ is closed



$b)$ $A$ is bounded



$c)$Closure of $B$ is compact ,for every $B subseteq A$



$d)$ $A$ is compact



I thinks option d) will correct .



is it true??










share|cite|improve this question













For a subset $A$ of a metric space which of the following implies the other three ?



$a)$ $A$ is closed



$b)$ $A$ is bounded



$c)$Closure of $B$ is compact ,for every $B subseteq A$



$d)$ $A$ is compact



I thinks option d) will correct .



is it true??







general-topology






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asked Nov 23 at 13:44









Messi fifa

51611




51611








  • 1




    Yes, it's true. (d) is the strongest condition you can have among them, (d) implies all the other.
    – John_Wick
    Nov 23 at 13:50










  • @John_Wick thanks u
    – Messi fifa
    Nov 23 at 13:51










  • Obviously If $A$ is compact then it implies $A$ is closed and bounded then there is a subset B whose closure is compact in fact any closed closed subset of a compact set in a (metric) space X is compact.
    – John Nash
    Nov 23 at 13:54
















  • 1




    Yes, it's true. (d) is the strongest condition you can have among them, (d) implies all the other.
    – John_Wick
    Nov 23 at 13:50










  • @John_Wick thanks u
    – Messi fifa
    Nov 23 at 13:51










  • Obviously If $A$ is compact then it implies $A$ is closed and bounded then there is a subset B whose closure is compact in fact any closed closed subset of a compact set in a (metric) space X is compact.
    – John Nash
    Nov 23 at 13:54










1




1




Yes, it's true. (d) is the strongest condition you can have among them, (d) implies all the other.
– John_Wick
Nov 23 at 13:50




Yes, it's true. (d) is the strongest condition you can have among them, (d) implies all the other.
– John_Wick
Nov 23 at 13:50












@John_Wick thanks u
– Messi fifa
Nov 23 at 13:51




@John_Wick thanks u
– Messi fifa
Nov 23 at 13:51












Obviously If $A$ is compact then it implies $A$ is closed and bounded then there is a subset B whose closure is compact in fact any closed closed subset of a compact set in a (metric) space X is compact.
– John Nash
Nov 23 at 13:54






Obviously If $A$ is compact then it implies $A$ is closed and bounded then there is a subset B whose closure is compact in fact any closed closed subset of a compact set in a (metric) space X is compact.
– John Nash
Nov 23 at 13:54












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(d) is right, as a compact set is closed and bounded in any metric space.
and also if $A$ is compact then for $B subseteq A$, $overline{B}$ is closed in $overline{A}=A$ and hence compact too.






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    1 Answer
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    accepted










    (d) is right, as a compact set is closed and bounded in any metric space.
    and also if $A$ is compact then for $B subseteq A$, $overline{B}$ is closed in $overline{A}=A$ and hence compact too.






    share|cite|improve this answer

























      up vote
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      accepted










      (d) is right, as a compact set is closed and bounded in any metric space.
      and also if $A$ is compact then for $B subseteq A$, $overline{B}$ is closed in $overline{A}=A$ and hence compact too.






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        (d) is right, as a compact set is closed and bounded in any metric space.
        and also if $A$ is compact then for $B subseteq A$, $overline{B}$ is closed in $overline{A}=A$ and hence compact too.






        share|cite|improve this answer












        (d) is right, as a compact set is closed and bounded in any metric space.
        and also if $A$ is compact then for $B subseteq A$, $overline{B}$ is closed in $overline{A}=A$ and hence compact too.







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        answered Nov 23 at 17:07









        Henno Brandsma

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